FP1 June 2010 Q2
2. \(\mathbf{M} = \begin{pmatrix} 2a & 3 \\ 6 & a \end{pmatrix}\), where \(a\) is a real constant.
(a) Given that \(a = 2\), find \(\mathbf{M}^{-1}\). (3)
(b) Find the values of \(a\) for which \(\mathbf{M}\) is singular. (2)
| Scheme | Marks |
|---|---|
| \(\mathbf{M} = \begin{pmatrix} 4 & 3 \\ 6 & 2 \end{pmatrix}\) Determinant: \((8 - 18) = -10\) | B1 |
| \(\mathbf{M}^{-1} = \dfrac{1}{-10}\begin{pmatrix} 2 & -3 \\ -6 & 4 \end{pmatrix} \quad \left[= \begin{pmatrix} -0.2 & 0.3 \\ 0.6 & -0.4 \end{pmatrix}\right]\) | M1 A1 |
| (3) |
Notes
(a) B1: must be −10
M1: for correct attempt at changing elements in major diagonal and changing signs in minor diagonal. Three or four of the numbers in the matrix should be correct – eg allow one slip
A1: for any form of the correct answer, with correct determinant then isw.
Special case: \(a\) not replaced is B0M1A0
| Scheme | Marks |
|---|---|
| Setting \(\Delta = 0\) and using \(2a^2 \pm 18 = 0\) to obtain \(a = \). | M1 |
| \(a = \pm 3\) | A1 cao |
| (2) | |
| 5 marks |
Notes
(b) Two correct answers, \(a = \pm 3\), with no working is M1A1
Just \(a = 3\) is M1A0, and also one of these answers rejected is A0.
Need 3 to be simplified (not \(\sqrt{9}\)).