CP-3 Matrices — printable coverage pack

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

Matrices can be added or subtracted only when they have the same order; combine corresponding entries.
The product $AB$ exists when the number of columns of $A$ equals the number of rows of $B$ ; each entry is a row-column scalar product.
For example, a $2\times3$ matrix multiplied by a $3\times2$ matrix gives a $2\times2$ matrix, while reversing the order gives a $3\times3$ matrix.
A common error is to multiply corresponding entries or assume $AB=BA$ ; matrix multiplication is generally not commutative.

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 marks]
Answer
- $\begin{pmatrix}5&-6\\4&-5\end{pmatrix}$
Method: $2A=\begin{pmatrix}4&-2\\6&0\end{pmatrix}$ . Subtract corresponding entries of $B$ : $2A-B=\begin{pmatrix}4-(-1)&-2-4\\6-2&0-5\end{pmatrix}=\begin{pmatrix}5&-6\\4&-5\end{pmatrix}$ .

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 marks]
Answer
- $CD=\begin{pmatrix}-3&9\\3&12\end{pmatrix}$ , of order $2\times2$
Method: $C$ is $2\times3$ and $D$ is $3\times2$ , so $CD$ exists and is $2\times2$ . Taking row-column products gives $CD=\begin{pmatrix}1(2)+2(-1)+(-1)3&1(1)+2(4)+(-1)0\\0(2)+3(-1)+2(3)&0(1)+3(4)+2(0)\end{pmatrix}=\begin{pmatrix}-3&9\\3&12\end{pmatrix}$ .

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 marks]
Answer
- $p=4$ , $q=3$
Method: $AB=\begin{pmatrix}p+2&1-2q\\3p&3\end{pmatrix}$ , 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 $3(4)+6=18$ .

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

The zero matrix $O$ has every entry equal to zero and is the additive identity: $A+O=A$ for matrices of the same order.
The identity matrix $I$ is square, has ones on its leading diagonal and zeros elsewhere, and satisfies $AI=IA=A$ .
Matrix polynomial identities are simplified just like algebraic ones, but each scalar constant is represented by a scalar multiple of $I$ .
A common error is to replace $I$ by the scalar $1$ or to use an identity matrix of the wrong order.

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 marks]
Answer
- $A+O=IA=\begin{pmatrix}2&-1\\0&3\end{pmatrix}$
Method: The $2\times2$ zero matrix leaves $A$ unchanged under addition, and the $2\times2$ identity leaves $A$ unchanged under multiplication. Hence both results equal $A$ .

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 marks]
Answer
- $A^2-3A+2I=O$
Method: $A-I=\begin{pmatrix}0&1\\0&1\end{pmatrix}$ 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 $A^2-3A+2I=O$ .

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 marks]
Answer
- $-2I$
Method: Multiply the given identity by $M$ : $M^3=3M^2+2M=3(3M+2I)+2M=11M+6I$ . Therefore $M^3-4M^2+M=(11M+6I)-4(3M+2I)+M=-2I$ .

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

For a linear transformation, the columns of its matrix are the images of the standard basis vectors.
If transformation $A$ is followed by transformation $B$ , the combined matrix is $BA$ , because $\mathbf{x}\mapsto A\mathbf{x}\mapsto BA\mathbf{x}$ .
In three dimensions, use a $3\times3$ matrix acting on $\begin{pmatrix}x\\y\\z\end{pmatrix}$ ; the specification confines 3-D transformations to reflection in one of the planes $x=0$ , $y=0$ , $z=0$ or rotation about one of the coordinate axes — for example, negating $z$ reflects points in the plane $z=0$ .
A common error is to reverse the order of multiplication for successive transformations.

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 marks]
Answer
- $P'=(2,3)$
- Rotation through $90^\circ$ anticlockwise about the origin
Method: $R\begin{pmatrix}3\\-2\end{pmatrix}=\begin{pmatrix}2\\3\end{pmatrix}$ . 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.

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 marks]
Answer
- $\begin{pmatrix}0&2\\1&0\end{pmatrix}$
- Image $(-2,2)$
Method: 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 $\begin{pmatrix}-2\\2\end{pmatrix}$ .

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.[4 marks]
Answer
- $\begin{pmatrix}1&0&0\\0&0&-1\\0&1&0\end{pmatrix}$
- Image $(2,-4,-1)$
- Rotation through $90^\circ$ about the $x$ -axis (taking the positive $y$ -axis to the positive $z$ -axis)
Method: 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.

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

An invariant point with position vector $\mathbf{x}$ satisfies $A\mathbf{x}=\mathbf{x}$ , so solve $(A-I)\mathbf{x}=\mathbf{0}$ .
An invariant line is mapped onto itself as a set; its individual points need not all remain fixed.
For a line $y=mx+c$ , transform a general point $(x,mx+c)$ and require the image $(X,Y)$ to satisfy $Y=mX+c$ for every $x$ .
A common error is to test only one point on a proposed invariant line or to confuse an invariant line with a line of invariant points.

Tier 1 · Easy

1. Find all invariant points of the transformation with matrix $A=\begin{pmatrix}1&0\\0&3\end{pmatrix}$ .[3 marks]
Answer
- All points $(x,0)$ on the $x$ -axis
Method: 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.

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 marks]
Answer
- $y=0$ and $y=-x$
Method: 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 $y=-x$ .

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 marks]
Answer
- $y=0$ and the family $y=x+c$ , where $c$ is any real constant
- $y=0$ consists entirely of invariant points
Method: 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.

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

For $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ , $\det A=ad-bc$ ; a $3\times3$ determinant can be found by cofactor expansion or row reduction.
In two dimensions, areas are multiplied by $|\det A|$ ; in three dimensions, volumes are multiplied by $|\det A|$ .
The sign is geometric: a negative determinant reverses orientation, while a positive determinant preserves it.
A common error is to use the signed determinant as a negative area or volume instead of taking its absolute value for size.

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 marks]
Answer
- Image area $=60$
- Orientation is reversed
Method: $\det A=1(3)-4(2)=-5$ . The area scale factor is $|-5|=5$ , so the image area is $12\times5=60$ . The negative determinant means orientation is reversed.

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 marks]
Answer
- $\det M=9$
- Volume scale factor $9$
- Orientation is preserved
Method: 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.

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 marks]
Answer
- $k=-3,0,1,4$
- Orientation is reversed for $k=0$ and $k=1$
Method: 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.

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.

A square matrix is singular when its determinant is zero; otherwise it is non-singular and has a unique inverse.
For a non-singular $2\times2$ matrix, use the determinant formula; for a $3\times3$ matrix, row-reduce $(A\mid I)$ to $(I\mid A^{-1})$ or use $A^{-1}=\frac1{\det A}\operatorname{adj}A$ .
Inverse products reverse order: $(AB)^{-1}=B^{-1}A^{-1}$ , and $AA^{-1}=A^{-1}A=I$ .
A common error is to divide every entry by the determinant without interchanging the diagonal entries and changing the signs of the off-diagonal entries.

Tier 1 · Easy

1. Show that $A=\begin{pmatrix}3&1\\2&1\end{pmatrix}$ is non-singular and find $A^{-1}$ .[3 marks]
Answer
- $A^{-1}=\begin{pmatrix}1&-1\\-2&3\end{pmatrix}$
Method: $\det A=3(1)-1(2)=1\ne0$ , so $A$ is non-singular. Hence $A^{-1}=\frac1{1}\begin{pmatrix}1&-1\\-2&3\end{pmatrix}$ .

Tier 2 · Standard

1. Find the inverse of $C=\begin{pmatrix}1&1&0\\0&1&1\\1&0&1\end{pmatrix}$ .[5 marks]
Answer
- $C^{-1}=\frac12\begin{pmatrix}1&-1&1\\1&1&-1\\-1&1&1\end{pmatrix}$
Method: $\det C=2$ . 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 $I$ .

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 $D^{-1}$ .[4 marks]
Answer
- $D^{-1}=\begin{pmatrix}2&-1\\-3&2\end{pmatrix}$
Method: 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 $D^{-1}=4I-D=\begin{pmatrix}2&-1\\-3&2\end{pmatrix}$ .

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

Write three simultaneous equations as $A\mathbf{x}=\mathbf{b}$ , keeping coefficients, variables and constants in a consistent order.
When $A$ is non-singular, left-multiply by $A^{-1}$ to obtain $\mathbf{x}=A^{-1}\mathbf{b}$ .
For example, the rows of $A$ contain the coefficients of $x,y,z$ , while $\mathbf{b}$ contains the three right-hand sides.
A common error is to write $\mathbf{x}=\mathbf{b}A^{-1}$ ; column vectors require multiplication by $A^{-1}$ on the left.

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 marks]
Answer
- $x=4$ , $y=0$ , $z=2$
Method: $\begin{pmatrix}x\\y\\z\end{pmatrix}=A^{-1}\begin{pmatrix}4\\2\\6\end{pmatrix}=\frac12\begin{pmatrix}4-2+6\\4+2-6\\-4+2+6\end{pmatrix}=\begin{pmatrix}4\\0\\2\end{pmatrix}$ .

Tier 2 · Standard

1. Use an inverse matrix to solve $x+2y=5$ , $y+z=4$ and $z=3$ .[5 marks]
Answer
- $x=3$ , $y=1$ , $z=3$
Method: 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 $\mathbf{x}=A^{-1}\mathbf{b}=\begin{pmatrix}5-8+6\\4-3\\3\end{pmatrix}=\begin{pmatrix}3\\1\\3\end{pmatrix}$ .

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 marks]
Answer
- $p=4$
- $(x,y,z)=(3,1,4)$
Method: 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 $(3,1,4)$ .

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

Each linear equation in $x,y,z$ represents a plane in three-dimensional space.
A unique solution is one point common to all three planes; infinitely many solutions arise when the planes share a common line (a sheaf) or are coincident.
No solution means there is no point common to all three planes — the planes form a prism or are otherwise inconsistent, often because one equation contradicts a linear combination of the others.
A common error is to say that a singular coefficient matrix always means no solution; it may instead give infinitely many solutions.

Tier 1 · Easy

1. Interpret geometrically the solution of the three equations $x=1$ , $y=2$ and $z=-3$ .[2 marks]
Answer
- The three planes meet at the single point $(1,2,-3)$
Method: 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.

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 marks]
Answer
- For $k=2$ , there are infinitely many solutions on a common line
- For $k\ne2$ , there is no common point
Method: 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.

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 marks]
Answer
- $k=3$ gives a common line and infinitely many solutions
- For $k\ne3$ there is no common point
Method: 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.