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Edexcel A-level Further Maths revision notes

Matrices

Section CP-3
Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.
8 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section CP-3

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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In the exam: Formulae booklet provided · calculator allowed in every paper

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CP-3.1

Add, subtract and multiply conformable matrices. Multiply a matrix by a scalar.

Notes
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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 ABAB exists when the number of columns of AA equals the number of rows of BB.
  • Its (i,j)(i,j) entry is the scalar product of row ii of AA with column jj of BB, and the result has the number of rows of AA and columns of BB.
  • Matrix multiplication is generally not commutative, so ABAB and BABA may have different orders or only one may exist.
  • Dimensions should be checked before any arithmetic.
Worked example

Let A=(1213)A=\begin{pmatrix}1&2\\-1&3\end{pmatrix} and B=(4022)B=\begin{pmatrix}4&0\\2&-2\end{pmatrix}. Calculate ABAB.

  1. 1.Both matrices are 2×22\times2, so ABAB is 2×22\times2.
  2. 2.The first row gives 1(4)+2(2)=81(4)+2(2)=8 and 1(0)+2(2)=41(0)+2(-2)=-4.
  3. 3.The second row gives 1(4)+3(2)=2-1(4)+3(2)=2 and 1(0)+3(2)=6-1(0)+3(-2)=-6.

Answer: AB=(8426)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=BAAB=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

ORIGINAL

1.

Given A=(2130)A=\begin{pmatrix}2&-1\\3&0\end{pmatrix} and B=(1425)B=\begin{pmatrix}-1&4\\2&5\end{pmatrix}, calculate 2AB2A-B.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Let C=(121032)C=\begin{pmatrix}1&2&-1\\0&3&2\end{pmatrix} and D=(211430)D=\begin{pmatrix}2&1\\-1&4\\3&0\end{pmatrix}. Calculate CDCD and state its order.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Given A=(1230)A=\begin{pmatrix}1&-2\\3&0\end{pmatrix} and B=(p11q)B=\begin{pmatrix}p&1\\-1&q\end{pmatrix}, the matrix AB+2AAB+2A is (89183)\begin{pmatrix}8&-9\\18&3\end{pmatrix}. Determine pp and qq.

(4)

(Total for Question 1 is 4 marks)

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CP-3.2

Understand and use zero and identity matrices.

Notes
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Explanation

  • The zero matrix OO has every entry zero and is the additive identity: A+O=AA+O=A for matrices of the same order.
  • An identity matrix II is square, with ones on the leading diagonal and zeros elsewhere, and satisfies AI=IA=AAI=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 II; for example, the matrix counterpart of x25x+6x^2-5x+6 is A25A+6IA^2-5A+6I.
  • The symbols OO and II therefore represent matrices, not the scalar numbers 00 and 11.
Worked example

A square matrix AA satisfies A25A+6I=OA^2-5A+6I=O. Express A3A^3 in the form pA+qIpA+qI.

  1. 1.Rearrange the identity to obtain A2=5A6IA^2=5A-6I.
  2. 2.Multiply by AA: A3=5A26AA^3=5A^2-6A.
  3. 3.Substitute again: A3=5(5A6I)6AA^3=5(5A-6I)-6A.

Answer: A3=19A30IA^3=19A-30I.

Common mistakes

  • Don't fall into the trap of writing a scalar constant such as 66 where the matrix expression requires 6I6I.
  • Don't fall into the trap of using an identity matrix whose order does not match the square matrix AA.
  • Don't fall into the trap of treating OO as though it can be added to a matrix of any order.

Exam tip

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

Tier 1 · Easy

ORIGINAL

1.

For A=(2103)A=\begin{pmatrix}2&-1\\0&3\end{pmatrix}, write down A+OA+O and IAIA, where OO and II have the appropriate order.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Let A=(1102)A=\begin{pmatrix}1&1\\0&2\end{pmatrix}. Verify that (AI)(A2I)=O(A-I)(A-2I)=O and hence state a quadratic matrix equation satisfied by AA.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

A square matrix MM satisfies M2=3M+2IM^2=3M+2I. Without finding the entries of MM, simplify M34M2+MM^3-4M^2+M.

(4)

(Total for Question 1 is 4 marks)

CP-3.3

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

Notes
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Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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=±xy=\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 AA acts first and matrix BB second, the combined matrix is BABA, because xAxBAx\mathbf x\mapsto A\mathbf x\mapsto BA\mathbf x.
  • In three dimensions, use a 3×33\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 9090^\circ anticlockwise is followed by reflection in the xx-axis. Find the combined matrix.

  1. 1.The rotation matrix is R=(0110)R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.
  2. 2.The reflection matrix is F=(1001)F=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.
  3. 3.Since the rotation acts first, the combined matrix is FRFR.

Answer: FR=(0110)FR=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}.

Common mistakes

  • Don't fall into the trap of writing ABAB when transformation AA acts first and transformation BB 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 x\mathbf x from right to left before multiplying the matrices.

Tier 1 · Easy

ORIGINAL

1.

The matrix R=(0110)R=\begin{pmatrix}0&-1\\1&0\end{pmatrix} acts on the point P(3,2)P(3,-2). Find the image of PP and describe the transformation.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

A reflection in the line y=xy=x is followed by a stretch parallel to the xx-axis with scale factor 22. Find the matrix of the combined transformation and the image of (2,1)(2,-1).

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

A transformation in three dimensions maps (x,y,z)(x,y,z) to (x,z,y)(x,-z,y). Write down its matrix, find the image of (2,1,4)(2,-1,4), and explain geometrically what the transformation does.

(5)

(Total for Question 1 is 5 marks)

CP-3.4

Find invariant points and lines for a linear transformation.

Notes
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Your confidence:

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Explanation

  • An invariant point is fixed by the transformation. For matrix AA, its position vector satisfies Ax=xA\mathbf x=\mathbf x, so solve (AI)x=0(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+cy=mx+c, transform a general point (x,mx+c)(x,mx+c) to (X,Y)(X,Y) and require Y=mX+cY=mX+c for every xx; 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)(x,y) to (2x,y)(2x,y). Find its invariant points and invariant coordinate axes.

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

Answer: All invariant points lie on x=0x=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+cy=mx+c.

Exam tip

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

Tier 1 · Easy

ORIGINAL

1.

Find all invariant points of the transformation with matrix A=(1003)A=\begin{pmatrix}1&0\\0&3\end{pmatrix}.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Find the invariant lines through the origin for the transformation with matrix B=(2101)B=\begin{pmatrix}2&1\\0&1\end{pmatrix}.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

The transformation TT maps (x,y)(x,y) to (x+y,2y)(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)

CP-3.5

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

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For A=(abcd)A=\begin{pmatrix}a&b\\c&d\end{pmatrix}, $\det A=ad-bc$. A 3×33\times3 determinant may be evaluated by cofactor expansion, keeping the alternating signs.
  • Under a two-dimensional linear transformation, area is multiplied by detA|\det A|; in three dimensions, volume is multiplied by detA|\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=(3121)A=\begin{pmatrix}3&1\\2&-1\end{pmatrix}. Find the image area of a region of area 99 and describe orientation.

  1. 1.detA=3(1)1(2)=5\det A=3(-1)-1(2)=-5.
  2. 2.The area scale factor is 5=5|-5|=5.
  3. 3.The image area is 9×5=459\times5=45.
  4. 4.The negative determinant reverses orientation.

Answer: Image area 4545 square units; orientation is reversed.

Common mistakes

  • Don't fall into the trap of using ad+bcad+bc instead of adbcad-bc for a 2×22\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 3×33\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

ORIGINAL

1.

A planar transformation has matrix A=(1423)A=\begin{pmatrix}1&4\\2&3\end{pmatrix}. A region has area 1212 before transformation. Find its image area and state what happens to orientation.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Calculate the determinant of M=(120131201)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)

Tier 3 · Hard

ORIGINAL

1.

A transformation has matrix P=(k23k1)P=\begin{pmatrix}k&2\\3&k-1\end{pmatrix}. A triangle of area 77 is mapped to a triangle of area 4242. Determine all possible values of kk and state which values reverse orientation.

(5)

(Total for Question 1 is 5 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.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 AA1=A1A=IAA^{-1}=A^{-1}A=I.
  • For A=(abcd)A=\begin{pmatrix}a&b\\c&d\end{pmatrix}, A1=1adbc(dbca)A^{-1}=\dfrac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}. A 3×33\times3 inverse can be found by row-reducing (AI)(A\mid I) to (IA1)(I\mid A^{-1}), by an adjugate method or by calculator when permitted, but the process must be understood.
  • Useful properties include (AB)1=B1A1(AB)^{-1}=B^{-1}A^{-1} and (A1)1=A(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=(4131)A=\begin{pmatrix}4&1\\3&1\end{pmatrix} and verify one product.

  1. 1.detA=4(1)1(3)=1\det A=4(1)-1(3)=1, so AA is non-singular.
  2. 2.Interchange the diagonal entries and negate the off-diagonal entries.
  3. 3.A1=(1134)A^{-1}=\begin{pmatrix}1&-1\\-3&4\end{pmatrix}.
  4. 4.Multiplication gives AA1=IAA^{-1}=I.

Answer: A1=(1134)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=A1B1(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

ORIGINAL

1.

Show that A=(3121)A=\begin{pmatrix}3&1\\2&1\end{pmatrix} is non-singular and find A1A^{-1}.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Find the inverse of C=(110011101)C=\begin{pmatrix}1&1&0\\0&1&1\\1&0&1\end{pmatrix}.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

The matrix D=(2132)D=\begin{pmatrix}2&1\\3&2\end{pmatrix} satisfies D24D+I=OD^2-4D+I=O. Use this identity, rather than the 2×22\times2 inverse formula, to find D1D^{-1}.

(5)

(Total for Question 1 is 5 marks)

CP-3.7

Solve three linear simultaneous equations in three variables by use of the inverse matrix.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Write three linear equations in the form Ax=bA\mathbf x=\mathbf b, where each row of AA contains the coefficients of x,y,zx,y,z in the same order and b\mathbf b contains the corresponding right-hand sides.
  • If AA is non-singular, left-multiply by A1A^{-1} to obtain x=A1b\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 A1=(111011001)A^{-1}=\begin{pmatrix}1&-1&1\\0&1&-1\\0&0&1\end{pmatrix} to solve Ax=(641)A\mathbf x=\begin{pmatrix}6\\4\\1\end{pmatrix}.

  1. 1.Left-multiply by A1A^{-1}: x=A1b\mathbf x=A^{-1}\mathbf b.
  2. 2.x=64+1=3x=6-4+1=3.
  3. 3.y=41=3y=4-1=3 and z=1z=1.

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

Common mistakes

  • Don't fall into the trap of writing x=bA1\mathbf x=\mathbf bA^{-1} even though b\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 AA, x\mathbf x and b\mathbf b before using the inverse so coefficient placement earns method credit.

Tier 1 · Easy

ORIGINAL

1.

A system is written as A(xyz)=(426)A\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\2\\6\end{pmatrix}, where A1=12(111111111)A^{-1}=\frac12\begin{pmatrix}1&-1&1\\1&1&-1\\-1&1&1\end{pmatrix}. Find x,y,zx,y,z.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Use an inverse matrix to solve x+2y=5x+2y=5, y+z=4y+z=4 and z=3z=3.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

The equations x+y=px+y=p, y+z=5y+z=5 and 2z=82z=8 have a solution satisfying x+y=zx+y=z. Use an inverse matrix to determine pp and the solution.

(6)

(Total for Question 1 is 6 marks)

CP-3.8

Interpret geometrically the solution and failure of solution of three simultaneous linear equations.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

1.

Interpret geometrically the solution of the three equations x=1x=1, y=2y=2 and z=3z=-3.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Consider x+y+z=1x+y+z=1, 2x+2y+2z=k2x+2y+2z=k and xy+z=3x-y+z=3. Describe the solution set geometrically when k=2k=2 and when k2k\ne2.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

The planes x+y+z=2x+y+z=2, 2xy+z=12x-y+z=1 and 3x+2z=k3x+2z=k are given. Determine the value of kk for which the three planes contain a common line. State what happens for every other value of kk.

(5)

(Total for Question 1 is 5 marks)

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