AS October 2020 Paper 1 Q4
4 You are given the system of equations
\[\begin{aligned} a^2x - 2y &= 1 \\ x + b^2y &= 3 \end{aligned}\]where \(a\) and \(b\) are real numbers.
(a) Use a matrix method to find \(x\) and \(y\) in terms of \(a\) and \(b\). [4]
(b) Explain why the method used in part (a) works for all values of \(a\) and \(b\). [2]
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} a^2 & -2 \\ 1 & b^2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}\) | B1 | 3.1a |
| \(\begin{pmatrix} a^2 & -2 \\ 1 & b^2 \end{pmatrix}^{-1} = \dfrac{1}{a^2b^2 + 2}\begin{pmatrix} b^2 & 2 \\ -1 & a^2 \end{pmatrix}\) | M1 | 1.1 |
| \(\begin{pmatrix} x \\ y \end{pmatrix} = \dfrac{1}{a^2b^2 + 2}\begin{pmatrix} b^2 & 2 \\ -1 & a^2 \end{pmatrix}\begin{pmatrix} 1 \\ 3 \end{pmatrix}\) | M1 | 1.1 |
| \(x = \dfrac{b^2 + 6}{a^2b^2 + 2},\ y = \dfrac{3a^2 - 1}{a^2b^2 + 2}\) | A1 | 1.1 |
| [4] |
Notes
B1: Rewriting LHS in matrix form
M1: (1st) correct process for finding the inverse
M1: (2nd) Multiplication by their inverse
A1: Need to see \(x = \ldots,\ y = \ldots\)
| Scheme | Marks | AO |
|---|---|---|
| Since \(a^2b^2 = (ab)^2 \geqslant 0\) then \(a^2b^2 + 2 \gt 0\) for all values of \(a\) and \(b\) the determinant of the matrix cannot be 0 (so the matrix is never singular) | B1 | 2.4 |
| so the inverse always exists and the method always works. | B1 | 2.4 |
| [2] |
Notes
B1: (1st) Argument must be complete and correct. eg \(a^2b^2 + 2 \geqslant 0\) is B0.