A2 June 2019 Paper 1 Q10
10 You are given the matrix \(\mathbf{A}\) where \(\mathbf{A} = \begin{pmatrix} a & 2 & 0 \\ 0 & a & 2 \\ 4 & 5 & 1 \end{pmatrix}\).
(a) Find, in terms of \(a\), the determinant of \(\mathbf{A}\), simplifying your answer. [2]
(b) Hence find the values of \(a\) for which \(\mathbf{A}\) is singular. [2]
You are given the following equations which are to be solved simultaneously.
\[\begin{array}{rcrcrcr} ax & + & 2y & & & = & 6 \\ & & ay & + & 2z & = & 8 \\ 4x & + & 5y & + & z & = & 16 \end{array}\](c) For each of the values of \(a\) found in part (b) determine whether the equations have
- a unique solution, which should be found, or
- an infinite set of solutions or
- no solution.
| Scheme | Marks | AO |
|---|---|---|
| \(\det\mathbf{A} = a^2 - 10a + 16\) | M1 A1 | 1.1a 1.1 |
| [2] |
Notes
M1: Attempt to work out the determinant
| Scheme | Marks | AO |
|---|---|---|
| \(a^2 - 10a + 16 = 0 \Rightarrow (a - 2)(a - 8) = 0 \Rightarrow a = 2,\ 8\) | M1 A1 | 1.1a 1.1 |
| [2] |
Notes
M1: Solving their quadratic soi
| Scheme | Marks | AO |
|---|---|---|
| For both values there is no unique solution as \(\det\mathbf{A} = 0\) | B1 | 2.4 |
| For \(a = 2\), equations are: \(p_1 : 2x + 2y = 6\) \(p_2 : 2y + 2z = 8\) \(p_3 : 4x + 5y + z = 16\) | M1 | 2.1 |
| \(2p_1 + \frac{1}{2}p_2 = p_3\) So there is an infinite set of solutions. | A1 A1 | 1.1 2.2a |
| For \(a = 8\), equations are: \(p_1 : 8x + 2y = 6\) \(p_2 : 8y + 2z = 8\) \(p_3 : 4x + 5y + z = 16\) | M1 | 2.1 |
| \(\frac{1}{2}p_1 + \frac{1}{2}p_2 \neq p_3\) as it gives \(4x + 5y + z = 7\) | A1 | 1.1 |
| and \(16 \neq 7\) so no solution | A1 | 2.2a |
| [7] |
Notes
B1: Soi by correct answers. “correct answers” means solns are either infinite or non-existent.
M1: Substitute one of their values and solve
M1: Substitute the other one of their values and solve