CP-3 Matrices — revision question pack

8 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section CP-3. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

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CP-3.1 · Add, subtract and multiply conformable matrices. Multiply a matrix by a scalar.

Explanation

Matrices may be added or subtracted only when they have the same order; combine entries in corresponding positions. Scalar multiplication multiplies every entry by the scalar.
A product $AB$ exists when the number of columns of $A$ equals the number of rows of $B$ .
Its $(i,j)$ entry is the scalar product of row $i$ of $A$ with column $j$ of $B$ , and the result has the number of rows of $A$ and columns of $B$ .
Matrix multiplication is generally not commutative, so $AB$ and $BA$ may have different orders or only one may exist.
Dimensions should be checked before any arithmetic.

Worked example

Let $A=\begin{pmatrix}1&2\\-1&3\end{pmatrix}$ and $B=\begin{pmatrix}4&0\\2&-2\end{pmatrix}$ . Calculate $AB$ .

1.Both matrices are $2\times2$ , so $AB$ is $2\times2$ .
2.The first row gives $1(4)+2(2)=8$ and $1(0)+2(-2)=-4$ .
3.The second row gives $-1(4)+3(2)=2$ and $-1(0)+3(-2)=-6$ .

Answer: $AB=\begin{pmatrix}8&-4\\2&-6\end{pmatrix}$ .

Common mistakes

Don't fall into the trap of multiplying entries in matching positions instead of taking row-column scalar products.
Don't fall into the trap of assuming $AB=BA$ even though matrix multiplication is not generally commutative.
Don't fall into the trap of giving the product the order of the two inner dimensions instead of the outer dimensions.

Exam tip

Write the matrix orders first; the equal inner dimensions establish conformability and the outer dimensions give the product's order.

Tier 1 · Easy

1.
Given $A=\begin{pmatrix}2&-1\\3&0\end{pmatrix}$ and $B=\begin{pmatrix}-1&4\\2&5\end{pmatrix}$ , calculate $2A-B$ .
(2)
(Total for Question 1 is 2 marks)
2.
Let $U=\begin{pmatrix}2\\-1\end{pmatrix}$ and $V=\begin{pmatrix}3&4\end{pmatrix}$ . Calculate both $UV$ and $VU$ , stating the order of each product.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Let $C=\begin{pmatrix}1&2&-1\\0&3&2\end{pmatrix}$ and $D=\begin{pmatrix}2&1\\-1&4\\3&0\end{pmatrix}$ . Calculate $CD$ and state its order.
(4)
(Total for Question 1 is 4 marks)
2.
Given $A=\begin{pmatrix}1&0&2\\-1&3&1\end{pmatrix}$ , $B=\begin{pmatrix}2\\-1\\1\end{pmatrix}$ and $C=\begin{pmatrix}1&4\end{pmatrix}$ , calculate $(AB)C$ and $A(BC)$ .
(4)
(Total for Question 2 is 4 marks)
3.
Let $A=\begin{pmatrix}1&2\\0&-1\end{pmatrix}$ and $B=\begin{pmatrix}3&-1\\2&0\end{pmatrix}$ . Calculate $C=AB-BA$ and hence find $C^2$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Given $A=\begin{pmatrix}1&-2\\3&0\end{pmatrix}$ and $B=\begin{pmatrix}p&1\\-1&q\end{pmatrix}$ , the matrix $AB+2A$ is $\begin{pmatrix}8&-9\\18&3\end{pmatrix}$ . Determine $p$ and $q$ .
(4)
(Total for Question 1 is 4 marks)
2.
Let $A=\begin{pmatrix}1&0&1\\0&1&1\end{pmatrix}$ and $B=\begin{pmatrix}1&u\\v&1\\1&-1\end{pmatrix}$ . Given that $AB=\begin{pmatrix}2&1\\-4&0\end{pmatrix}$ , determine the value of $u$ and the value of $v$ , then state the order of $BA$ and of $AB$ .
(5)
(Total for Question 2 is 5 marks)
3.
Let $A=\begin{pmatrix}1&0\\0&2\end{pmatrix}$ , $B=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ and $X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ . Given that $AX+XB=\begin{pmatrix}3&1\\5&5\end{pmatrix}$ , determine $X$ .
(6)
(Total for Question 3 is 6 marks)
4.
Let $A=\begin{pmatrix}1&2\\-1&0\end{pmatrix}$ , $B=\begin{pmatrix}0&1\\3&-2\end{pmatrix}$ and $C=\begin{pmatrix}2&3\\-5&2\end{pmatrix}$ . Given that $C=pA+qB$ , find the values of $p$ and $q$ . Hence calculate $A(B+C)$ .
(6)
(Total for Question 4 is 6 marks)
5.
Let $A=\begin{pmatrix}2&-1\\1&1\end{pmatrix}$ and $B=\begin{pmatrix}0&3\\-2&1\end{pmatrix}$ . Calculate $(A+B)^2$ and $A^2+2AB+B^2$ , and explain why they differ.
(6)
(Total for Question 5 is 6 marks)

CP-3.2 · Understand and use zero and identity matrices.

Explanation

The zero matrix $O$ has every entry zero and is the additive identity: $A+O=A$ for matrices of the same order.
An identity matrix $I$ is square, with ones on the leading diagonal and zeros elsewhere, and satisfies $AI=IA=A$ whenever the products are defined.
Its order must match the matrix calculation.
Polynomial identities in a square matrix are manipulated algebraically, but a scalar constant becomes that scalar multiplied by $I$ ; for example, the matrix counterpart of $x^2-5x+6$ is $A^2-5A+6I$ .
The symbols $O$ and $I$ therefore represent matrices, not the scalar numbers $0$ and $1$ .

Worked example

A square matrix $A$ satisfies $A^2-5A+6I=O$ . Express $A^3$ in the form $pA+qI$ .

1.Rearrange the identity to obtain $A^2=5A-6I$ .
2.Multiply by $A$ : $A^3=5A^2-6A$ .
3.Substitute again: $A^3=5(5A-6I)-6A$ .

Answer: $A^3=19A-30I$ .

Common mistakes

Don't fall into the trap of writing a scalar constant such as $6$ where the matrix expression requires $6I$ .
Don't fall into the trap of using an identity matrix whose order does not match the square matrix $A$ .
Don't fall into the trap of treating $O$ as though it can be added to a matrix of any order.

Exam tip

In a matrix polynomial, attach $I$ to every constant term and preserve the order of matrix factors.

Tier 1 · Easy

1.
For $A=\begin{pmatrix}2&-1\\0&3\end{pmatrix}$ , write down $A+O$ and $IA$ , where $O$ and $I$ have the appropriate order.
(2)
(Total for Question 1 is 2 marks)
2.
A square matrix $N$ satisfies $N^2=O$ . Simplify $(I+N)^3$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Let $A=\begin{pmatrix}1&1\\0&2\end{pmatrix}$ . Verify that $(A-I)(A-2I)=O$ and hence state a quadratic matrix equation satisfied by $A$ .
(3)
(Total for Question 1 is 3 marks)
2.
A square matrix $P$ satisfies $P^2=P-I$ . Show that $P^6=I$ .
(4)
(Total for Question 2 is 4 marks)
3.
A square matrix $E$ satisfies $E^2=E$ . Let $S=2E-I$ . Find $S^2$ and hence simplify $S^{2027}$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A square matrix $M$ satisfies $M^2=3M+2I$ . Without finding the entries of $M$ , simplify $M^3-4M^2+M$ .
(4)
(Total for Question 1 is 4 marks)
2.
A square matrix $Q$ satisfies $Q^2=4Q-I$ . (a) Given that $(Q+I)^2=uQ+vI$ , where $u$ and $v$ are integers, find the value of $u$ and the value of $v$ . (b) Hence find the value of $w$ and the value of $t$ , given that $(Q+I)^4=wQ+tI$ .
(6)
(Total for Question 2 is 6 marks)
3.
A square matrix $K$ satisfies $K^3=K+I$ . Express $K^5$ in the form $aK^2+bK+cI$ , where $a$ , $b$ and $c$ are integers.
(6)
(Total for Question 3 is 6 marks)
4.
Let $A=\begin{pmatrix}1&-1\\2&-2\end{pmatrix}$ and $B=\begin{pmatrix}q&3\\2&r\end{pmatrix}$ , where $q$ and $r$ are real. Given that $AB=O$ , find the values of $q$ and $r$ . Calculate $BA$ and state what the result shows.
(7)
(Total for Question 4 is 7 marks)
5.
Let $A=\begin{pmatrix}a&b\\c&-a\end{pmatrix}$ , where $a$ , $b$ and $c$ are real. Show that $A^2=(a^2+bc)I$ . Taking $a=2$ and $b=1$ , find the value of $c$ when $a^2+bc=0$ and when $a^2+bc=1$ , and state $A^2$ in each case.
(7)
(Total for Question 5 is 7 marks)

CP-3.3 · Use matrices to represent linear transformations in 2-D. Successive transformations. Single transformations in 3-D.

Explanation

A linear transformation is determined by the images of the standard basis vectors, which form the columns of its matrix.
In two dimensions, required matrices include reflections in coordinate axes or $y=\pm x$ , rotations through any angle about the origin, stretches parallel to an axis and enlargements about the origin with non-zero real scale factor.
If matrix $A$ acts first and matrix $B$ second, the combined matrix is $BA$ , because $\mathbf x\mapsto A\mathbf x\mapsto BA\mathbf x$ .
In three dimensions, use a $3\times3$ matrix; transformations are confined to reflections in coordinate planes and rotations about coordinate axes.
Basis-vector images also identify these transformations.

The columns of a two-dimensional transformation matrix are the images of the two standard basis vectors.

Worked example

A rotation through $90^\circ$ anticlockwise is followed by reflection in the $x$ -axis. Find the combined matrix.

1.The rotation matrix is $R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ .
2.The reflection matrix is $F=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ .
3.Since the rotation acts first, the combined matrix is $FR$ .

Answer: $FR=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$ .

Common mistakes

Don't fall into the trap of writing $AB$ when transformation $A$ acts first and transformation $B$ acts second.
Don't fall into the trap of putting basis-vector images into rows rather than columns.
Don't fall into the trap of using a two-dimensional transformation matrix on a three-dimensional coordinate vector.

Exam tip

For successive transformations, write the action on $\mathbf x$ from right to left before multiplying the matrices.

Tier 1 · Easy

1.
The matrix $R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ acts on the point $P(3,-2)$ . Find the image of $P$ and describe the transformation.
(2)
(Total for Question 1 is 2 marks)
2.
The matrix $E=\begin{pmatrix}-2&0\\0&-2\end{pmatrix}$ represents a transformation. Find the image of $P(3,-1)$ and describe the transformation.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A reflection in the line $y=x$ is followed by a stretch parallel to the $x$ -axis with scale factor $2$ . Find the matrix of the combined transformation and the image of $(2,-1)$ .
(4)
(Total for Question 1 is 4 marks)
2.
Let $R$ be rotation through $90^\circ$ anticlockwise and let $H$ be a stretch parallel to the $x$ -axis with scale factor $3$ . A point $(1,2)$ is mapped to $(-2,3)$ . Determine which of the two transformations is applied first, and find the matrix of the combined transformation.
(5)
(Total for Question 2 is 5 marks)
3.
A rotation through the smallest positive angle $\theta$ about the origin maps $(2,0)$ to $(\sqrt3,1)$ . Find $\theta$ , write down the matrix of the rotation, and find the image of $(-1,\sqrt3)$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
A transformation in three dimensions maps $(x,y,z)$ to $(x,-z,y)$ . Write down its matrix, find the image of $(2,-1,4)$ , and explain geometrically what the transformation does.
(5)
(Total for Question 1 is 5 marks)
2.
A linear transformation in three dimensions maps $(1,1,0)$ to $(-1,1,0)$ , $(0,1,1)$ to $(-1,0,1)$ and $(1,0,1)$ to $(0,1,1)$ . Determine its matrix, describe the transformation, and find the image of $(2,-1,3)$ .
(5)
(Total for Question 2 is 5 marks)
3.
The matrix $T=\begin{pmatrix}0&3\\3&0\end{pmatrix}$ can be written as $T=ER$ , where $R$ represents reflection in either $y=x$ or $y=-x$ , and $E$ is an enlargement with centre the origin and positive scale factor. Determine $R$ and $E$ , state the transformations in the order in which they act, and find the image of $(2,-1)$ under $T$ .
(6)
(Total for Question 3 is 6 marks)
4.
A reflection in the line $y=-x$ is followed by a reflection in the $y$ -axis. Find the matrix of the combined transformation, stating the order in which the matrices are multiplied. Describe the single equivalent transformation and find the image of $(4,1)$ .
(6)
(Total for Question 4 is 6 marks)
5.
A reflection in the line $y=x$ is followed by a stretch parallel to the $y$ -axis with scale factor $4$ , and then by a rotation through $180^\circ$ about the origin. Find the combined matrix. Find also the matrix when the three transformations are applied in reverse order, and compare the images of $(5,-2)$ .
(7)
(Total for Question 5 is 7 marks)

CP-3.4 · Find invariant points and lines for a linear transformation.

Explanation

An invariant point is fixed by the transformation. For matrix $A$ , its position vector satisfies $A\mathbf x=\mathbf x$ , so solve $(A-I)\mathbf x=\mathbf0$ .
An invariant line is mapped onto itself as a set, although individual points on it may move.
To test $y=mx+c$ , transform a general point $(x,mx+c)$ to $(X,Y)$ and require $Y=mX+c$ for every $x$ ; equate coefficients and constants.
Vertical lines must be checked separately because they cannot be written in gradient-intercept form.
A line consisting entirely of invariant points is stronger than an invariant line, so the two descriptions must not be confused.

Worked example

The transformation maps $(x,y)$ to $(2x,y)$ . Find its invariant points and invariant coordinate axes.

1.Invariant points satisfy $(2x,y)=(x,y)$ , so $x=0$ .
2.Hence every point on the $y$ -axis is fixed.
3.A point $(x,0)$ maps to $(2x,0)$ , so the $x$ -axis is invariant as a set.
4.The $y$ -axis is both invariant and a line of invariant points.

Answer: All invariant points lie on $x=0$ ; both coordinate axes are invariant lines.

Common mistakes

Don't fall into the trap of calling every point on an invariant line fixed when points may move along that line.
Don't fall into the trap of testing one convenient point instead of a general point on a proposed invariant line.
Don't fall into the trap of omitting possible vertical invariant lines after considering only $y=mx+c$ .

Exam tip

State separately whether the result is an invariant line or a line whose every point is invariant.

Tier 1 · Easy

1.
Find all invariant points of the transformation with matrix $A=\begin{pmatrix}1&0\\0&3\end{pmatrix}$ .
(3)
(Total for Question 1 is 3 marks)
2.
The matrix $A_k=\begin{pmatrix}3&k\\1&2\end{pmatrix}$ , where $k$ is a constant, represents a transformation. (a) Find the value of $k$ for which the line $y=x$ is invariant. (b) Determine whether the line consists entirely of invariant points, justifying your answer.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Find the invariant lines through the origin for the transformation with matrix $B=\begin{pmatrix}2&1\\0&1\end{pmatrix}$ .
(4)
(Total for Question 1 is 4 marks)
2.
The matrix $A_t=\begin{pmatrix}0&-1\\1&t\end{pmatrix}$ , where $t$ is a constant, has exactly one invariant line through the origin. Determine the possible values of $t$ and give the equation of the line in each case.
(4)
(Total for Question 2 is 4 marks)
3.
The matrix $A_k=\begin{pmatrix}k&2\\3&4\end{pmatrix}$ represents a transformation, where $k$ is real. Determine the value of $k$ for which there are non-zero invariant points. For this value of $k$ , find all invariant lines through the origin and state which line consists entirely of invariant points.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
The transformation $T$ maps $(x,y)$ to $(x+y,2y)$ . Find all invariant straight lines and identify any line consisting entirely of invariant points.
(6)
(Total for Question 1 is 6 marks)
2.
Determine the equations of all the invariant lines of the transformation that maps $(x,y)$ to $(kx+2y,5y)$ , where $k$ is a real constant.
(7)
(Total for Question 2 is 7 marks)
3.
The transformation $T_k$ has matrix $\begin{pmatrix}6&k\\-1&2\end{pmatrix}$ , where $k$ is real. For every value of $k$ , determine the number and equations of the invariant lines through the origin. State when any such line consists entirely of invariant points.
(7)
(Total for Question 3 is 7 marks)
4.
A linear transformation $T$ multiplies every vector parallel to $y=3x$ by $2$ and every vector parallel to $y=-2x$ by $3$ . Determine the matrix of $T$ . The line $L$ has equation $y=x+4$ . Determine whether any point of $L$ is invariant, justifying your answer.
(7)
(Total for Question 4 is 7 marks)
5.
The singular transformation $T$ has matrix $A=\begin{pmatrix}2&6\\1&3\end{pmatrix}$ . Determine all invariant straight lines. For each invariant line, describe how $T$ acts on its points.
(8)
(Total for Question 5 is 8 marks)

CP-3.5 · Calculate determinants of 2x2 and 3x3 matrices and interpret as scale factors, including the effect on orientation.

Explanation

For $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ , $\det A=ad-bc$. A $3\times3$ determinant may be evaluated by cofactor expansion, keeping the alternating signs.
Under a two-dimensional linear transformation, area is multiplied by $|\det A|$ ; in three dimensions, volume is multiplied by $|\det A|$ .
The determinant's sign records orientation: a positive determinant preserves orientation and a negative determinant reverses it.
A zero determinant collapses dimension and gives zero area or volume scale factor.
Geometric sizes are never negative, so use the absolute value when calculating image area or volume.

Worked example

A transformation has matrix $A=\begin{pmatrix}3&1\\2&-1\end{pmatrix}$ . Find the image area of a region of area $9$ and describe orientation.

1. $\det A=3(-1)-1(2)=-5$ .
2.The area scale factor is $|-5|=5$ .
3.The image area is $9\times5=45$ .
4.The negative determinant reverses orientation.

Answer: Image area $45$ square units; orientation is reversed.

Common mistakes

Don't fall into the trap of using $ad+bc$ instead of $ad-bc$ for a $2\times2$ determinant.
Don't fall into the trap of reporting a negative area because the determinant is negative.
Don't fall into the trap of ignoring the alternating signs during a $3\times3$ cofactor expansion.

Exam tip

Keep the signed determinant for orientation, but use its absolute value for an area or volume scale factor.

Tier 1 · Easy

1.
A planar transformation has matrix $A=\begin{pmatrix}1&4\\2&3\end{pmatrix}$ . A region has area $12$ before transformation. Find its image area and state what happens to orientation.
(3)
(Total for Question 1 is 3 marks)
2.
The matrix $\begin{pmatrix}2&3\\4&6\end{pmatrix}$ represents a planar transformation. Find the image area of a region whose original area is $7$ , and describe the image geometrically.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Calculate the determinant of $M=\begin{pmatrix}1&2&0\\-1&3&1\\2&0&1\end{pmatrix}$ . Hence state the volume scale factor and the effect on orientation.
(4)
(Total for Question 1 is 4 marks)
2.
A transformation $R$ has determinant $-2$ and is followed by a transformation $S$ . Their combination maps an area of $5$ to an area of $30$ and reverses orientation. Determine $\det S$ and state whether $S$ itself preserves orientation.
(4)
(Total for Question 2 is 4 marks)
3.
The matrix $B=\begin{pmatrix}2&1&0\\1&-3&2\\-4&0&1\end{pmatrix}$ represents a three-dimensional transformation. Calculate $\det B$ by cofactor expansion, state the effect on orientation, and find the image volume of a solid whose original volume is $4$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A transformation has matrix $P=\begin{pmatrix}k&2\\3&k-1\end{pmatrix}$ . A triangle of area $7$ is mapped to a triangle of area $42$ . Determine all possible values of $k$ and state which values reverse orientation.
(5)
(Total for Question 1 is 5 marks)
2.
A three-dimensional transformation has matrix $A_k=\begin{pmatrix}k&1&0\\2&1&1\\0&1&2\end{pmatrix}$ and is followed by $B=\begin{pmatrix}1&0&0\\0&-1&0\\0&0&2\end{pmatrix}$ . A solid of volume $3$ is mapped to volume $30$ , with orientation preserved. Determine the real value of $k$ .
(6)
(Total for Question 2 is 6 marks)
3.
The matrix $A_h=\begin{pmatrix}1&0&h\\0&h&4\\-1&2&-2\end{pmatrix}$ represents a three-dimensional transformation, where $h$ is real. Find the set of values of $h$ for which the transformation is singular. Find the value of $h>0$ for which a solid of volume $3$ is mapped to a solid of volume $24$ with its orientation reversed.
(6)
(Total for Question 3 is 6 marks)
4.
A planar linear transformation $T$ maps $\mathbf u=\begin{pmatrix}1\\2\end{pmatrix}$ to $\begin{pmatrix}3\\-2\end{pmatrix}$ and $\mathbf v=\begin{pmatrix}3\\-1\end{pmatrix}$ to $\begin{pmatrix}1\\3\end{pmatrix}$ . Find the determinant of the matrix of $T$ . Hence find the image area of a region of area $14$ and state the effect on orientation.
(6)
(Total for Question 4 is 6 marks)
5.
A planar transformation has matrix $M_k=\begin{pmatrix}k&2\\3&k\end{pmatrix}$ , where $k$ is real. Find the set of possible values of $k$ for which every finite region has a smaller image area and orientation is reversed.
(6)
(Total for Question 5 is 6 marks)

CP-3.6 · Understand and use singular and non-singular matrices. Properties of inverse matrices. Calculate and use the inverse of non-singular 2x2 and 3x3 matrices.

Explanation

A square matrix is singular when its determinant is zero; it then has no inverse. A non-singular matrix has non-zero determinant and a unique inverse satisfying $AA^{-1}=A^{-1}A=I$ .
For $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ , $A^{-1}=\dfrac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ . A $3\times3$ inverse can be found by row-reducing $(A\mid I)$ to $(I\mid A^{-1})$ , by an adjugate method or by calculator when permitted, but the process must be understood.
Useful properties include $(AB)^{-1}=B^{-1}A^{-1}$ and $(A^{-1})^{-1}=A$ .
Matrix order matters, so inverse factors reverse order.
Calculator output should still be checked by multiplying by the original matrix.

Worked example

Find the inverse of $A=\begin{pmatrix}4&1\\3&1\end{pmatrix}$ and verify one product.

1. $\det A=4(1)-1(3)=1$ , so $A$ is non-singular.
2.Interchange the diagonal entries and negate the off-diagonal entries.
3. $A^{-1}=\begin{pmatrix}1&-1\\-3&4\end{pmatrix}$ .
4.Multiplication gives $AA^{-1}=I$ .

Answer: $A^{-1}=\begin{pmatrix}1&-1\\-3&4\end{pmatrix}$ .

Common mistakes

Don't fall into the trap of attempting to invert a matrix whose determinant is zero.
Don't fall into the trap of dividing the original entries by the determinant without swapping and negating the required entries.
Don't fall into the trap of writing $(AB)^{-1}=A^{-1}B^{-1}$ instead of reversing the factor order.

Exam tip

Calculate and display the determinant before applying an inverse formula, then check the result against the identity matrix.

Tier 1 · Easy

1.
Show that $A=\begin{pmatrix}3&1\\2&1\end{pmatrix}$ is non-singular and find $A^{-1}$ .
(3)
(Total for Question 1 is 3 marks)
2.
The matrix $M=\begin{pmatrix}k+1&4\\2&k-1\end{pmatrix}$ is singular. Find the possible values of $k$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Find the inverse of $C=\begin{pmatrix}1&1&0\\0&1&1\\1&0&1\end{pmatrix}$ .
(5)
(Total for Question 1 is 5 marks)
2.
The non-singular matrix $A$ has $A^{-1}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}$ . Solve the matrix equation $XA=\begin{pmatrix}2&3\\1&4\end{pmatrix}$ for $X$ .
(4)
(Total for Question 2 is 4 marks)
3.
Non-singular matrices $A$ and $B$ satisfy $A^{-1}=\begin{pmatrix}1&2\\0&1\end{pmatrix}$ and $B^{-1}=\begin{pmatrix}2&0\\-1&1\end{pmatrix}$ . Use the inverse of a product to solve $AB\mathbf x=\begin{pmatrix}3\\-1\end{pmatrix}$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
The matrix $D=\begin{pmatrix}2&1\\3&2\end{pmatrix}$ satisfies $D^2-4D+I=O$ . Use this identity, rather than the $2\times2$ inverse formula, to find $D^{-1}$ .
(5)
(Total for Question 1 is 5 marks)
2.
Let $A=\begin{pmatrix}2&1\\1&1\end{pmatrix}$ and $B=\begin{pmatrix}1&0\\1&1\end{pmatrix}$ . Find the matrix $X$ satisfying $AXB=\begin{pmatrix}4&-1\\3&0\end{pmatrix}$ .
(6)
(Total for Question 2 is 6 marks)
3.
Let $A=\begin{pmatrix}1&1\\0&2\end{pmatrix}$ and $B=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ . Define $X=I-AB$ and $Y=I-BA$ . Show that $XA=AY$ , and hence find $X^{-1}A$ .
(5)
(Total for Question 3 is 5 marks)
4.
Let $A=\begin{pmatrix}2&1&1\\1&2&3\\3&1&2\end{pmatrix}$ . Find $A^{-1}$ and hence solve $AX=\begin{pmatrix}7&1\\14&4\\11&3\end{pmatrix}$ for the $3\times2$ matrix $X$ . Verify your answer.
(6)
(Total for Question 4 is 6 marks)
5.
Let $A=\begin{pmatrix}3&1\\1&1\end{pmatrix}$ and $B=\begin{pmatrix}2&1\\0&1\end{pmatrix}$ . Calculate $(A^{-1}+B^{-1})^{-1}$ . Calculate also $A(A+B)^{-1}B$ and verify that the two matrices are equal.
(7)
(Total for Question 5 is 7 marks)

CP-3.7 · Solve three linear simultaneous equations in three variables by use of the inverse matrix.

Explanation

Write three linear equations in the form $A\mathbf x=\mathbf b$ , where each row of $A$ contains the coefficients of $x,y,z$ in the same order and $\mathbf b$ contains the corresponding right-hand sides.
If $A$ is non-singular, left-multiply by $A^{-1}$ to obtain $\mathbf x=A^{-1}\mathbf b$ .
The inverse may be calculated exactly or by calculator as permitted, but the matrix equation and multiplication order should be shown.
A zero coefficient must still occupy its correct position.
The resulting values should be checked in the original equations, since a transposed coefficient or misplaced constant can produce a plausible but incorrect matrix calculation.

Worked example

Use $A^{-1}=\begin{pmatrix}1&-1&1\\0&1&-1\\0&0&1\end{pmatrix}$ to solve $A\mathbf x=\begin{pmatrix}6\\4\\1\end{pmatrix}$ .

1.Left-multiply by $A^{-1}$ : $\mathbf x=A^{-1}\mathbf b$ .
2. $x=6-4+1=3$ .
3. $y=4-1=3$ and $z=1$ .

Answer: $(x,y,z)=(3,3,1)$ .

Common mistakes

Don't fall into the trap of writing $\mathbf x=\mathbf bA^{-1}$ even though $\mathbf b$ is a column vector.
Don't fall into the trap of changing the order of variables between different coefficient rows.
Don't fall into the trap of omitting a zero entry for a variable absent from one equation.

Exam tip

Display $A$ , $\mathbf x$ and $\mathbf b$ before using the inverse so coefficient placement earns method credit.

Tier 1 · Easy

1.
A system is written as $A\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\2\\6\end{pmatrix}$ , where $A^{-1}=\frac12\begin{pmatrix}1&-1&1\\1&1&-1\\-1&1&1\end{pmatrix}$ . Find $x,y,z$ .
(3)
(Total for Question 1 is 3 marks)
2.
(a) Write $2x+z=5$ , $x-y=0$ and $3y+2z=7$ in the form $A\mathbf{x}=\mathbf{b}$ , where $\mathbf{x}=(x,y,z)^T$ , stating $A$ and $\mathbf{b}$ . (b) Given $A^{-1}=\begin{pmatrix}2&-3&-1\\2&-4&-1\\-3&6&2\end{pmatrix}$ , find $x$ , $y$ and $z$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
Use an inverse matrix to solve $x+2y=5$ , $y+z=4$ and $z=3$ .
(5)
(Total for Question 1 is 5 marks)
2.
Let $A_k=\begin{pmatrix}1&1&1\\4&k&2\\1&2&3\end{pmatrix}$ . $A_k$ maps $(1,1,1)^T$ to $(3,10,6)^T$ . (a) Find the value of $k$ . (b) Given that $A_k$ maps $(x,y,z)^T$ to $(6,22,10)^T$ , use an inverse matrix to find the values of $x$ , $y$ and $z$ .
(6)
(Total for Question 2 is 6 marks)
3.
A notebook costs $x$ pounds, a pen costs $y$ pounds and a folder costs $z$ pounds. Two notebooks, one pen and one folder cost £13. One notebook, three pens and two folders cost £19. Three notebooks, two pens and one folder cost £18. Use an inverse matrix to find $x$ , $y$ and $z$ . Hence find the cost of one notebook, two pens and three folders.
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
The equations $x+y=p$ , $y+z=5$ and $2z=8$ have a solution satisfying $x+y=z$ . Use an inverse matrix to determine $p$ and the solution.
(6)
(Total for Question 1 is 6 marks)
2.
The quadratic $q(t)=x+yt+zt^2$ satisfies $q(0)=2$ , $q(2)=9$ and $q(3)=20$ . Use an inverse matrix to find $x$ , $y$ and $z$ , and hence determine $q(1)$ .
(7)
(Total for Question 2 is 7 marks)
3.
The equations $x+y+z=4$ , $x+2y=11-3p$ and $y+2z=1$ are given, where $p$ is real. Use an inverse matrix to express $x$ , $y$ and $z$ in terms of $p$ . Hence find the set of possible values of $p$ for which $x$ , $y$ and $z$ are all positive.
(7)
(Total for Question 3 is 7 marks)
4.
Let $A=\begin{pmatrix}0&1&1\\1&0&1\\1&1&1\end{pmatrix}$ . Find $A^{-1}$ and hence solve both $A\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}30\\28\\42\end{pmatrix}$ and $A\begin{pmatrix}p\\q\\r\end{pmatrix}=\begin{pmatrix}44\\40\\61\end{pmatrix}$ .
(7)
(Total for Question 4 is 7 marks)
5.
The plane $z=ax+by+c$ contains the points $(1,0,5)$ , $(0,2,1)$ and $(2,-1,8)$ . Use an inverse matrix to find the values of $a$ , $b$ and $c$ . Find the value of $z$ at $(x,y)=(3,4)$ and find where the plane meets the line $x=y$ , $z=0$ .
(7)
(Total for Question 5 is 7 marks)

CP-3.8 · Interpret geometrically the solution and failure of solution of three simultaneous linear equations.

Explanation

Each linear equation in three variables represents a plane in three-dimensional space. Three independent planes meeting at one point give a unique solution.
If all three contain the same line, they form a sheaf and every point on that line is a solution. Coincident planes can also produce infinitely many solutions.
A prism arrangement has pairwise intersection lines but no point common to all three, so the system is inconsistent; parallel distinct planes are another no-solution case.
Algebraically, a contradictory row after elimination signals inconsistency, while a dependent equation without contradiction leaves free variables.
A singular coefficient matrix therefore does not by itself distinguish no solution from infinitely many solutions.

Sketches of a unique solution, a sheaf of planes and an inconsistent prism.

Worked example

The third equation of a system is the sum of the first two. Interpret the solution geometrically if the first two planes intersect.

1.Every point satisfying the first two equations also satisfies their sum.
2.The first two non-parallel planes meet in a line.
3.The third plane contains that complete intersection line.

Answer: The planes form a sheaf with infinitely many solutions along their common line.

Common mistakes

Don't fall into the trap of saying a singular coefficient matrix always means no solution.
Don't fall into the trap of calling three pairwise-intersecting planes a unique solution without checking for one common point.
Don't fall into the trap of describing coincident planes as distinct parallel planes and therefore inconsistent.

Exam tip

Link the algebraic outcome explicitly to one point, a common line or no common point in the geometric interpretation.

Tier 1 · Easy

1.
Interpret geometrically the solution of the three equations $x=1$ , $y=2$ and $z=-3$ .
(2)
(Total for Question 1 is 2 marks)
2.
The planes $x=1$ , $y=2$ and $x+y+kz=3$ , where $k$ is a constant, are given. Determine when they form a sheaf and when they meet at a unique point. Give the common line or point in each case.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Consider $x+y+z=1$ , $2x+2y+2z=k$ and $x-y+z=3$ . Describe the solution set geometrically when $k=2$ and when $k\ne2$ .
(4)
(Total for Question 1 is 4 marks)
2.
The planes $x+y=2$ , $y+z=1$ and $x+ky+3z=4$ , where $k$ is a constant, are given. Determine the coordinates of the point where the three planes intersect, giving each coordinate in simplest form in terms of $k$ . State the value of $k$ for which no such point exists and interpret that case geometrically.
(5)
(Total for Question 2 is 5 marks)
3.
Find the set of solutions of the three plane equations $x+2y-z=3$ , $3x-y+2z=2$ and $-x+5y-4z=4$ . Give a vector equation for the solution set and interpret the result geometrically.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
The planes $x+y+z=2$ , $2x-y+z=1$ and $3x+2z=k$ are given. Determine the value of $k$ for which the three planes contain a common line. State what happens for every other value of $k$ .
(5)
(Total for Question 1 is 5 marks)
2.
The planes $x+z=2$ , $y+z=1$ and $x+(3-2k)y+(k^2-5k+6)z=k+2$ , where $k$ is a constant, are given. Determine whether they meet at a unique point, form a sheaf or form a prism.
(6)
(Total for Question 2 is 6 marks)
3.
The planes $\Pi_1$ , $\Pi_2$ and $\Pi_3$ have equations $x+2z=4$ , $2y-z=-1$ and $x+2y+z=8$ respectively. Find a vector equation for each of the three pairwise intersection lines. Hence determine whether the planes have a common point and interpret their geometric arrangement.
(7)
(Total for Question 3 is 7 marks)
4.
The planes $\Pi_1$ , $\Pi_2$ and $\Pi_3$ have equations $x+y=8$ , $y+z=-3$ and $(k^2-13k+43)x+(k-5)y+(k-6)z=c$ respectively, where $k$ and $c$ are real. Describe how the three planes meet for all values of $k$ and $c$ . Give a vector equation whenever their common solution is a line.
(8)
(Total for Question 4 is 8 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

CP-3.1 · Add, subtract and multiply conformable matrices. Multiply a matrix by a scalar.

Tier 1 · Easy

Mark scheme for CP-3.1 Tier 1 · Easy
Question	Scheme	Marks
1	$\begin{pmatrix}5&-6\\4&-5\end{pmatrix}$	2
(2 marks)		2
Notes
$2A=\begin{pmatrix}4&-2\\6&0\end{pmatrix}$ . Subtract corresponding entries of $B$ : $2A-B=\begin{pmatrix}4-(-1)&-2-4\\6-2&0-5\end{pmatrix}=\begin{pmatrix}5&-6\\4&-5\end{pmatrix}$ .
2	$UV=\begin{pmatrix}6&8\\-3&-4\end{pmatrix}$ $UV$ has order $2\times2$ $VU=\begin{pmatrix}2\end{pmatrix}$ , of order $1\times1$	3
(3 marks)		3
Notes
$U$ is $2\times1$ and $V$ is $1\times2$ , so both products exist but have different outer dimensions. The outer product is $UV=\begin{pmatrix}2(3)&2(4)\\-1(3)&-1(4)\end{pmatrix}=\begin{pmatrix}6&8\\-3&-4\end{pmatrix}$ . Reversing the order gives the scalar product $VU=\begin{pmatrix}3(2)+4(-1)\end{pmatrix}=\begin{pmatrix}2\end{pmatrix}$ , so the products cannot be equal.

Tier 2 · Standard

Mark scheme for CP-3.1 Tier 2 · Standard
Question	Scheme	Marks
1	$CD=\begin{pmatrix}-3&9\\3&12\end{pmatrix}$ , of order $2\times2$	4
(4 marks)		4
Notes
$C$ is $2\times3$ and $D$ is $3\times2$ , so $CD$ exists and is $2\times2$ . Taking row-column products gives $CD=\begin{pmatrix}1(2)+2(-1)+(-1)3&1(1)+2(4)+(-1)0\\0(2)+3(-1)+2(3)&0(1)+3(4)+2(0)\end{pmatrix}=\begin{pmatrix}-3&9\\3&12\end{pmatrix}$ .
2	$AB=\begin{pmatrix}4\\-4\end{pmatrix}$ $(AB)C=\begin{pmatrix}4&16\\-4&-16\end{pmatrix}$ $BC=\begin{pmatrix}2&8\\-1&-4\\1&4\end{pmatrix}$ $A(BC)=\begin{pmatrix}4&16\\-4&-16\end{pmatrix}$	4
(4 marks)		4
Notes
$AB=\begin{pmatrix}4\\-4\end{pmatrix}$ , so $(AB)C=\begin{pmatrix}4\\-4\end{pmatrix}\begin{pmatrix}1&4\end{pmatrix}=\begin{pmatrix}4&16\\-4&-16\end{pmatrix}$ . Alternatively, $BC=\begin{pmatrix}2&8\\-1&-4\\1&4\end{pmatrix}$ and row-column multiplication gives $A(BC)=\begin{pmatrix}4&16\\-4&-16\end{pmatrix}$ . The intermediate orders differ, but both complete products are conformable $2\times2$ matrices.
3	$AB=\begin{pmatrix}7&-1\\-2&0\end{pmatrix}$ $BA=\begin{pmatrix}3&7\\2&4\end{pmatrix}$ $C=\begin{pmatrix}4&-8\\-4&-4\end{pmatrix}$ $C^2=\begin{pmatrix}48&0\\0&48\end{pmatrix}=48I$	4
(4 marks)		4
Notes
Row-column multiplication gives $AB=\begin{pmatrix}7&-1\\-2&0\end{pmatrix}$ and $BA=\begin{pmatrix}3&7\\2&4\end{pmatrix}$ . Therefore $C=AB-BA=\begin{pmatrix}4&-8\\-4&-4\end{pmatrix}$ . Multiplying this matrix by itself gives $C^2=\begin{pmatrix}48&0\\0&48\end{pmatrix}=48I$ .

Tier 3 · Hard

Mark scheme for CP-3.1 Tier 3 · Hard
Question	Scheme	Marks
1	$p=4$ , $q=3$	4
(4 marks)		4
Notes
$AB=\begin{pmatrix}p+2&1-2q\\3p&3\end{pmatrix}$ , so $AB+2A=\begin{pmatrix}p+4&-3-2q\\3p+6&3\end{pmatrix}$ . Equating entries gives $p+4=8$ , so $p=4$ , and $-3-2q=-9$ , so $q=3$ . The lower-left entry checks this because $3(4)+6=18$ .
2	$u-1=1$ $u=2$ $v+1=-4$ $v=-5$ $BA$ has order $3\times3$ and $AB$ has order $2\times2$	5
(5 marks)		5
Notes
Multiplication in the stated order gives $AB=\begin{pmatrix}2&u-1\\v+1&0\end{pmatrix}$ . Comparing entries with the given matrix yields $u-1=1$ and $v+1=-4$ , so $u=2$ and $v=-5$ . Since $B$ is $3\times2$ and $A$ is $2\times3$ , $BA$ has order $3\times3$ ; the stated product $AB$ has order $2\times2$ .
3	$AX=\begin{pmatrix}a&b\\2c&2d\end{pmatrix}$ $XB=\begin{pmatrix}-b&a\\-d&c\end{pmatrix}$ $a-b=3$ and $a+b=1$ $a=2$ and $b=-1$ $2c-d=5$ and $c+2d=5$ $X=\begin{pmatrix}2&-1\\3&1\end{pmatrix}$	6
(6 marks)		6
Notes
The two products are $AX=\begin{pmatrix}a&b\\2c&2d\end{pmatrix}$ and $XB=\begin{pmatrix}-b&a\\-d&c\end{pmatrix}$ . Equating the first row of their sum gives $a-b=3$ and $a+b=1$ , so $a=2$ and $b=-1$ . The second row gives $2c-d=5$ and $c+2d=5$ , whose solution is $c=3$ and $d=1$ . Hence $X=\begin{pmatrix}2&-1\\3&1\end{pmatrix}$ .
4	$pA+qB=\begin{pmatrix}p&2p+q\\-p+3q&-2q\end{pmatrix}$ $p=2$ $2p+q=3$ $q=-1$ $B+C=\begin{pmatrix}2&4\\-2&0\end{pmatrix}$ $A(B+C)=\begin{pmatrix}-2&4\\-2&-4\end{pmatrix}$	6
(6 marks)		6
Notes
The linear combination is $pA+qB=\begin{pmatrix}p&2p+q\\-p+3q&-2q\end{pmatrix}$ . Comparing top-left entries with $C$ gives $p=2$ , and the top-right entries then give $2p+q=3$ , so $q=-1$ . The lower entries confirm these values, since $-p+3q=-5$ and $-2q=2$ . Also $B+C=\begin{pmatrix}2&4\\-2&0\end{pmatrix}$ , so $A(B+C)=\begin{pmatrix}-2&4\\-2&-4\end{pmatrix}$ .
5	$A+B=\begin{pmatrix}2&2\\-1&2\end{pmatrix}$ $(A+B)^2=\begin{pmatrix}2&8\\-4&2\end{pmatrix}$ $AB=\begin{pmatrix}2&5\\-2&4\end{pmatrix}$ $A^2+2AB+B^2=\begin{pmatrix}1&10\\-3&3\end{pmatrix}$ $(A+B)^2=A^2+AB+BA+B^2$ $BA=\begin{pmatrix}3&3\\-3&3\end{pmatrix}\ne AB$ , so $AB+BA\ne2AB$	6
(6 marks)		6
Notes
First, $A+B=\begin{pmatrix}2&2\\-1&2\end{pmatrix}$ , so $(A+B)^2=\begin{pmatrix}2&8\\-4&2\end{pmatrix}$ . Direct calculation gives $AB=\begin{pmatrix}2&5\\-2&4\end{pmatrix}$ and $A^2+2AB+B^2=\begin{pmatrix}1&10\\-3&3\end{pmatrix}$ . The correct expansion is $(A+B)^2=A^2+AB+BA+B^2$ . Here $BA=\begin{pmatrix}3&3\\-3&3\end{pmatrix}\ne AB$ , so the two middle terms cannot be replaced by $2AB$ .

CP-3.2 · Understand and use zero and identity matrices.

Tier 1 · Easy

Mark scheme for CP-3.2 Tier 1 · Easy
Question	Scheme	Marks
1	$A+O=IA=\begin{pmatrix}2&-1\\0&3\end{pmatrix}$	2
(2 marks)		2
Notes
The $2\times2$ zero matrix leaves $A$ unchanged under addition, and the $2\times2$ identity leaves $A$ unchanged under multiplication. Hence both results equal $A$ .
2	$(I+N)^2=I+2N$ $(I+N)^3=I+3N+2N^2$ $(I+N)^3=I+3N$	3
(3 marks)		3
Notes
Since $N^2=O$ , $(I+N)^2=I+N+N+N^2=I+2N$ . Then $(I+N)^3=(I+2N)(I+N)=I+3N+2N^2=I+3N$ .

Tier 2 · Standard

Mark scheme for CP-3.2 Tier 2 · Standard
Question	Scheme	Marks
1	$A^2-3A+2I=O$	3
(3 marks)		3
Notes
$A-I=\begin{pmatrix}0&1\\0&1\end{pmatrix}$ and $A-2I=\begin{pmatrix}-1&1\\0&0\end{pmatrix}$ . Their product is $O$ . Expanding $(A-I)(A-2I)$ gives $A^2-3A+2I$ , so $A^2-3A+2I=O$ .
2	$P^3=-I$ $P^4=-P$ $P^5=I-P$ $P^6=I$	4
(4 marks)		4
Notes
From $P^2=P-I$ , $P^3=P^2-P=(P-I)-P=-I$ . Squaring this exact matrix identity gives $P^6=(-I)^2=I$ . Equivalently, continuing the recurrence gives $P^4=-P$ , $P^5=I-P$ and $P^6=I$ , which checks the reduction.
3	$S^2=4E^2-4E+I$ $S^2=I$ $S^{2027}=(S^2)^{1013}S$ $S^{2027}=2E-I$	4
(4 marks)		4
Notes
$S^2=(2E-I)^2=4E^2-4E+I$ . Using $E^2=E$ gives $S^2=I$ . Since $2027=2(1013)+1$ , $S^{2027}=(S^2)^{1013}S=I^{1013}S=2E-I$ .

Tier 3 · Hard

Mark scheme for CP-3.2 Tier 3 · Hard
Question	Scheme	Marks
1	$-2I$	4
(4 marks)		4
Notes
Multiply the given identity by $M$ : $M^3=3M^2+2M=3(3M+2I)+2M=11M+6I$ . Therefore $M^3-4M^2+M=(11M+6I)-4(3M+2I)+M=-2I$ .
2	$(Q+I)^2=6Q$ $u=6$ $v=0$ $(Q+I)^4=144Q-36I$ $w=144$ $t=-36$	6
(6 marks)		6
Notes
Using $Q^2=4Q-I$ , $(Q+I)^2=Q^2+2Q+I=6Q$ , so $u=6$ and $v=0$ . Hence $(Q+I)^4=(6Q)^2=36Q^2=36(4Q-I)=144Q-36I$ , giving $w=144$ and $t=-36$ .
3	$K^4=K(K+I)$ $K^4=K^2+K$ $K^5=K(K^2+K)$ $K^5=K^3+K^2$ $K^5=K^2+K+I$ $a=1$ , $b=1$ and $c=1$	6
(6 marks)		6
Notes
Multiplying $K^3=K+I$ by $K$ gives $K^4=K^2+K$ . Multiplying again gives $K^5=K^3+K^2$ . Substituting the original relation then gives $K^5=K^2+K+I$ , so $a=b=c=1$ .
4	$AB=\begin{pmatrix}q-2&3-r\\2q-4&6-2r\end{pmatrix}$ $q-2=0$ $q=2$ $3-r=0$ $r=3$ $BA=\begin{pmatrix}8&-8\\8&-8\end{pmatrix}$ $AB=O$ does not imply that $BA=O$	7
(7 marks)		7
Notes
Multiplication in the stated order gives $AB=\begin{pmatrix}q-2&3-r\\2q-4&6-2r\end{pmatrix}$ . Equating its entries to zero gives $q=2$ and $r=3$ , with the second row confirming both values. Thus $B=\begin{pmatrix}2&3\\2&3\end{pmatrix}$ . Reversing the order gives $BA=\begin{pmatrix}8&-8\\8&-8\end{pmatrix}$ , which is not the zero matrix. Therefore a zero product in one order need not give a zero product in the reverse order.
5	The diagonal entries of $A^2$ are both $a^2+bc$ The off-diagonal entries of $A^2$ are $ab-ab=0$ and $ac-ac=0$ $A^2=\begin{pmatrix}a^2+bc&0\\0&a^2+bc\end{pmatrix}=(a^2+bc)I$ When $a^2+bc=0$ , $4+c=0$ , so $c=-4$ $A=\begin{pmatrix}2&1\\-4&-2\end{pmatrix}\ne O$ and $A^2=O$ When $a^2+bc=1$ , $4+c=1$ , so $c=-3$ $A=\begin{pmatrix}2&1\\-3&-2\end{pmatrix}$ and $A^2=I$	7
(7 marks)		7
Notes
Multiplication gives $A^2=\begin{pmatrix}a^2+bc&ab-ab\\ac-ac&bc+a^2\end{pmatrix}=(a^2+bc)I$ . With $a=2$ and $b=1$ , the scalar factor is $4+c$ . It is zero when $c=-4$ , giving the non-zero matrix $A=\begin{pmatrix}2&1\\-4&-2\end{pmatrix}$ with $A^2=O$ . It is one when $c=-3$ , giving $A=\begin{pmatrix}2&1\\-3&-2\end{pmatrix}$ with $A^2=I$ .

CP-3.3 · Use matrices to represent linear transformations in 2-D. Successive transformations. Single transformations in 3-D.

Tier 1 · Easy

Mark scheme for CP-3.3 Tier 1 · Easy
Question	Scheme	Marks
1	$P'=(2,3)$ Rotation through $90^\circ$ anticlockwise about the origin	2
(2 marks)		2
Notes
$R\begin{pmatrix}3\\-2\end{pmatrix}=\begin{pmatrix}2\\3\end{pmatrix}$ . The basis vector $\begin{pmatrix}1\\0\end{pmatrix}$ maps to $\begin{pmatrix}0\\1\end{pmatrix}$ , identifying a $90^\circ$ anticlockwise rotation about the origin.
2	$P'=(-6,2)$ Enlargement with centre the origin and scale factor $-2$	2
(2 marks)		2
Notes
$E\begin{pmatrix}3\\-1\end{pmatrix}=\begin{pmatrix}-6\\2\end{pmatrix}$ , so $P'=(-6,2)$ . Since both coordinates are multiplied by $-2$ , the transformation is an enlargement with centre the origin and scale factor $-2$ .

Tier 2 · Standard

Mark scheme for CP-3.3 Tier 2 · Standard
Question	Scheme	Marks
1	$\begin{pmatrix}0&2\\1&0\end{pmatrix}$ Image $(-2,2)$	4
(4 marks)		4
Notes
The reflection matrix is $R=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ and the stretch matrix is $S=\begin{pmatrix}2&0\\0&1\end{pmatrix}$ . Since the reflection occurs first, the combined matrix is $SR=\begin{pmatrix}0&2\\1&0\end{pmatrix}$ . Multiplying by $\begin{pmatrix}2\\-1\end{pmatrix}$ gives $\begin{pmatrix}-2\\2\end{pmatrix}$ .
2	$R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ $H=\begin{pmatrix}3&0\\0&1\end{pmatrix}$ $RH=\begin{pmatrix}0&-1\\3&0\end{pmatrix}$ $RH\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}-2\\3\end{pmatrix}$ , whereas $HR\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}-6\\1\end{pmatrix}$ $H$ acts first and $R$ acts second	5
(5 marks)		5
Notes
$R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ and $H=\begin{pmatrix}3&0\\0&1\end{pmatrix}$ . If $H$ acts first, the product is $RH=\begin{pmatrix}0&-1\\3&0\end{pmatrix}$ and $(1,2)$ maps to $(-2,3)$ , matching the observation. The reverse order gives $HR=\begin{pmatrix}0&-3\\1&0\end{pmatrix}$ and maps the same point to $(-6,1)$ , so that ordering is ruled out.
3	$2\cos\theta=\sqrt3$ $2\sin\theta=1$ $\theta=30^\circ$ $R=\begin{pmatrix}\sqrt3/2&-1/2\\1/2&\sqrt3/2\end{pmatrix}$ $(-1,\sqrt3)$ maps to $(-\sqrt3,1)$	5
(5 marks)		5
Notes
The image of $(2,0)$ under a rotation is $(2\cos\theta,2\sin\theta)$ . Hence $\cos\theta=\sqrt3/2$ and $\sin\theta=1/2$ , so the smallest positive angle is $30^\circ$ . Thus $R=\begin{pmatrix}\sqrt3/2&-1/2\\1/2&\sqrt3/2\end{pmatrix}$ . Multiplication gives $R\begin{pmatrix}-1\\\sqrt3\end{pmatrix}=\begin{pmatrix}-\sqrt3\\1\end{pmatrix}$ .

Tier 3 · Hard

Mark scheme for CP-3.3 Tier 3 · Hard
Question	Scheme	Marks
1	$\begin{pmatrix}1&0&0\\0&0&-1\\0&1&0\end{pmatrix}$ Image $(2,-4,-1)$ Rotation through $90^\circ$ about the $x$ -axis (taking the positive $y$ -axis to the positive $z$ -axis)	5
(5 marks)		5
Notes
The images of the basis vectors form the columns: $\begin{pmatrix}1\\0\\0\end{pmatrix}\mapsto\begin{pmatrix}1\\0\\0\end{pmatrix}$ , $\begin{pmatrix}0\\1\\0\end{pmatrix}\mapsto\begin{pmatrix}0\\0\\1\end{pmatrix}$ , $\begin{pmatrix}0\\0\\1\end{pmatrix}\mapsto\begin{pmatrix}0\\-1\\0\end{pmatrix}$ , giving $T=\begin{pmatrix}1&0&0\\0&0&-1\\0&1&0\end{pmatrix}$ . Thus $T\begin{pmatrix}2\\-1\\4\end{pmatrix}=\begin{pmatrix}2\\-4\\-1\end{pmatrix}$ . Every point on the $x$ -axis is fixed and the $y$ -axis maps onto the $z$ -axis, so this is a rotation through $90^\circ$ about the $x$ -axis.
2	The basis vectors map to $(0,1,0)$ , $(-1,0,0)$ and $(0,0,1)$ $\begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix}$ $(x,y,z)$ maps to $(-y,x,z)$ Rotation through $90^\circ$ anticlockwise about the $z$ -axis $(2,-1,3)$ maps to $(1,2,3)$	5
(5 marks)		5
Notes
Write the three given vectors as $\mathbf u_1,\mathbf u_2,\mathbf u_3$ , in the order stated. Then $(1,0,0)=\tfrac12(\mathbf u_1+\mathbf u_3-\mathbf u_2)$ , $(0,1,0)=\tfrac12(\mathbf u_1+\mathbf u_2-\mathbf u_3)$ and $(0,0,1)=\tfrac12(\mathbf u_2+\mathbf u_3-\mathbf u_1)$ . Applying the same combinations to their images gives $(0,1,0)$ , $(-1,0,0)$ and $(0,0,1)$ respectively. These are the columns of $T=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix}$ . Thus $(x,y,z)$ maps to $(-y,x,z)$ , a rotation through $90^\circ$ anticlockwise about the $z$ -axis, and $(2,-1,3)$ maps to $(1,2,3)$ .
3	$T\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}0\\3\end{pmatrix}$ and $T\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}3\\0\end{pmatrix}$ $R$ is reflection in $y=x$ $R=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ $E=\begin{pmatrix}3&0\\0&3\end{pmatrix}$ , an enlargement with centre the origin and scale factor $3$ $T=ER$ , so the reflection acts first and the enlargement acts second $(2,-1)$ maps to $(-3,6)$	6
(6 marks)		6
Notes
The columns show that the positive coordinate directions are exchanged and their lengths are multiplied by $3$ . Hence $R=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ is reflection in $y=x$ , while $E=3I$ is enlargement with centre the origin and scale factor $3$ . Since $T=ER$ , $R$ acts first and $E$ acts second, in agreement with the right-to-left convention. Finally, $T\begin{pmatrix}2\\-1\end{pmatrix}=\begin{pmatrix}-3\\6\end{pmatrix}$ .
4	Reflection in $y=-x$ has matrix $R=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$ Reflection in the $y$ -axis has matrix $F=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ Since $R$ acts first and $F$ second, the combined matrix is $FR$ $FR=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ The single equivalent transformation is rotation through $90^\circ$ clockwise about the origin $(4,1)$ maps to $(1,-4)$	6
(6 marks)		6
Notes
The reflection matrices are $R=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$ for $y=-x$ and $F=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ for the $y$ -axis. The first transformation is the right-hand factor, so the combined matrix is $FR=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ . This is rotation through $90^\circ$ clockwise about the origin. It maps $\begin{pmatrix}4\\1\end{pmatrix}$ to $\begin{pmatrix}1\\-4\end{pmatrix}$ .
5	$F=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ $H=\begin{pmatrix}1&0\\0&4\end{pmatrix}$ $R=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ $RHF=\begin{pmatrix}0&-1\\-4&0\end{pmatrix}$ $FHR=\begin{pmatrix}0&-4\\-1&0\end{pmatrix}$ In the stated order, $(5,-2)$ maps to $(2,-20)$ In reverse order, $(5,-2)$ maps to $(8,-5)$ , so reversing the order changes the transformation	7
(7 marks)		7
Notes
The matrices are $F=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ , $H=\begin{pmatrix}1&0\\0&4\end{pmatrix}$ and $R=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ . In the stated order the product is $RHF=\begin{pmatrix}0&-1\\-4&0\end{pmatrix}$ , which maps $(5,-2)$ to $(2,-20)$ . Reversing the transformations reverses the factor order, giving $FHR=\begin{pmatrix}0&-4\\-1&0\end{pmatrix}$ and image $(8,-5)$ . The different images confirm that order matters.

CP-3.4 · Find invariant points and lines for a linear transformation.

Tier 1 · Easy

Mark scheme for CP-3.4 Tier 1 · Easy
Question	Scheme	Marks
1	All points $(x,0)$ on the $x$ -axis	3
(3 marks)		3
Notes
Set $A\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x\\y\end{pmatrix}$ . This gives $x=x$ and $3y=y$ , so $2y=0$ and $y=0$ . Hence every point $(x,0)$ is invariant.
2	$k=0$ No: the line does not consist entirely of invariant points When $k=0$ , $(t,t)\mapsto(3t,3t)$ , so only the origin is fixed	3
(3 marks)		3
Notes
A direction vector for $y=x$ is $(1,1)$ . Its image is $(3+k,3)$ , which is parallel to $(1,1)$ precisely when $3+k=3$ , so $k=0$ . At this value $A_0(t,t)=(3t,3t)$ , which remains on $y=x$ ; hence the line is invariant. Except at $t=0$ , the image differs from $(t,t)$ , so the line is not pointwise invariant.

Tier 2 · Standard

Mark scheme for CP-3.4 Tier 2 · Standard
Question	Scheme	Marks
1	$y=0$ and $y=-x$	4
(4 marks)		4
Notes
A point $(x,mx)$ maps to $((2+m)x,mx)$ . For the image to remain on $y=m x$ , require $mx=m(2+m)x$ . Thus $m=0$ or $1=2+m$ , giving $m=-1$ . The line $x=0$ is not invariant because $(0,y)$ maps to $(y,y)$ . Therefore the invariant lines are $y=0$ and $y=-x$ .
2	$m^2+tm+1=0$ $t^2-4=0$ $t=\pm2$ $y=-x$ when $t=2$ ; $y=x$ when $t=-2$	4
(4 marks)		4
Notes
For a non-vertical line with direction $(1,m)$ , the image is $(-m,1+tm)$ . Parallelism requires $1+tm=-m^2$ , hence $m^2+tm+1=0$ . There is exactly one real gradient when the discriminant $t^2-4$ is zero, so $t=\pm2$ . The repeated gradient is $m=-1$ when $t=2$ , giving $y=-x$ , and $m=1$ when $t=-2$ , giving $y=x$ . The vertical direction maps to $(-1,t)$ and supplies no further invariant line.
3	$\det(A_k-I)=3(k-3)$ $k=3$ The non-zero invariant points lie on $y=-x$ The invariant lines through the origin are $y=-x$ and $y=\dfrac32x$ $y=-x$ consists entirely of invariant points	5
(5 marks)		5
Notes
Non-zero invariant points require $(A_k-I)\mathbf x=\mathbf0$ to have a non-zero solution. Since $\det(A_k-I)=3(k-3)$ , this occurs when $k=3$ . At this value, solving $(A_3-I)\mathbf x=\mathbf0$ gives $y=-x$ , so this line is pointwise fixed. The other eigendirection of $A_3$ satisfies $y=3x/2$ . Hence the two invariant lines through the origin are $y=-x$ and $y=3x/2$ , with only the first consisting entirely of invariant points.

Tier 3 · Hard

Mark scheme for CP-3.4 Tier 3 · Hard
Question	Scheme	Marks
1	$y=0$ and the family $y=x+c$ , where $c$ is any real constant $y=0$ consists entirely of invariant points	6
(6 marks)		6
Notes
Take $y=mx+c$ . Then $X=(1+m)x+c$ and $Y=2mx+2c$ . Requiring $Y=mX+c$ for every $x$ gives $2m=m(1+m)$ and $2c=mc+c$ . Hence either $m=0,c=0$ , giving $y=0$ , or $m=1$ with any $c$ , giving $y=x+c$ . No vertical line is invariant because $X=x+y$ varies with $y$ . On $y=0$ , $(x,0)$ maps to itself, so this is a line of invariant points; points on $y=x+c$ are generally moved along the line.
2	$m(5-k-2m)=0$ $2c(2-m)=0$ $y=0$ is invariant for every value of $k$ If $k\ne1,5$ , the other invariant line is $y=\dfrac{5-k}{2}x$ If $k=5$ , only $y=0$ is invariant If $k=1$ , the invariant lines are $y=0$ and every line $y=2x+c$ , where $c\in\mathbb R$ There is no invariant vertical line	7
(7 marks)		7
Notes
Let $y=mx+c$ . Its image has $X=(k+2m)x+2c$ and $Y=5mx+5c$ . Requiring $Y=mX+c$ for every real $x$ gives $m(5-k-2m)=0$ and $2c(2-m)=0$ . Thus $m=0,c=0$ always gives $y=0$ . The other gradient is $m=(5-k)/2$ : it gives an additional line through the origin unless $k=5$ , while $m=2$ at $k=1$ permits every real $c$ , producing the family $y=2x+c$ . A vertical line is not invariant because $X=kx+2y$ varies with $y$ .
3	A non-vertical invariant line $y=mx$ satisfies $km^2+4m+1=0$ The vertical line $x=0$ is invariant only when $k=0$ If $k>4$ , there are no invariant lines through the origin If $k=4$ , the only invariant line is $y=-\dfrac12x$ If $k<4$ and $k\ne0$ , the two lines are $y=\dfrac{-2+\sqrt{4-k}}k x$ and $y=\dfrac{-2-\sqrt{4-k}}k x$ If $k=0$ , the two invariant lines are $y=-\dfrac14x$ and $x=0$ Only when $k=-5$ does an invariant line consist entirely of invariant points; that line is $y=x$	7
(7 marks)		7
Notes
The direction $(1,m)$ maps to $(6+km,-1+2m)$ . Parallelism gives $-1+2m=m(6+km)$ , so $km^2+4m+1=0$ . Its discriminant is $16-4k$ . Thus there are no real gradients when $k>4$ , one repeated gradient $m=-1/2$ when $k=4$ , and two real gradients when $k<4$ , except that $k=0$ must be solved as the linear equation $4m+1=0$ . The vertical direction maps to $(k,2)$ , so $x=0$ supplies the second invariant line precisely when $k=0$ . A line is pointwise fixed only if $1$ is an eigenvalue: $\det(T_k-I)=k+5$ , so $k=-5$ , when the fixed line is $y=x$ .
4	$T\begin{pmatrix}1\\3\end{pmatrix}=2\begin{pmatrix}1\\3\end{pmatrix}$ $T\begin{pmatrix}1\\-2\end{pmatrix}=3\begin{pmatrix}1\\-2\end{pmatrix}$ $T=\begin{pmatrix}13/5&-1/5\\-6/5&12/5\end{pmatrix}$ $T-I=\begin{pmatrix}8/5&-1/5\\-6/5&7/5\end{pmatrix}$ $\det(T-I)=2$ Since $\det(T-I)\ne0$ , the origin is the only invariant point The origin is not on $L$ , so no point of $L$ is invariant	7
(7 marks)		7
Notes
Using the two independent direction vectors as columns gives $T\begin{pmatrix}1&1\\3&-2\end{pmatrix}=\begin{pmatrix}2&3\\6&-6\end{pmatrix}$ . Hence $T=\begin{pmatrix}2&3\\6&-6\end{pmatrix}\begin{pmatrix}1&1\\3&-2\end{pmatrix}^{-1}=\begin{pmatrix}13/5&-1/5\\-6/5&12/5\end{pmatrix}$ . Invariant points satisfy $(T-I)\mathbf x=\mathbf0$ . Here $T-I=\begin{pmatrix}8/5&-1/5\\-6/5&7/5\end{pmatrix}$ and $\det(T-I)=2\ne0$ , so the origin is the only invariant point. Since $(0,0)$ does not satisfy $y=x+4$ , no point of $L$ is invariant.
5	A point $(x,mx+c)$ maps to $X=(2+6m)x+6c$ The corresponding second coordinate is $Y=(1+3m)x+3c$ Equating coefficients in $Y=mX+c$ gives $6m^2-m-1=0$ Equating constants gives $c(1-3m)=0$ $m=\dfrac12$ or $m=-\dfrac13$ For both gradients, $c=0$ No vertical line is invariant The invariant lines are $y=\dfrac12x$ , on which every image lies, and $y=-\dfrac13x$ , whose points map to the origin	8
(8 marks)		8
Notes
For a general point on $y=mx+c$ , the image is $(X,Y)=((2+6m)x+6c,(1+3m)x+3c)$ . Requiring $Y=mX+c$ for every $x$ gives $6m^2-m-1=0$ and $c(1-3m)=0$ . The gradients are $m=1/2$ and $m=-1/3$ , and each forces $c=0$ . A vertical line is not invariant because $X=2x+6y$ varies with $y$ . Since $X=2Y$ for every image, every image lies on $y=x/2$ . A point $(x,-x/3)$ maps to the origin, so $y=-x/3$ is also invariant.

CP-3.5 · Calculate determinants of 2x2 and 3x3 matrices and interpret as scale factors, including the effect on orientation.

Tier 1 · Easy

Mark scheme for CP-3.5 Tier 1 · Easy
Question	Scheme	Marks
1	Image area $=60$ Orientation is reversed	3
(3 marks)		3
Notes
$\det A=1(3)-4(2)=-5$ . The area scale factor is $\|-5\|=5$ , so the image area is $12\times5=60$ . The negative determinant means orientation is reversed.
2	$\det\begin{pmatrix}2&3\\4&6\end{pmatrix}=0$ The region collapses onto the line $y=2x$ . Image area $0$	3
(3 marks)		3
Notes
The determinant is $2(6)-3(4)=0$ , so the area scale factor is $0$ and the image area is $7\times0=0$ . The two columns are parallel, so every image vector lies on the line $y=2x$ through the origin.

Tier 2 · Standard

Mark scheme for CP-3.5 Tier 2 · Standard
Question	Scheme	Marks
1	$\det M=9$ Volume scale factor $9$ Orientation is preserved	4
(4 marks)		4
Notes
Expanding along the first row, $\det M=1\begin{vmatrix}3&1\\0&1\end{vmatrix}-2\begin{vmatrix}-1&1\\2&1\end{vmatrix}=3-2(-3)=9$ . Therefore volumes are multiplied by $\|9\|=9$ , and the positive sign preserves orientation.
2	$\|\det(SR)\|=30/5=6$ $\det(SR)=-6$ $\det S=3$ $S$ preserves orientation	4
(4 marks)		4
Notes
The magnitude of the combined determinant is the area ratio $30/5=6$ . Because the combination reverses orientation, its signed determinant is $-6$ . Since $R$ acts first, the combined matrix is $SR$ , and $\det(SR)=(\det S)(-2)=-6$ . Therefore $\det S=3$ , whose positive sign shows that $S$ preserves orientation.
3	$\det B=2(-3)-1(1+8)$ $\det B=-15$ Orientation is reversed The image volume is $4\times\|-15\|=60$	4
(4 marks)		4
Notes
Expanding along the first row gives $\det B=2((-3)(1)-2(0))-1(1(1)-2(-4))+0=-6-9=-15$ . The negative determinant reverses orientation, while volumes are multiplied by $\|-15\|=15$ , so the image volume is $4\times15=60$ .

Tier 3 · Hard

Mark scheme for CP-3.5 Tier 3 · Hard
Question	Scheme	Marks
1	$k=-3,0,1,4$ Orientation is reversed for $k=0$ and $k=1$	5
(5 marks)		5
Notes
The area scale factor is $42/7=6$ , so $\|\det P\|=6$ . Now $\det P=k(k-1)-6=k^2-k-6$ . Solving $k^2-k-6=6$ gives $k^2-k-12=0$ , so $k=4$ or $-3$ . Solving $k^2-k-6=-6$ gives $k(k-1)=0$ , so $k=0$ or $1$ . The determinant is negative in the second case, so $k=0,1$ reverse orientation.
2	$\det A_k=k-4$ $\det B=-2$ $\det(BA_k)=-2(k-4)$ $\|-2(k-4)\|=10$ , so $k=9$ or $k=-1$ Preserved orientation requires $-2(k-4)>0$ , so $k<4$ $k=-1$	6
(6 marks)		6
Notes
Cofactor expansion gives $\det A_k=k-4$ , while $\det B=-2$ . The combined matrix is $BA_k$ , so its determinant is $-2(k-4)$ . The volume ratio is $30/3=10$ , hence $\|-2(k-4)\|=10$ and $k=9$ or $k=-1$ . Preserved orientation requires $-2(k-4)>0$ , which holds only when $k<4$ . Therefore the value of $k$ satisfying the stated condition is $k=-1$ ; indeed the combined determinant is then $10$ .
3	$\det A_h=h^2-2h-8$ $\det A_h=(h-4)(h+2)$ The transformation is singular for $h\in\{-2,4\}$ Reversed orientation and a volume scale factor of $8$ require $\det A_h=-8$ $(h-4)(h+2)=-8$ , so $h(h-2)=0$ $h=0$ or $h=2$ , and $h>0$ gives $h=2$	6
(6 marks)		6
Notes
Cofactor expansion gives $\det A_h=-2h-8+h^2=h^2-2h-8=(h-4)(h+2)$ , so the transformation is singular precisely when $h=-2$ or $h=4$ . The image-to-original volume ratio is $24/3=8$ . Because orientation is reversed, the signed determinant must be $-8$ . Thus $(h-4)(h+2)=-8$ , which simplifies to $h(h-2)=0$ . The candidates are $h=0$ and $h=2$ , and the condition $h>0$ selects $h=2$ .
4	$\det\begin{pmatrix}1&3\\2&-1\end{pmatrix}=-7$ $\det\begin{pmatrix}3&1\\-2&3\end{pmatrix}=11$ $11=(\det T)(-7)$ $\det T=-\dfrac{11}7$ The image area is $14\left\|-\dfrac{11}7\right\|=22$ Orientation is reversed	6
(6 marks)		6
Notes
Place the two original vectors in a matrix as columns. Its determinant is $\det\begin{pmatrix}1&3\\2&-1\end{pmatrix}=-7$ . The corresponding image-vector matrix has determinant $\det\begin{pmatrix}3&1\\-2&3\end{pmatrix}=11$ . Since the image matrix is $T\begin{pmatrix}1&3\\2&-1\end{pmatrix}$ , determinants give $11=(\det T)(-7)$ , so $\det T=-11/7$ . The area scale factor is $11/7$ , giving image area $14(11/7)=22$ , and the negative sign reverses orientation.
5	$\det M_k=k^2-6$ Reversed orientation requires $k^2-6<0$ A smaller image area requires $\|k^2-6\|<1$ Together, $-1<k^2-6<0$ $5<k^2<6$ $k\in(-\sqrt6,-\sqrt5)\cup(\sqrt5,\sqrt6)$	6
(6 marks)		6
Notes
The signed area scale factor is $\det M_k=k^2-6$ . Reversed orientation requires this determinant to be negative, while a smaller image area requires its magnitude to be less than $1$ . Combining the strict inequalities gives $-1<k^2-6<0$ , or $5<k^2<6$ . Therefore $-\sqrt6<k<-\sqrt5$ or $\sqrt5<k<\sqrt6$ ; the strict inequalities exclude all four endpoints.

CP-3.6 · Understand and use singular and non-singular matrices. Properties of inverse matrices. Calculate and use the inverse of non-singular 2x2 and 3x3 matrices.

Tier 1 · Easy

Mark scheme for CP-3.6 Tier 1 · Easy
Question	Scheme	Marks
1	$A^{-1}=\begin{pmatrix}1&-1\\-2&3\end{pmatrix}$	3
(3 marks)		3
Notes
$\det A=3(1)-1(2)=1\ne0$ , so $A$ is non-singular. Hence $A^{-1}=\frac1{1}\begin{pmatrix}1&-1\\-2&3\end{pmatrix}$ .
2	$(k+1)(k-1)-4(2)=0$ $k^2-9=0$ $k=\pm3$	3
(3 marks)		3
Notes
A singular matrix has determinant zero. Hence $(k+1)(k-1)-4(2)=0$ , so $k^2-9=0$ and $k=\pm3$ .

Tier 2 · Standard

Mark scheme for CP-3.6 Tier 2 · Standard
Question	Scheme	Marks
1	$C^{-1}=\frac12\begin{pmatrix}1&-1&1\\1&1&-1\\-1&1&1\end{pmatrix}$	5
(5 marks)		5
Notes
$\det C=2$ . The signed minors give the cofactor matrix $\begin{pmatrix}1&1&-1\\-1&1&1\\1&-1&1\end{pmatrix}$ , so $\operatorname{adj}C=\begin{pmatrix}1&-1&1\\1&1&-1\\-1&1&1\end{pmatrix}$ . Hence $C^{-1}=\frac1{\det C}\operatorname{adj}C=\frac12\begin{pmatrix}1&-1&1\\1&1&-1\\-1&1&1\end{pmatrix}$ . Multiplying by $C$ gives $I$ .
2	$XAA^{-1}=\begin{pmatrix}2&3\\1&4\end{pmatrix}A^{-1}$ $X=\begin{pmatrix}2&3\\1&4\end{pmatrix}\begin{pmatrix}1&-1\\0&1\end{pmatrix}$ $X=\begin{pmatrix}2&1\\1&3\end{pmatrix}$ $XA=\begin{pmatrix}2&3\\1&4\end{pmatrix}$	4
(4 marks)		4
Notes
Right-multiply both sides by $A^{-1}$ : $XAA^{-1}=\begin{pmatrix}2&3\\1&4\end{pmatrix}A^{-1}$ . Hence $X=\begin{pmatrix}2&3\\1&4\end{pmatrix}\begin{pmatrix}1&-1\\0&1\end{pmatrix}=\begin{pmatrix}2&1\\1&3\end{pmatrix}$ . Since $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ , the check $XA=\begin{pmatrix}2&3\\1&4\end{pmatrix}$ confirms the side on which the inverse was used.
3	$(AB)^{-1}=B^{-1}A^{-1}$ $(AB)^{-1}=\begin{pmatrix}2&0\\-1&1\end{pmatrix}\begin{pmatrix}1&2\\0&1\end{pmatrix}$ $(AB)^{-1}=\begin{pmatrix}2&4\\-1&-1\end{pmatrix}$ $\mathbf x=(AB)^{-1}\begin{pmatrix}3\\-1\end{pmatrix}$ $\mathbf x=\begin{pmatrix}2\\-2\end{pmatrix}$	5
(5 marks)		5
Notes
The factors reverse when a product is inverted, so $(AB)^{-1}=B^{-1}A^{-1}=\begin{pmatrix}2&0\\-1&1\end{pmatrix}\begin{pmatrix}1&2\\0&1\end{pmatrix}=\begin{pmatrix}2&4\\-1&-1\end{pmatrix}$ . Left-multiplying the equation by this inverse gives $\mathbf x=\begin{pmatrix}2&4\\-1&-1\end{pmatrix}\begin{pmatrix}3\\-1\end{pmatrix}=\begin{pmatrix}2\\-2\end{pmatrix}$ .

Tier 3 · Hard

Mark scheme for CP-3.6 Tier 3 · Hard
Question	Scheme	Marks
1	$D^{-1}=\begin{pmatrix}2&-1\\-3&2\end{pmatrix}$	5
(5 marks)		5
Notes
From $D^2-4D+I=O$ , rearrange to $D(4I-D)=I$ . Since the same expression is a polynomial in $D$ , $(4I-D)D=I$ as well. Therefore $D^{-1}=4I-D=\begin{pmatrix}2&-1\\-3&2\end{pmatrix}$ .
2	$A^{-1}=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}$ $B^{-1}=\begin{pmatrix}1&0\\-1&1\end{pmatrix}$ $X=A^{-1}\begin{pmatrix}4&-1\\3&0\end{pmatrix}B^{-1}$ $A^{-1}\begin{pmatrix}4&-1\\3&0\end{pmatrix}=\begin{pmatrix}1&-1\\2&1\end{pmatrix}$ $X=\begin{pmatrix}2&-1\\1&1\end{pmatrix}$ $AXB=\begin{pmatrix}4&-1\\3&0\end{pmatrix}$	6
(6 marks)		6
Notes
$A$ and $B$ have determinant $1$ , with $A^{-1}=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}$ and $B^{-1}=\begin{pmatrix}1&0\\-1&1\end{pmatrix}$ . Left-multiplying by $A^{-1}$ and right-multiplying by $B^{-1}$ gives $X=A^{-1}\begin{pmatrix}4&-1\\3&0\end{pmatrix}B^{-1}=\begin{pmatrix}2&-1\\1&1\end{pmatrix}$ . Direct multiplication gives $AX=\begin{pmatrix}5&-1\\3&0\end{pmatrix}$ and $(AX)B=\begin{pmatrix}4&-1\\3&0\end{pmatrix}$ , as required.
3	$X=\begin{pmatrix}0&-1\\-2&1\end{pmatrix}$ and $Y=\begin{pmatrix}1&-2\\-1&0\end{pmatrix}$ $XA=\begin{pmatrix}0&-2\\-2&0\end{pmatrix}$ $AY=\begin{pmatrix}0&-2\\-2&0\end{pmatrix}$ , so $XA=AY$ $X^{-1}A=AY^{-1}$ , or directly $X^{-1}=\begin{pmatrix}-1/2&-1/2\\-1&0\end{pmatrix}$ $X^{-1}A=\begin{pmatrix}-1/2&-3/2\\-1&-1\end{pmatrix}$	5
(5 marks)		5
Notes
Calculation gives $X=I-AB=\begin{pmatrix}0&-1\\-2&1\end{pmatrix}$ and $Y=I-BA=\begin{pmatrix}1&-2\\-1&0\end{pmatrix}$ . Both $XA$ and $AY$ equal $\begin{pmatrix}0&-2\\-2&0\end{pmatrix}$ . One route is to use $Y^{-1}=\begin{pmatrix}0&-1\\-1/2&-1/2\end{pmatrix}$ : from $XA=AY$ , left-multiplication by $X^{-1}$ and right-multiplication by $Y^{-1}$ give $X^{-1}A=AY^{-1}=\begin{pmatrix}-1/2&-3/2\\-1&-1\end{pmatrix}$ . Directly calculating $X^{-1}$ and multiplying by $A$ earns the same result.
4	$\det A=4$ $\operatorname{adj}A=\begin{pmatrix}1&-1&1\\7&1&-5\\-5&1&3\end{pmatrix}$ $A^{-1}=\dfrac14\begin{pmatrix}1&-1&1\\7&1&-5\\-5&1&3\end{pmatrix}$ $X=A^{-1}\begin{pmatrix}7&1\\14&4\\11&3\end{pmatrix}$ $X=\begin{pmatrix}1&0\\2&-1\\3&2\end{pmatrix}$ $\begin{pmatrix}2&1&1\\1&2&3\\3&1&2\end{pmatrix}\begin{pmatrix}1&0\\2&-1\\3&2\end{pmatrix}=\begin{pmatrix}7&1\\14&4\\11&3\end{pmatrix}$	6
(6 marks)		6
Notes
Cofactor expansion gives $\det A=4$ and $\operatorname{adj}A=\begin{pmatrix}1&-1&1\\7&1&-5\\-5&1&3\end{pmatrix}$ . Hence $A^{-1}=\frac14\begin{pmatrix}1&-1&1\\7&1&-5\\-5&1&3\end{pmatrix}$ . Left-multiplying the equation by $A^{-1}$ gives $X=A^{-1}\begin{pmatrix}7&1\\14&4\\11&3\end{pmatrix}=\begin{pmatrix}1&0\\2&-1\\3&2\end{pmatrix}$ . Multiplication by $A$ reproduces the stated $3\times2$ matrix.
5	$A^{-1}=\dfrac12\begin{pmatrix}1&-1\\-1&3\end{pmatrix}$ $B^{-1}=\begin{pmatrix}1/2&-1/2\\0&1\end{pmatrix}$ $A^{-1}+B^{-1}=\begin{pmatrix}1&-1\\-1/2&5/2\end{pmatrix}$ $(A^{-1}+B^{-1})^{-1}=\begin{pmatrix}5/4&1/2\\1/4&1/2\end{pmatrix}$ $A+B=\begin{pmatrix}5&2\\1&2\end{pmatrix}$ $(A+B)^{-1}=\begin{pmatrix}1/4&-1/4\\-1/8&5/8\end{pmatrix}$ $A(A+B)^{-1}B=\begin{pmatrix}5/4&1/2\\1/4&1/2\end{pmatrix}$	7
(7 marks)		7
Notes
The inverses are $A^{-1}=\frac12\begin{pmatrix}1&-1\\-1&3\end{pmatrix}$ and $B^{-1}=\begin{pmatrix}1/2&-1/2\\0&1\end{pmatrix}$ . Their sum is $\begin{pmatrix}1&-1\\-1/2&5/2\end{pmatrix}$ , whose inverse is $\begin{pmatrix}5/4&1/2\\1/4&1/2\end{pmatrix}$ . Also $A+B=\begin{pmatrix}5&2\\1&2\end{pmatrix}$ , so $(A+B)^{-1}=\begin{pmatrix}1/4&-1/4\\-1/8&5/8\end{pmatrix}$ . Multiplying in the stated order gives $A(A+B)^{-1}B=\begin{pmatrix}5/4&1/2\\1/4&1/2\end{pmatrix}$ , verifying the equality.

CP-3.7 · Solve three linear simultaneous equations in three variables by use of the inverse matrix.

Tier 1 · Easy

Mark scheme for CP-3.7 Tier 1 · Easy
Question	Scheme	Marks
1	$x=4$ , $y=0$ , $z=2$	3
(3 marks)		3
Notes
$\begin{pmatrix}x\\y\\z\end{pmatrix}=A^{-1}\begin{pmatrix}4\\2\\6\end{pmatrix}=\frac12\begin{pmatrix}4-2+6\\4+2-6\\-4+2+6\end{pmatrix}=\begin{pmatrix}4\\0\\2\end{pmatrix}$ .
2	$\mathbf{x}=\begin{pmatrix}x\\y\\z\end{pmatrix}$ $A=\begin{pmatrix}2&0&1\\1&-1&0\\0&3&2\end{pmatrix}$ $\mathbf{b}=\begin{pmatrix}5\\0\\7\end{pmatrix}$ $(x,y,z)=(3,3,-1)$	4
(4 marks)		4
Notes
Use the variable order $x,y,z$ in every row, including zeros for missing variables. This gives $\begin{pmatrix}2&0&1\\1&-1&0\\0&3&2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}5\\0\\7\end{pmatrix}$ . Then $\mathbf{x}=A^{-1}\mathbf{b}=\begin{pmatrix}2&-3&-1\\2&-4&-1\\-3&6&2\end{pmatrix}\begin{pmatrix}5\\0\\7\end{pmatrix}=\begin{pmatrix}3\\3\\-1\end{pmatrix}$ .

Tier 2 · Standard

Mark scheme for CP-3.7 Tier 2 · Standard
Question	Scheme	Marks
1	$x=3$ , $y=1$ , $z=3$	5
(5 marks)		5
Notes
The system is $A\mathbf{x}=\mathbf{b}$ with $A=\begin{pmatrix}1&2&0\\0&1&1\\0&0&1\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix}5\\4\\3\end{pmatrix}$ . Row reduction gives $A^{-1}=\begin{pmatrix}1&-2&2\\0&1&-1\\0&0&1\end{pmatrix}$ . Thus $\mathbf{x}=A^{-1}\mathbf{b}=\begin{pmatrix}5-8+6\\4-3\\3\end{pmatrix}=\begin{pmatrix}3\\1\\3\end{pmatrix}$ .
2	$k+6=10$ $k=4$ $A_4^{-1}=\begin{pmatrix}4&-1/2&-1\\-5&1&1\\2&-1/2&0\end{pmatrix}$ $x=4(6)-\tfrac12(22)-10=3$ $y=-5(6)+22+10=2$ $z=2(6)-\tfrac12(22)=1$	6
(6 marks)		6
Notes
$A_k\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}3\\k+6\\6\end{pmatrix}$ . Comparing this with $(3,10,6)^T$ gives $k+6=10$ , so $k=4$ . For this value, $A_4^{-1}=\begin{pmatrix}4&-1/2&-1\\-5&1&1\\2&-1/2&0\end{pmatrix}$ . Hence $\begin{pmatrix}x\\y\\z\end{pmatrix}=A_4^{-1}\begin{pmatrix}6\\22\\10\end{pmatrix}=\begin{pmatrix}3\\2\\1\end{pmatrix}$ . Multiplying $A_4$ by this vector reproduces $(6,22,10)^T$ .
3	$\begin{pmatrix}2&1&1\\1&3&2\\3&2&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}13\\19\\18\end{pmatrix}$ $\det\begin{pmatrix}2&1&1\\1&3&2\\3&2&1\end{pmatrix}=-4$ $A^{-1}=\begin{pmatrix}1/4&-1/4&1/4\\-5/4&1/4&3/4\\7/4&1/4&-5/4\end{pmatrix}$ $\begin{pmatrix}x\\y\\z\end{pmatrix}=A^{-1}\begin{pmatrix}13\\19\\18\end{pmatrix}$ $(x,y,z)=(3,2,5)$ One notebook, two pens and three folders cost £22	6
(6 marks)		6
Notes
The information gives $A\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}13\\19\\18\end{pmatrix}$ with $A=\begin{pmatrix}2&1&1\\1&3&2\\3&2&1\end{pmatrix}$ . Its determinant is $-4$ and $A^{-1}=\begin{pmatrix}1/4&-1/4&1/4\\-5/4&1/4&3/4\\7/4&1/4&-5/4\end{pmatrix}$ . Multiplication gives $(x,y,z)=(3,2,5)$ . The requested cost is $x+2y+3z=3+4+15=22$ pounds.

Tier 3 · Hard

Mark scheme for CP-3.7 Tier 3 · Hard
Question	Scheme	Marks
1	$p=4$ $(x,y,z)=(3,1,4)$	6
(6 marks)		6
Notes
Write $T\mathbf{x}=\begin{pmatrix}p\\5\\8\end{pmatrix}$ with $T=\begin{pmatrix}1&1&0\\0&1&1\\0&0&2\end{pmatrix}$ . Its inverse is $T^{-1}=\begin{pmatrix}1&-1&1/2\\0&1&-1/2\\0&0&1/2\end{pmatrix}$ . Therefore $(x,y,z)=(p-1,1,4)$ . The condition $x+y=z$ gives $(p-1)+1=4$ , so $p=4$ and the solution is $(3,1,4)$ .
2	$\begin{pmatrix}1&0&0\\1&2&4\\1&3&9\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2\\9\\20\end{pmatrix}$ $\begin{pmatrix}1&0&0\\1&2&4\\1&3&9\end{pmatrix}^{-1}=\begin{pmatrix}1&0&0\\-5/6&3/2&-2/3\\1/6&-1/2&1/3\end{pmatrix}$ $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1&0&0\\-5/6&3/2&-2/3\\1/6&-1/2&1/3\end{pmatrix}\begin{pmatrix}2\\9\\20\end{pmatrix}$ $x=2$ $y=-\tfrac32$ $z=\tfrac52$ $q(1)=3$	7
(7 marks)		7
Notes
The three given values give $A\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2\\9\\20\end{pmatrix}$ , where $A=\begin{pmatrix}1&0&0\\1&2&4\\1&3&9\end{pmatrix}$ and $A^{-1}=\begin{pmatrix}1&0&0\\-5/6&3/2&-2/3\\1/6&-1/2&1/3\end{pmatrix}$ . Therefore $\begin{pmatrix}x\\y\\z\end{pmatrix}=A^{-1}\begin{pmatrix}2\\9\\20\end{pmatrix}=\begin{pmatrix}2\\-3/2\\5/2\end{pmatrix}$ . Substitution into the quadratic then gives $q(1)=2-3/2+5/2=3$ .
3	$\begin{pmatrix}1&1&1\\1&2&0\\0&1&2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\11-3p\\1\end{pmatrix}$ $A^{-1}=\begin{pmatrix}4/3&-1/3&-2/3\\-2/3&2/3&1/3\\1/3&-1/3&1/3\end{pmatrix}$ $\begin{pmatrix}x\\y\\z\end{pmatrix}=A^{-1}\begin{pmatrix}4\\11-3p\\1\end{pmatrix}$ $x=p+1$ $y=5-2p$ $z=p-2$ $2<p<\dfrac52$	7
(7 marks)		7
Notes
The coefficient matrix is $A=\begin{pmatrix}1&1&1\\1&2&0\\0&1&2\end{pmatrix}$ , with $A^{-1}=\begin{pmatrix}4/3&-1/3&-2/3\\-2/3&2/3&1/3\\1/3&-1/3&1/3\end{pmatrix}$ . Multiplying $A^{-1}\begin{pmatrix}4\\11-3p\\1\end{pmatrix}$ gives $x=p+1$ , $y=5-2p$ and $z=p-2$ . Positivity requires $p>-1$ , $p<5/2$ and $p>2$ respectively. Their intersection is $2<p<5/2$ .
4	$\det A=1$ $\operatorname{adj}A=\begin{pmatrix}-1&0&1\\0&-1&1\\1&1&-1\end{pmatrix}$ $A^{-1}=\begin{pmatrix}-1&0&1\\0&-1&1\\1&1&-1\end{pmatrix}$ $\begin{pmatrix}x\\y\\z\end{pmatrix}=A^{-1}\begin{pmatrix}30\\28\\42\end{pmatrix}$ $(x,y,z)=(12,14,16)$ $\begin{pmatrix}p\\q\\r\end{pmatrix}=A^{-1}\begin{pmatrix}44\\40\\61\end{pmatrix}$ $(p,q,r)=(17,21,23)$	7
(7 marks)		7
Notes
The determinant of $A$ is $1$ and its adjugate is $\begin{pmatrix}-1&0&1\\0&-1&1\\1&1&-1\end{pmatrix}$ , so this is also $A^{-1}$ . Reusing the inverse gives $A^{-1}(30,28,42)^T=(12,14,16)^T$ and $A^{-1}(44,40,61)^T=(17,21,23)^T$ . Multiplication by $A$ reproduces the two stated right-hand sides.
5	$\begin{pmatrix}1&0&1\\0&2&1\\2&-1&1\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}5\\1\\8\end{pmatrix}$ $\det\begin{pmatrix}1&0&1\\0&2&1\\2&-1&1\end{pmatrix}=-1$ $A^{-1}=\begin{pmatrix}-3&1&2\\-2&1&1\\4&-1&-2\end{pmatrix}$ $(a,b,c)=(2,-1,3)$ $z=2x-y+3$ At $(3,4)$ , $z=5$ The plane meets $x=y$ , $z=0$ at $(-3,-3,0)$	7
(7 marks)		7
Notes
Substitution of the three points gives $A(a,b,c)^T=(5,1,8)^T$ , where $A=\begin{pmatrix}1&0&1\\0&2&1\\2&-1&1\end{pmatrix}$ . Its determinant is $-1$ and $A^{-1}=\begin{pmatrix}-3&1&2\\-2&1&1\\4&-1&-2\end{pmatrix}$ . Hence $(a,b,c)^T=A^{-1}(5,1,8)^T=(2,-1,3)^T$ , so the plane is $z=2x-y+3$ . At $(3,4)$ this gives $z=5$ . On $x=y=t$ and $z=0$ , the equation becomes $0=t+3$ , so the meeting point is $(-3,-3,0)$ .

CP-3.8 · Interpret geometrically the solution and failure of solution of three simultaneous linear equations.

Tier 1 · Easy

Mark scheme for CP-3.8 Tier 1 · Easy
Question	Scheme	Marks
1	The three planes meet at the single point $(1,2,-3)$	2
(2 marks)		2
Notes
Each equation fixes one coordinate and represents a plane parallel to a coordinate plane. All three conditions hold only at $(1,2,-3)$ , so the planes have one common point and the solution is unique.
2	The first two planes give $kz=0$ If $k=0$ , the planes form a sheaf with common line $(x,y,z)=(1,2,t)$ , $t\in\mathbb R$ If $k\ne0$ , the planes meet at the unique point $(1,2,0)$	3
(3 marks)		3
Notes
The first two planes force $x=1$ and $y=2$ . The third then becomes $kz=0$ . If $k=0$ , every real $z$ is allowed, so the planes form a sheaf with common line $(1,2,t)$ . If $k\ne0$ , then $z=0$ , so the planes meet at the unique point $(1,2,0)$ .

Tier 2 · Standard

Mark scheme for CP-3.8 Tier 2 · Standard
Question	Scheme	Marks
1	For $k=2$ , there are infinitely many solutions on a common line For $k\ne2$ , there is no common point	4
(4 marks)		4
Notes
The left side of the second equation is twice the left side of the first. If $k=2$ , the first two planes coincide, and this plane meets the third plane in a line, giving infinitely many solutions. If $k\ne2$ , the first two planes are distinct and parallel, so no point can satisfy both and the system has no solution.
2	For $k\ne4$ , $x=\dfrac{2k-7}{k-4}$ For $k\ne4$ , $y=-\dfrac1{k-4}$ For $k\ne4$ , $z=\dfrac{k-3}{k-4}$ $k=4$ gives no common point At $k=4$ , the planes form a prism because they intersect pairwise but have no common point	5
(5 marks)		5
Notes
The first two equations give $x=2-y$ and $z=1-y$ . Substitution in the third gives $5+(k-4)y=4$ . Hence, for $k\ne4$ , $y=-1/(k-4)$ , so $x=(2k-7)/(k-4)$ and $z=(k-3)/(k-4)$ ; the planes meet at this unique point. For $k=4$ , the third normal is the first normal plus three times the second, but the corresponding right-hand sides would require $4=2+3(1)=5$ . The system is inconsistent, and the three normals are pairwise non-parallel, so the planes form a prism.
3	The third equation is twice the first equation minus the second equation Eliminating $x$ from the first two equations gives $7y-5z=7$ $x=1+3t$ , $y=1-5t$ and $z=-7t$ , where $t\in\mathbb R$ $\mathbf r=\begin{pmatrix}1\\1\\0\end{pmatrix}+t\begin{pmatrix}3\\-5\\-7\end{pmatrix}$ , $t\in\mathbb R$ The three planes form a sheaf with infinitely many solutions on their common line	5
(5 marks)		5
Notes
Twice the first equation minus the second gives $-x+5y-4z=4$ , so the third equation is dependent. Eliminating $x$ from the first two equations gives $7y-5z=7$ . Taking $z=-7t$ gives $y=1-5t$ and then $x=1+3t$ . Hence the solution set is $\mathbf r=\begin{pmatrix}1\\1\\0\end{pmatrix}+t\begin{pmatrix}3\\-5\\-7\end{pmatrix}$ . All three planes contain this line, so they form a sheaf.

Tier 3 · Hard

Mark scheme for CP-3.8 Tier 3 · Hard
Question	Scheme	Marks
1	$k=3$ gives a common line and infinitely many solutions For $k\ne3$ there is no common point	5
(5 marks)		5
Notes
Adding the first two equations gives $3x+2z=3$ . Therefore, when $k=3$ , the third equation is the sum of the first two and contains their line of intersection, so all three planes share that line. When $k\ne3$ , any point on the first two planes must satisfy $3x+2z=3$ and so cannot lie on the third plane; there is no common point.
2	The coefficient determinant is $(k-1)(k-2)$ If $k\ne1,2$ , the unique point is $\left(\dfrac{2k-7}{k-2},\dfrac{k-5}{k-2},\dfrac3{k-2}\right)$ At $k=1$ , the third equation is the sum of the first two If $k=1$ , the planes form a sheaf with common line $(x,y,z)=(2-t,1-t,t)$ , $t\in\mathbb R$ At $k=2$ , the first two planes force $x-y=1$ but the third plane requires $x-y=4$ If $k=2$ , the planes form a prism, so there is no common point	6
(6 marks)		6
Notes
The coefficient determinant is $(k-1)(k-2)$ . For $k\ne1,2$ , the first two equations give $x=2-z$ and $y=1-z$ ; substitution yields $(k-1)(k-2)z=3(k-1)$ , so $z=3/(k-2)$ and the stated $x,y$ follow. At $k=1$ , the third equation is the sum of the first two, so the planes form a sheaf with common line $(2-t,1-t,t)$ . At $k=2$ , the third equation is $x-y=4$ , whereas the first two force $x-y=1$ . The normals are coplanar but the system is inconsistent, so the planes form a prism.
3	$\Pi_1\cap\Pi_2: \mathbf r=\begin{pmatrix}2\\0\\1\end{pmatrix}+t\begin{pmatrix}-4\\1\\2\end{pmatrix}$ $\Pi_1\cap\Pi_3: \mathbf r=\begin{pmatrix}4\\2\\0\end{pmatrix}+t\begin{pmatrix}-4\\1\\2\end{pmatrix}$ $\Pi_2\cap\Pi_3: \mathbf r=\begin{pmatrix}7\\0\\1\end{pmatrix}+t\begin{pmatrix}-4\\1\\2\end{pmatrix}$ All three pairwise intersection lines have direction $\begin{pmatrix}-4\\1\\2\end{pmatrix}$ The three pairwise intersection lines are distinct $\Pi_1$ and $\Pi_2$ imply $x+2y+z=3$ , which contradicts $\Pi_3$ The planes have no common point and form a prism	7
(7 marks)		7
Notes
Solving each pair gives $\Pi_1\cap\Pi_2: \mathbf r=\begin{pmatrix}2\\0\\1\end{pmatrix}+t\begin{pmatrix}-4\\1\\2\end{pmatrix}$ , $\Pi_1\cap\Pi_3: \mathbf r=\begin{pmatrix}4\\2\\0\end{pmatrix}+t\begin{pmatrix}-4\\1\\2\end{pmatrix}$ and $\Pi_2\cap\Pi_3: \mathbf r=\begin{pmatrix}7\\0\\1\end{pmatrix}+t\begin{pmatrix}-4\\1\\2\end{pmatrix}$ . Their common direction and different base points show that the three lines are parallel and distinct. Adding the first two plane equations gives $x+2y+z=3$ , whereas $\Pi_3$ requires this expression to equal $8$ . Hence there is no common point, and the planes form a prism.
4	The coefficient determinant is $(k-6)(k-7)$ If $k\ne6,7$ , the three planes meet in a point When $k=6$ , the left side of $\Pi_3$ is $x+y$ If $k=6$ and $c=8$ , $\Pi_1$ and $\Pi_3$ are coincident, and the common line is $\mathbf r=\begin{pmatrix}8\\0\\-3\end{pmatrix}+t\begin{pmatrix}-1\\1\\-1\end{pmatrix}$ If $k=6$ and $c\ne8$ , the planes are otherwise inconsistent When $k=7$ , the left side of $\Pi_3$ is the sum of the left sides of $\Pi_1$ and $\Pi_2$ If $k=7$ and $c=5$ , the planes form a sheaf with common line $\mathbf r=\begin{pmatrix}8\\0\\-3\end{pmatrix}+t\begin{pmatrix}-1\\1\\-1\end{pmatrix}$ If $k=7$ and $c\ne5$ , the planes form a prism	8
(8 marks)		8
Notes
The coefficient determinant is $(k-6)(k-7)$ , so for $k\ne6,7$ the planes meet in a point. At $k=6$ , $\Pi_3$ has left side $x+y$ : it is coincident with $\Pi_1$ when $c=8$ , giving the common line $(x,y,z)=(8-t,t,-3-t)$ , and otherwise the planes are inconsistent because $\Pi_1$ and $\Pi_3$ are distinct and parallel. At $k=7$ , the third left side is $(x+y)+(y+z)$ . The first two right sides sum to $5$ , so $c=5$ gives a sheaf on the same common line. If $c\ne5$ , each pair of planes meets but no point lies on all three, so they form a prism.