A2 June 2022 Paper 1 Q7
7 The matrix \(\mathbf{M}\) is defined as
\[\mathbf{M} = \begin{bmatrix} 1 & 7 & -3 \\ 3 & 6 & k + 1 \\ 1 & 3 & 2 \end{bmatrix}\]where \(k\) is a constant.
(a)
(i) Given that \(\mathbf{M}\) is a non-singular matrix, find \(\mathbf{M}^{-1}\) in terms of \(k\) [5 marks]
(ii) State any restrictions on the value of \(k\) [1 mark]
(b) Using your answer to part (a)(i), solve\[\begin{aligned} x + 7y - 3z &= 6 \\ 3x + 6y + 6z &= 3 \\ x + 3y + 2z &= 1 \end{aligned}\] [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| (i) Expands \(|\mathbf{M}|\) to get a linear expression in \(k\), using any row or column. | M1 | 1.1a |
| Obtains matrix of minors/cofactors with at least four correct elements PI transposed form. Condone overall sign error on each element. | B1 | 1.1b |
| Obtains matrix of minors/cofactors with at least seven correct elements PI transposed form. Condone overall sign error on each element. | B1 | 1.1b |
| Obtains correct matrix of minors/cofactors PI transposed form. Condone overall sign error on each element. | B1 | 1.1b |
| Obtains fully correct, simplified answer | A1 | 1.1b |
| (5) | ||
| (ii) Obtains correct answer for their linear expression for \(|\mathbf{M}|\) | B1F | 1.1b |
| (1) |
Typical solution
(i)
\[\begin{aligned}|\mathbf{M}| &= (12 - 3k - 3) - 7(6 - k - 1) - 3(9 - 6) \\ &= 4k - 35\end{aligned}\]Cofactors are
\[\begin{bmatrix} 9 - 3k & k - 5 & 3 \\ -23 & 5 & 4 \\ 7k + 25 & -k - 10 & -15 \end{bmatrix}\]\[\mathbf{M}^{-1} = \frac{1}{4k - 35}\begin{bmatrix} 9 - 3k & -23 & 7k + 25 \\ k - 5 & 5 & -k - 10 \\ 3 & 4 & -15 \end{bmatrix}\](ii)
\[k \neq \frac{35}{4}\]| Scheme | Marks | AO |
|---|---|---|
| Obtains their correct \(\mathbf{M}^{-1}\), need not be simplified. | B1F | 3.1a |
| Forms their product \(\mathbf{M}^{-1}\begin{bmatrix}6 \\ 3 \\ 1\end{bmatrix}\) | M1 | 1.1a |
| Obtains \(x = 3,\ y = 0,\ z = -1\) ACF CSO | A1 | 1.1b |
| (3) | ||
| (9 marks) |
Typical solution
\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{-1}{15}\begin{bmatrix} -6 & -23 & 60 \\ 0 & 5 & -15 \\ 3 & 4 & -15 \end{bmatrix}\begin{bmatrix}6 \\ 3 \\ 1\end{bmatrix}\]\[= \frac{-1}{15}\begin{bmatrix}-45 \\ 0 \\ 15\end{bmatrix}\]\[= \begin{bmatrix}3 \\ 0 \\ -1\end{bmatrix}\]\(x = 3,\ y = 0,\ z = -1\)