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AS & A Level · Practice questions · Matrix operations

Check the orders, then multiply.

Six original Cambridge-style questions on matrix operations: sums and scalar multiples, products of different orders, matrix equations with unknown entries, powers of a square matrix and the failure of commutativity.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Orders first, then row by column, then compare entries.

01
Recall
[2 marks]

Given \(A = \begin{pmatrix} 3 & -1 \\ 2 & 0 \\ 1 & 4 \end{pmatrix}\) and \(B = \begin{pmatrix} 1 & 2 \\ -1 & 1 \\ 0 & 3 \end{pmatrix}\), find \(A - 2B\) and state its order.

  • \(2B = \begin{pmatrix} 2 & 4 \\ -2 & 2 \\ 0 & 6 \end{pmatrix}\), and both matrices are \(3 \times 2\). ✓
  • \(A - 2B = \begin{pmatrix} 3 - 2 & -1 - 4 \\ 2 + 2 & 0 - 2 \\ 1 - 0 & 4 - 6 \end{pmatrix} = \begin{pmatrix} 1 & -5 \\ 4 & -2 \\ 1 & -2 \end{pmatrix}\), of order \(3 \times 2\). ✓
02
Calculation
[4 marks]

\(P = \begin{pmatrix} 1 & 0 & 2 \\ -1 & 3 & 1 \end{pmatrix}\) and \(Q = \begin{pmatrix} 2 & 1 \\ 0 & -1 \\ 1 & 4 \end{pmatrix}\).

(a) State the orders of \(PQ\) and \(QP\).

(b) Find \(PQ\) and \(QP\).

  • (a) \((2 \times 3)(3 \times 2)\) gives \(PQ\) of order \(2 \times 2\); \((3 \times 2)(2 \times 3)\) gives \(QP\) of order \(3 \times 3\). ✓
  • (b) Row 1 of \(P\): \(2 + 0 + 2 = 4\), \(1 + 0 + 8 = 9\). Row 2: \(-2 + 0 + 1 = -1\), \(-1 - 3 + 4 = 0\). So \(PQ = \begin{pmatrix} 4 & 9 \\ -1 & 0 \end{pmatrix}\). ✓
  • Row 1 of \(Q\), \((2, 1)\): \(2 - 1 = 1\), \(0 + 3 = 3\), \(4 + 1 = 5\). Row 2, \((0, -1)\): \(1\), \(-3\), \(-1\). Row 3, \((1, 4)\): \(1 - 4 = -3\), \(0 + 12 = 12\), \(2 + 4 = 6\). ✓
  • \[QP = \begin{pmatrix} 1 & 3 & 5 \\ 1 & -3 & -1 \\ -3 & 12 & 6 \end{pmatrix}\] ✓
03
Calculation
[4 marks]

\(A = \begin{pmatrix} a & 2 \\ 1 & b \end{pmatrix}\), where \(a\) and \(b\) are constants, and \(A^2 = \begin{pmatrix} 11 & 10 \\ 5 & 6 \end{pmatrix}\). Find \(a\) and \(b\).

  • \[A^2 = \begin{pmatrix} a & 2 \\ 1 & b \end{pmatrix}\begin{pmatrix} a & 2 \\ 1 & b \end{pmatrix}\] \[= \begin{pmatrix} a^2 + 2 & 2a + 2b \\ a + b & 2 + b^2 \end{pmatrix}\] ✓
  • Equate entries: \(a^2 + 2 = 11\), \(2a + 2b = 10\), \(a + b = 5\), \(2 + b^2 = 6\). ✓
  • So \(a = \pm 3\) and \(b = \pm 2\) with \(a + b = 5\): only \(a = 3\), \(b = 2\) works. ✓
  • Check: \(A = \begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix}\) gives \(A^2 = \begin{pmatrix} 9 + 2 & 6 + 4 \\ 3 + 2 & 2 + 4 \end{pmatrix} = \begin{pmatrix} 11 & 10 \\ 5 & 6 \end{pmatrix}\). ✓
04
Analysis
[6 marks]

\(A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}\).

(a) Find \(A^2\), and find the constants \(p\) and \(q\) such that \(A^2 = pA + qI\).

(b) Hence express \(A^3\) and \(A^4\) in the form \(rA + sI\).

(c) Write \(A^4\) out in full.

  • (a) \(A^2 = \begin{pmatrix} 1 + 4 & 2 + 2 \\ 2 + 2 & 4 + 1 \end{pmatrix} = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix}\). ✓
  • Compare \((1, 2)\) entries: \(4 = 2p\), so \(p = 2\); \((1, 1)\) entries: \(5 = p + q\), so \(q = 3\). Thus \(A^2 = 2A + 3I\). ✓
  • (b) \(A^3 = A(2A + 3I) = 2A^2 + 3A = 2(2A + 3I) + 3A = 7A + 6I\). ✓
  • \(A^4 = A(7A + 6I) = 7A^2 + 6A = 7(2A + 3I) + 6A = 20A + 21I\). ✓
  • (c) \(A^4 = \begin{pmatrix} 20 & 40 \\ 40 & 20 \end{pmatrix} + \begin{pmatrix} 21 & 0 \\ 0 & 21 \end{pmatrix} = \begin{pmatrix} 41 & 40 \\ 40 & 41 \end{pmatrix}\). (Check: \((A^2)^2 = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix}^2 = \begin{pmatrix} 25 + 16 & 20 + 20 \\ 20 + 20 & 16 + 25 \end{pmatrix}\), the same.) ✓
05
Analysis
[4 marks]

\(A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\) and \(B = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}\).

(a) Find \((A + B)^2\).

(b) Show that \((A + B)^2 \neq A^2 + 2AB + B^2\), and explain why.

  • (a) \(A + B = \begin{pmatrix} 2 & 2 \\ 3 & 2 \end{pmatrix}\), so \((A + B)^2 = \begin{pmatrix} 4 + 6 & 4 + 4 \\ 6 + 6 & 6 + 4 \end{pmatrix} = \begin{pmatrix} 10 & 8 \\ 12 & 10 \end{pmatrix}\). ✓
  • (b) \(A^2 = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix}\), \(B^2 = \begin{pmatrix} 1 & 0 \\ 6 & 1 \end{pmatrix}\) and \(AB = \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix}\). ✓
  • \[A^2 + 2AB + B^2 = \begin{pmatrix} 1 + 14 + 1 & 4 + 4 + 0 \\ 0 + 6 + 6 & 1 + 2 + 1 \end{pmatrix}\] \[= \begin{pmatrix} 16 & 8 \\ 12 & 4 \end{pmatrix} \neq \begin{pmatrix} 10 & 8 \\ 12 & 10 \end{pmatrix}\] ✓
  • \((A + B)^2 = (A + B)(A + B) = A^2 + AB + BA + B^2\). This equals \(A^2 + 2AB + B^2\) only if \(BA = AB\), and here \(BA = \begin{pmatrix} 1 & 2 \\ 3 & 7 \end{pmatrix} \neq AB\). ✓
06
Analysis
[6 marks]

\(A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}\).

(a) Find \(A^2\) and \(A^3\).

(b) Find the constants \(p\), \(q\) and \(r\) such that \(A^3 = pA^2 + qA + rI\).

  • (a) \[A^2 = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}\] \[= \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix}\] ✓
  • \[A^3 = A^2A = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}\] \[= \begin{pmatrix} 1 & 3 & 3 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix}\] ✓
  • (b) \(pA^2 + qA + rI\) has \((1, 3)\) entry \(p\), \((1, 2)\) entry \(2p + q\) and diagonal entries \(p + q + r\). ✓
  • Equate with \(A^3\): \(p = 3\); \(2p + q = 3\), so \(q = -3\); \(p + q + r = 1\), so \(r = 1\). ✓
  • Check the \((2, 3)\) entry: \(2p + q = 3\), as in \(A^3\); every entry below the diagonal is \(0\) on both sides. So \(A^3 = 3A^2 - 3A + I\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.