AS June 2020 Paper 1 Q4
4 The matrices \(\mathbf{A}\) and \(\mathbf{B}\) are such that
\[\mathbf{A} = \begin{bmatrix} 2 & a & 3 \\ 0 & -2 & 1 \end{bmatrix} \quad \text{and} \quad \mathbf{B} = \begin{bmatrix} 1 & -3 \\ -2 & 4a \\ 0 & 5 \end{bmatrix}\](a) Find the product \(\mathbf{AB}\) in terms of \(a\). [2 marks]
(b) Find the determinant of \(\mathbf{AB}\) in terms of \(a\). [1 mark]
(c) Show that \(\mathbf{AB}\) is singular when \(a = -1\) [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains one correct element in terms of \(a\). Must be a \(2 \times 2\) matrix. | M1 | 1.1a |
| Obtains the correct product. Accept unsimplified. ISW | A1 | 1.1b |
Typical solution
\[\begin{bmatrix} 2 & a & 3 \\ 0 & -2 & 1 \end{bmatrix}\begin{bmatrix} 1 & -3 \\ -2 & 4a \\ 0 & 5 \end{bmatrix}\]\[= \begin{bmatrix} 2 - 2a + 0 & -6 + 4a^2 + 15 \\ 0 + 4 + 0 & 0 - 8a + 5 \end{bmatrix} = \begin{bmatrix} 2 - 2a & 4a^2 + 9 \\ 4 & 5 - 8a \end{bmatrix}\]| Scheme | Marks | AO |
|---|---|---|
| Obtains the correct determinant. Accept unsimplified. ISW Follow through their \(2 \times 2\) matrix with at least one element in terms of \(a\). | B1F | 1.1b |
Typical solution
\[\begin{aligned}&(2 - 2a)(5 - 8a) - 4(4a^2 + 9) \\ &= 10 - 16a - 10a + 16a^2 - 16a^2 - 36 \\ &= -26 - 26a\end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Selects a method to show that AB is singular. e.g. equates their expression for the determinant to zero or substitutes \(a = -1\) into their expression for the determinant. | M1 | 1.1a |
| Completes a fully correct reasoned argument to show that AB is singular, clearly referring to singular \(\Leftrightarrow\) determinant = 0. | R1 | 2.1 |
| (5 marks) |
Typical solution
AB is singular when \(\det \mathbf{AB} = 0\)
\[-26 - 26a = 0\]\[-26a = 26\]\[a = -1\]