AS June 2022 Paper 1 Q1
1
(a)
(i) Write the following simultaneous equations as a matrix equation.\[\begin{aligned} x + y + 2z &= 7 \\ 2x - 4y - 3z &= -5 \\ -5x + 3y + 5z &= 13 \end{aligned}\] [1]
(ii) Hence solve the equations. [2]
(b) Determine the set of values of the constant \(k\) for which the matrix equation\[\begin{pmatrix} k + 1 & 1 \\ 2 & k \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 23 \\ -17 \end{pmatrix}\]has a unique solution. [3]
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\begin{pmatrix} 1 & 1 & 2 \\ 2 & -4 & -3 \\ -5 & 3 & 5 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 7 \\ -5 \\ 13 \end{pmatrix}\) | B1 | 1.1 |
| [1] | ||
| (ii) \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 & 1 & 2 \\ 2 & -4 & -3 \\ -5 & 3 & 5 \end{pmatrix}^{-1}\begin{pmatrix} 7 \\ -5 \\ 13 \end{pmatrix}\) | M1 | 1.1a |
| \(x = \dfrac{1}{2},\ y = -\dfrac{3}{2},\ z = 4\) | A1 | 1.1 |
| [2] |
Notes
(a)(ii)
M1: (soi) allow M1 if order of matrices incorrect
A1: BC. Accept in vector form. Give full marks for a correct answer.
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{vmatrix} k + 1 & 1 \\ 2 & k \end{vmatrix} = k(k + 1) - 2\) | B1 | 1.1 |
| \(k^2 + k - 2 = 0\) when \(k = 1\) or \(-2\) | B1 | 2.1 |
| so unique solution provided \(k \ne 1\) or \(-2\) | B1 | 2.2a |
| [3] |
Notes
B1: (1st) or \(\begin{pmatrix} x \\ y \end{pmatrix} = \dfrac{1}{k(k + 1) - 2}\begin{pmatrix} \ldots \\ \ldots \end{pmatrix}\)
B1: (2nd) soi (e.g. from inequalities) or \(k = 1\) or \(-2\) found by inspection
B1: (3rd) oe, e.g. \(k \lt -2,\ -2 \lt k \lt 1,\ k \gt 1\).