A2 June 2024 Paper 2 Q14
14 The matrix \(\mathbf{M}\) is defined as
\[\mathbf{M} = \begin{bmatrix} 5 & 2 & 1 \\ 6 & 3 & 2k + 3 \\ 2 & 1 & 5 \end{bmatrix}\]where \(k\) is a constant.
(a) Given that \(\mathbf{M}\) is a non-singular matrix, find \(\mathbf{M}^{-1}\) in terms of \(k\) [5 marks]
(b) State any restrictions on the value of \(k\) [1 mark]
(c) Using your answer to part (a), show that the solution to the set of simultaneous equations below is independent of the value of \(k\)\[\begin{array}{rcrcrcl} 5x & + & 2y & + & z & = & 1 \\ 6x & + & 3y & + & (2k + 3)z & = & 4k + 3 \\ 2x & + & y & + & 5z & = & 9 \end{array}\] [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains \(12 - 2k\) | B1 | 1.1b |
| Obtains matrix of minors/cofactors with at least four correct elements. PI by transposed form. Condone overall sign error on each element. | M1 | 1.1a |
| Obtains matrix of minors/cofactors with at least seven correct elements. PI transposed form. Condone overall sign error on each element. | M1 | 1.1a |
| Obtains correct matrix of minors/cofactors. PI transposed form. Condone overall sign error on one element. | M1 | 1.1a |
| Obtains fully correct, simplified answer. | A1 | 1.1b |
| (5) |
Typical solution
\[\begin{aligned} |\mathbf{M}| &= 5(15 - 2k - 3) - 2(30 - 4k - 6) + (6 - 6) \\ &= 12 - 2k \end{aligned}\]Cofactors are
\[\begin{bmatrix} -2k + 12 & 4k - 24 & 0 \\ -9 & 23 & -1 \\ 4k + 3 & -10k - 9 & 3 \end{bmatrix}\]\[\mathbf{M}^{-1} = \frac{1}{12 - 2k}\begin{bmatrix} -2k + 12 & -9 & 4k + 3 \\ 4k - 24 & 23 & -10k - 9 \\ 0 & -1 & 3 \end{bmatrix}\]| Scheme | Marks | AO |
|---|---|---|
| Obtains \(k \neq 6\) Follow through their determinant. | B1F | 1.1b |
| (1) |
Typical solution
\[k \neq 6\]| Scheme | Marks | AO |
|---|---|---|
| Uses their \(\mathbf{M}^{-1}\) to form a product to find the solution set. Must include \(\begin{bmatrix} 1 \\ 4k + 3 \\ 9 \end{bmatrix}\) or Obtains \(\mathbf{M}^{-1}\begin{bmatrix} 1 \\ 4k + 3 \\ 9 \end{bmatrix}\) for a particular value of \(k\) | M1 | 3.1a |
| Obtains at least one correct component from their \(\mathbf{M}^{-1}\), can be unsimplified. or Obtains correct \(\mathbf{M}^{-1}\begin{bmatrix} 1 \\ 4k + 3 \\ 9 \end{bmatrix}\) for their \(k\) | A1F | 1.1b |
| Obtains at least two correct components from their \(\mathbf{M}^{-1}\) (can be unsimplified). or Correctly substitutes their \(\mathbf{M}^{-1}\begin{bmatrix} 1 \\ 4k + 3 \\ 9 \end{bmatrix}\) Into equation. | A1F | 1.1b |
| Uses correct reasoning to obtain the required result. | R1 | 2.1 |
| (4) | ||
| (10 marks) |
Typical solution
\[\begin{aligned} \begin{bmatrix} x \\ y \\ z \end{bmatrix} &= \frac{1}{12 - 2k}\begin{bmatrix} -2k + 12 & -9 & 4k + 3 \\ 4k - 24 & 23 & -10k - 9 \\ 0 & -1 & 3 \end{bmatrix}\begin{bmatrix} 1 \\ 4k + 3 \\ 9 \end{bmatrix} \\ &= \frac{1}{12 - 2k}\begin{bmatrix} -2k + 12 - 36k - 27 + 36k + 27 \\ 4k - 24 + 92k + 69 - 90k - 81 \\ -4k - 3 + 27 \end{bmatrix} \\ &= \frac{1}{12 - 2k}\begin{bmatrix} -2k + 12 \\ -36 + 6k \\ 24 - 4k \end{bmatrix} \\ &= \begin{bmatrix} 1 \\ -3 \\ 2 \end{bmatrix} \end{aligned}\]which is independent of \(k\)