AS June 2018 Paper 1 Q16
16 Two matrices \(\mathbf{A}\) and \(\mathbf{B}\) satisfy the equation
\[\mathbf{AB} = \boldsymbol{I} + 2\mathbf{A}\]where \(\boldsymbol{I}\) is the identity matrix and \(\mathbf{B} = \begin{bmatrix} 3 & -2 \\ -4 & 8 \end{bmatrix}\)
Find \(\mathbf{A}\). [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Uses factorisation or pre-multiplication to isolate \(\boldsymbol{A}\) | M1 | 3.1a |
| Deduces \(\boldsymbol{A}\) in terms of \(\boldsymbol{B}\) and \(\boldsymbol{I}\). Could be implied by sight of \(\begin{bmatrix} 1 & -2 \\ -4 & 6 \end{bmatrix}\) with attempt to invert. | A1 | 2.2a |
| Obtains correct matrix \(\boldsymbol{A}\). | A1 | 1.1b |
Typical solution
\[\boldsymbol{AB} - 2\boldsymbol{A} = \boldsymbol{I}\]\[\boldsymbol{A}(\boldsymbol{B} - 2\boldsymbol{I}) = \boldsymbol{I}\]\[\boldsymbol{A} = (\boldsymbol{B} - 2\boldsymbol{I})^{-1}\]\[\boldsymbol{A} = \begin{bmatrix} 1 & -2 \\ -4 & 6 \end{bmatrix}^{-1}\]\[\boldsymbol{A} = \frac{1}{-2}\begin{bmatrix} 6 & 2 \\ 4 & 1 \end{bmatrix}\]\[\boldsymbol{A} = \begin{bmatrix} -3 & -1 \\ -2 & -0.5 \end{bmatrix}\]ALT 16
| Scheme | Marks | AO |
|---|---|---|
| Sets up four equations with at least three correct. | M1 | 3.1a |
| Deduces at least two correct elements of \(\boldsymbol{A}\). Note: Two correct elements from just two correct equations can score M1A1. | A1 | 2.2a |
| Obtains correct matrix \(\boldsymbol{A}\) | A1 | 1.1b |
| (3 marks) |
Typical solution
Let \(\boldsymbol{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\)
\[\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 3 & -2 \\ -4 & 8 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + 2\begin{bmatrix} a & b \\ c & d \end{bmatrix}\]\[3a - 4b = 1 + 2a \quad \text{and} \quad 3c - 4d = 0 + 2c\]\[\text{and} \quad -2a + 8b = 0 + 2b\]\[\text{and} \quad -2c + 8d = 1 + 2d\]\[a = 4b + 1 \quad \text{and} \quad c = 4d\]\[6b = 2a \quad \text{and} \quad 6d = 2c + 1\]\[\therefore 3b = 4b + 1 \quad \text{and} \quad 6d = 2(4d) + 1\]\[-1 = b \quad \text{and} \quad -1 = 2d \Rightarrow d = -\frac{1}{2}\]\[\therefore a = 4 \times -1 + 1 \quad \text{and} \quad c = 4 \times -\frac{1}{2}\]\[a = -3 \quad \text{and} \quad c = -2\]\[\therefore \boldsymbol{A} = \begin{bmatrix} -3 & -1 \\ -2 & -0.5 \end{bmatrix}\]