AS October 2021 Paper 1 Q8
8 The matrix \(\mathbf{A}\) is given by \(\mathbf{A} = \begin{pmatrix} t - 1 & t - 1 & t - 1 \\ 1 - t & 6 & t \\ 2 - 2t & 2 - 2t & 1 \end{pmatrix}\).
You are given the following system of equations in \(x\), \(y\) and \(z\), where \(b\) is a real number.
\[\begin{aligned} (b^2 + 1)x + (b^2 + 1)y + (b^2 + 1)z &= 5 \\ (-b^2 - 1)x + 6y + (b^2 + 2)z &= 10 \\ (-2b^2 - 2)x + (-2b^2 - 2)y + z &= 15 \end{aligned}\]- There is a unique solution for all values of \(b\).
- There is a unique solution for some, but not all, values of \(b\).
- There is no unique solution for any value of \(b\). [2]
| Scheme | Marks | AO |
|---|---|---|
| \((t - 1)(6 - t(2 - 2t))\) \(- (t - 1)((1 - t) - t(2 - 2t))\) \(+ (t - 1)((1 - t)(2 - 2t) - 6(2 - 2t))\) | M1 | 1.1 |
| \((t - 1)[(6 - t(2 - 2t)) - ((1 - t) - t(2 - 2t)) + ((1 - t)(2 - 2t) - 6(2 - 2t))]\) | M1 | 1.1 |
| \((t - 1)(6 - 2t + 2t^2 - 1 + t + 2t - 2t^2 + 2 - 4t + 2t^2 - 12 + 12t)\) \(= (t - 1)(2t^2 + 9t - 5)\) \(= (t - 1)(2t - 1)(t + 5)\) | A1 | 1.1 |
| [3] |
Notes
M1: (1st) Correct process for expanding determinant.
Fully expanded form: \(2t^3 + 7t^2 - 14t + 5\)
M1: (2nd) Bringing \((t - 1)\) or \((t + 5)\) or \((2t - 1)\) oe out as factor of the entire expression
Factors may appear BC from no working
| Scheme | Marks | AO |
|---|---|---|
| \(-5,\ \tfrac{1}{2},\ 1\) | B1 | 1.1 |
| [1] |
Notes
B1: FT their complete factorisation of determinant into 3 linear factors.
| Scheme | Marks | AO |
|---|---|---|
| \(t = b^2 + 2\) | M1 | 2.1 |
| and so \(t \geqslant 2\) so cannot be \(-5\), \(\tfrac{1}{2}\) or 1 therefore \(\mathbf{A}^{-1}\) will exist (for all values of \(b\)) and so there will be a unique solution to the system for all values of \(b\). | A1 | 2.4 |
| [2] |
Notes
M1: So that the system is \(\mathbf{Ar} = \mathbf{c}\)
A1: Complete reasoning must be seen for A1.
Could test \(t = 1, \tfrac{1}{2}, -5\) in \(b^2 = t - 2\), and show that these do not give real values of \(b\)