FP1 January 2010 Q5
5. \(\mathbf{A} = \begin{pmatrix} a & -5 \\ 2 & a + 4 \end{pmatrix}\), where \(a\) is real.
(a) Find \(\det\mathbf{A}\) in terms of \(a\). (2)
(b) Show that the matrix \(\mathbf{A}\) is non-singular for all values of \(a\). (3)
Given that \(a = 0\),
(c) find \(\mathbf{A}^{-1}\). (3)
| Scheme | Marks |
|---|---|
| \(\det\mathbf{A} = a(a + 4) - (-5 \times 2) = a^2 + 4a + 10\) | M1 A1 |
| (2) |
Notes
(a) Correct use of \(ad - bc\) for M1
| Scheme | Marks |
|---|---|
| \(a^2 + 4a + 10 = (a + 2)^2 + 6\) | M1 A1ft |
| Positive for all values of \(a\), so \(\mathbf{A}\) is non-singular | A1cso |
| (3) |
Notes
(b) Attempt to complete square for M1
Alt 1
Attempt to establish turning point (e.g. calculus, graph) M1
Minimum value 6 for A1ft
Positive for all values of \(a\), so \(\mathbf{A}\) is non-singular for A1 cso
Alt 2
Attempt at \(b^2 - 4ac\) for M1. Can be part of quadratic formula
Their correct −24 for first A1
No real roots or equivalent, so \(\mathbf{A}\) is non-singular for final A1cso
| Scheme | Marks |
|---|---|
| \(\mathbf{A}^{-1} = \dfrac{1}{10}\begin{pmatrix} 4 & 5 \\ -2 & 0 \end{pmatrix}\) B1 for \(\dfrac{1}{10}\) | B1 M1 A1 |
| (3) | |
| [8] |
Notes
(c) Swap leading diagonal, and change sign of other diagonal, with numbers or \(a\) for M1
Correct matrix independent of ‘their \(\dfrac{1}{10}\) award’ final A1