Matrix equation solving (AB = C)

Questions where a matrix equation like AB = C or similar must be solved to find unknown matrices or parameters.

13 questions · Standard +0.3

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Edexcel F1 2017 June Q2
4 marks Moderate -0.5
2. Given that $$\mathbf { A } = \left( \begin{array} { r r r } 3 & 1 & - 2 \\ - 1 & 0 & 5 \end{array} \right) \text { and } \mathbf { B } = \left( \begin{array} { r r } 2 & 4 \\ - k & 2 k \\ 3 & 0 \end{array} \right) , \text { where } k \text { is a constant }$$
  1. find the matrix \(\mathbf { A B }\),
  2. find the exact value of \(k\) for which \(\operatorname { det } ( \mathbf { A B } ) = 0\)
Edexcel F3 2016 June Q4
9 marks Standard +0.3
4. $$\mathbf { M } = \left( \begin{array} { r r r } 1 & k & 0 \\ - 1 & 1 & 1 \\ 1 & k & 3 \end{array} \right) , \text { where } k \text { is a constant }$$
  1. Find \(\mathbf { M } ^ { - 1 }\) in terms of \(k\). Hence, given that \(k = 0\)
  2. find the matrix \(\mathbf { N }\) such that $$\mathbf { M N } = \left( \begin{array} { r r r } 3 & 5 & 6 \\ 4 & - 1 & 1 \\ 3 & 2 & - 3 \end{array} \right)$$
OCR FP1 2008 June Q10
11 marks Standard +0.8
10 The matrix \(\mathbf { A }\) is given by \(\mathbf { A } = \left( \begin{array} { r r r } a & 8 & 10 \\ 2 & 1 & 2 \\ 4 & 3 & 6 \end{array} \right)\). The matrix \(\mathbf { B }\) is such that \(\mathbf { A B } = \left( \begin{array} { l l l } a & 6 & 1 \\ 1 & 1 & 0 \\ 1 & 3 & 0 \end{array} \right)\).
  1. Show that \(\mathbf { A B }\) is non-singular.
  2. Find \(( \mathbf { A B } ) ^ { - 1 }\).
  3. Find \(\mathbf { B } ^ { - 1 }\).
OCR MEI FP1 2009 January Q10
12 marks Standard +0.3
10 You are given that \(\mathbf { A } = \left( \begin{array} { r r r } 3 & 4 & - 1 \\ 1 & - 1 & k \\ - 2 & 7 & - 3 \end{array} \right)\) and \(\mathbf { B } = \left( \begin{array} { r r c } 11 & - 5 & - 7 \\ 1 & 11 & 5 + k \\ - 5 & 29 & 7 \end{array} \right)\) and that \(\mathbf { A B }\) is of the form \(\mathbf { A B } = \left( \begin{array} { c c c } 42 & \alpha & 4 k - 8 \\ 10 - 5 k & - 16 + 29 k & - 12 + 6 k \\ 0 & 0 & \beta \end{array} \right)\).
  1. Show that \(\alpha = 0\) and \(\beta = 28 + 7 k\).
  2. Find \(\mathbf { A B }\) when \(k = 2\).
  3. For the case when \(k = 2\) write down the matrix \(\mathbf { A } ^ { - 1 }\).
  4. Use the result from part (iii) to solve the following simultaneous equations. $$\begin{aligned} 3 x + 4 y - z & = 1 \\ x - y + 2 z & = - 9 \\ - 2 x + 7 y - 3 z & = 26 \end{aligned}$$
OCR MEI FP1 2011 January Q9
12 marks Standard +0.3
\(\mathbf { 9 }\) You are given that \(\mathbf { A } = \left( \begin{array} { r r r } - 2 & 1 & - 5 \\ 3 & a & 1 \\ 1 & - 1 & 2 \end{array} \right)\) and \(\mathbf { B } = \left( \begin{array} { c c c } 2 a + 1 & 3 & 1 + 5 a \\ - 5 & 1 & - 13 \\ - 3 - a & - 1 & - 2 a - 3 \end{array} \right)\).
  1. Show that \(\mathbf { A B } = ( 8 + a ) \mathbf { I }\).
  2. State the value of \(a\) for which \(\mathbf { A } ^ { - 1 }\) does not exist. Write down \(\mathbf { A } ^ { - 1 }\) in terms of \(a\), when \(\mathbf { A } ^ { - 1 }\) exists.
  3. Use \(\mathbf { A } ^ { - 1 }\) to solve the following simultaneous equations. $$\begin{aligned} - 2 x + y - 5 z & = - 55 \\ 3 x + 4 y + z & = - 9 \\ x - y + 2 z & = 26 \end{aligned}$$
  4. What can you say about the solutions of the following simultaneous equations? $$\begin{aligned} - 2 x + y - 5 z & = p \\ 3 x - 8 y + z & = q \\ x - y + 2 z & = r \end{aligned}$$
OCR MEI FP1 2013 January Q9
12 marks Standard +0.3
9 You are given that \(\mathbf { A } = \left( \begin{array} { r r r } 8 & - 7 & - 12 \\ - 10 & 5 & 15 \\ - 9 & 6 & 6 \end{array} \right)\) and \(\mathbf { A } ^ { - 1 } = k \left( \begin{array} { r r r } 4 & 2 & 3 \\ 5 & 4 & 0 \\ 1 & - 1 & 2 \end{array} \right)\).
  1. Find the exact value of \(k\).
  2. Using your answer to part (i), solve the following simultaneous equations. $$\begin{aligned} 8 x - 7 y - 12 z & = 14 \\ - 10 x + 5 y + 15 z & = - 25 \\ - 9 x + 6 y + 6 z & = 3 \end{aligned}$$ You are also given that \(\mathbf { B } = \left( \begin{array} { r r r } - 7 & 5 & 15 \\ a & - 8 & - 21 \\ 2 & - 1 & - 3 \end{array} \right)\) and \(\mathbf { B } ^ { - 1 } = \frac { 1 } { 3 } \left( \begin{array} { r r r } 1 & 0 & 5 \\ - 4 & - 3 & 1 \\ 2 & 1 & b \end{array} \right)\).
  3. Find the values of \(a\) and \(b\).
  4. Write down an expression for \(( \mathbf { A B } ) ^ { - 1 }\) in terms of \(\mathbf { A } ^ { - 1 }\) and \(\mathbf { B } ^ { - 1 }\). Hence find \(( \mathbf { A B } ) ^ { - 1 }\).
OCR MEI FP1 2012 June Q9
12 marks Standard +0.3
9 You are given that \(\mathbf { A } = \left( \begin{array} { r r r } - 3 & - 4 & 1 \\ 2 & 1 & k \\ 7 & - 1 & - 1 \end{array} \right) , \mathbf { B } = \left( \begin{array} { r r c } - 4 & - 5 & 11 \\ - 19 & - 4 & - 7 \\ - 9 & - 31 & 2 - k \end{array} \right)\) and \(\mathbf { A B } = \left( \begin{array} { c c c } 79 & 0 & - 3 - k \\ - 9 k - 27 & - 31 k - 14 & q \\ p & 0 & 82 + k \end{array} \right)\) where \(p\) and \(q\) are to be determined.
  1. Show that \(p = 0\) and \(q = 15 + 2 k - k ^ { 2 }\). It is now given that \(k = - 3\).
  2. Find \(\mathbf { A B }\) and hence write down the inverse matrix \(\mathbf { A } ^ { - 1 }\).
  3. Use a matrix method to find the values of \(x , y\) and \(z\) that satisfy the equation \(\mathbf { A } \left( \begin{array} { l } x \\ y \\ z \end{array} \right) = \left( \begin{array} { r } 14 \\ - 23 \\ 9 \end{array} \right)\). \section*{THERE ARE NO QUESTIONS WRITTEN ON THIS PAGE}
OCR MEI FP1 2014 June Q9
12 marks Standard +0.3
9 You are given that \(\mathbf { A } = \left( \begin{array} { r r r } 1 & 3 & - 1 \\ - 1 & \alpha & - 1 \\ - 2 & - 1 & 3 \end{array} \right) , \mathbf { B } = \left( \begin{array} { c c c } 3 \alpha - 1 & - 8 & \alpha - 3 \\ 5 & 1 & 2 \\ 2 \alpha + 1 & - 5 & \alpha + 3 \end{array} \right)\) and \(\mathbf { A B } = \left( \begin{array} { c c c } \gamma & 0 & 0 \\ \beta & \gamma & 0 \\ 0 & 0 & \gamma \end{array} \right)\).
  1. Show that \(\beta = 0\).
  2. Find \(\gamma\) in terms of \(\alpha\).
  3. Write down \(\mathbf { A } ^ { - 1 }\) for the case when \(\alpha = 2\). State the value of \(\alpha\) for which \(\mathbf { A } ^ { - 1 }\) does not exist.
  4. Use your answer to part (iii) to solve the following simultaneous equations. $$\begin{aligned} x + 3 y - z & = 25 \\ - x + 2 y - z & = 11 \\ - 2 x - y + 3 z & = - 23 \end{aligned}$$
OCR MEI FP1 2016 June Q3
6 marks Standard +0.3
3 You are given that \(\mathbf { A } = \left( \begin{array} { c c c } \lambda & 6 & - 4 \\ 2 & 5 & - 1 \\ - 1 & 4 & 3 \end{array} \right) , \mathbf { B } = \left( \begin{array} { c c c } - 19 & 34 & - 14 \\ 5 & - 5 & 5 \\ - 13 & 18 & - 3 \end{array} \right)\) and \(\mathbf { A B } = \mu \mathbf { I }\), where \(\mathbf { I }\) is the \(3 \times 3\) identity
matrix.
  1. Find the values of \(\lambda\) and \(\mu\).
  2. Hence find \(\mathbf { B } ^ { - 1 }\).
OCR Further Pure Core 2 2021 November Q1
3 marks Standard +0.8
1 Two matrices, \(\mathbf { A }\) and \(\mathbf { B }\), are given by \(\mathbf { A } = \left( \begin{array} { r r r } 1 & - 2 & - 1 \\ 2 & - 3 & 1 \\ a & 1 & 1 \end{array} \right)\) and \(\mathbf { B } = \left( \begin{array} { r r r } - 6 & 3 & - 4 \\ - 1 & 6 & - 4 \\ 8 & - 8 & - 1 \end{array} \right)\) where \(a\) is a constant. Find the value of \(a\) for which \(\mathbf { A B } = \mathbf { B A }\).
Edexcel F3 2018 Specimen Q4
9 marks Standard +0.3
$$\mathbf{M} = \begin{pmatrix} 1 & k & 0 \\ -1 & 1 & 1 \\ 1 & k & 3 \end{pmatrix}, \text{ where } k \text{ is a constant}$$
  1. Find \(\mathbf{M}^{-1}\) in terms of \(k\). [5]
Hence, given that \(k = 0\)
  1. find the matrix \(\mathbf{N}\) such that $$\mathbf{MN} = \begin{pmatrix} 3 & 5 & 6 \\ 4 & -1 & 1 \\ 3 & 2 & -3 \end{pmatrix}$$ [4]
Edexcel CP1 2021 June Q4
9 marks Standard +0.3
  1. \(\mathbf{A}\) is a 2 by 2 matrix and \(\mathbf{B}\) is a 2 by 3 matrix. Giving a reason for your answer, explain whether it is possible to evaluate
    1. \(\mathbf{AB}\)
    2. \(\mathbf{A} + \mathbf{B}\)
    [2]
  2. Given that $$\begin{pmatrix} -5 & 3 & 1 \\ a & 0 & 0 \\ b & a & b \end{pmatrix}\begin{pmatrix} 0 & 5 & 0 \\ 2 & 12 & -1 \\ -1 & -11 & 3 \end{pmatrix} = \lambda\mathbf{I}$$ where \(a\), \(b\) and \(\lambda\) are constants,
    1. determine
    2. Hence deduce the inverse of the matrix \(\begin{pmatrix} -5 & 3 & 1 \\ a & 0 & 0 \\ b & a & b \end{pmatrix}\)
    [3]
  3. Given that $$\mathbf{M} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & \sin\theta & \cos\theta \\ 0 & \cos 2\theta & \sin 2\theta \end{pmatrix} \quad \text{where } 0 \leq \theta < \pi$$ determine the values of \(\theta\) for which the matrix \(\mathbf{M}\) is singular. [4]
SPS SPS FM Pure 2022 February Q4
9 marks Standard +0.3
  1. \(\mathbf{A}\) is a 2 by 2 matrix and \(\mathbf{B}\) is a 2 by 3 matrix. Giving a reason for your answer, explain whether it is possible to evaluate
    1. \(\mathbf{AB}\)
    2. \(\mathbf{A} + \mathbf{B}\)
    [2]
  2. Given that $$\begin{pmatrix} -5 & 3 & 1 \\ a & 0 & 0 \\ b & a & b \end{pmatrix} \begin{pmatrix} 0 & 5 & 0 \\ 2 & 12 & -1 \\ -1 & -11 & 3 \end{pmatrix} = \lambda \mathbf{I}$$ where \(a\), \(b\) and \(\lambda\) are constants,
    1. determine • the value of \(\lambda\) • the value of \(a\) • the value of \(b\)
    2. Hence deduce the inverse of the matrix \(\begin{pmatrix} -5 & 3 & 1 \\ a & 0 & 0 \\ b & a & b \end{pmatrix}\)
    [3]
  3. Given that $$\mathbf{M} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & \sin\theta & \cos\theta \\ 0 & \cos 2\theta & \sin 2\theta \end{pmatrix} \quad \text{where } 0 \leqslant \theta < \pi$$ determine the values of \(\theta\) for which the matrix \(\mathbf{M}\) is singular. [4]