A2 June 2023 Paper 1 Q10
10 The matrix \(\mathbf{M}\) is defined as
\[\mathbf{M} = \begin{bmatrix} 2 & -1 & 1 \\ -1 & -1 & -2 \\ 1 & 2 & c \end{bmatrix}\]where \(c\) is a real number.
(a) The linear transformation T is represented by the matrix \(\mathbf{M}\)
Show that, for one particular value of \(c\), the image under T of every point lies in the plane
\[x + 5y + 3z = 0\]State the value of \(c\) for which this occurs. [3 marks]
(b) It is given that \(\mathbf{M}\) is a non-singular matrix.
(i) State any restrictions on the value of \(c\) [2 marks]
(ii) Find \(\mathbf{M}^{-1}\) in terms of \(c\) [4 marks]
(iii) Using your answer from part (b)(ii), solve\[\begin{aligned} 2x - y + z &= -3 \\ -x - y - 2z &= -6 \\ x + 2y + 4z &= 13 \end{aligned}\] [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Correctly obtains the three components of the image of a general point under \(\mathbf{M}\) in terms of \(c\) or using \(c = 3\) | B1 | 1.1b |
| Substitutes their components into \(x + 5y + 3z\) | M1 | 1.1a |
| Completes a reasoned argument to show that every image point lies in the plane \(x + 5y + 3z = 0\) when \(c = 3\) | R1 | 2.1 |
| (3) |
Typical solution
\[\begin{bmatrix} x^{\prime} \\ y^{\prime} \\ z^{\prime} \end{bmatrix} = \begin{bmatrix} 2 & -1 & 1 \\ -1 & -1 & -2 \\ 1 & 2 & c \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2x - y + z \\ -x - y - 2z \\ x + 2y + cz \end{bmatrix}\]\[\begin{aligned} x^{\prime} + 5y^{\prime} + 3z^{\prime} &= (2x - y + z) + 5(-x - y - 2z) + 3(x + 2y + cz) \\ &= (2 - 5 + 3)x + (-1 - 5 + 6)y + (1 - 10 + 3c)z \\ &= (-9 + 3c)z \end{aligned}\]So every image point lies in the plane \(x + 5y + 3z = 0\) when \(c = 3\)
| Scheme | Marks | AO |
|---|---|---|
| (i) Obtains an expression for \(|\mathbf{M}|\) Can be seen in (a) | M1 | 1.1a |
| (i) Deduces correct restriction on \(c\) | A1 | 2.2a |
| (2) | ||
| (ii) Obtains matrix of minors/cofactors with at least four correct elements PI by transposed form. Condone overall sign error on each element. | M1 | 1.1a |
| (ii) Obtains matrix of minors/cofactors with at least seven correct elements PI by transposed form. Condone overall sign error on each element. | M1 | 1.1a |
| (ii) Obtains correct matrix of minors/cofactors PI by transposed form. Condone overall sign error on each element. | A1 | 1.1b |
| (ii) Deduces fully correct, simplified answer | A1 | 2.2a |
| (4) | ||
| (iii) Obtains their correct \(\mathbf{M}^{-1}\) using \(c = 4\). Need not be simplified. | B1F | 3.1a |
| (iii) Forms their product \(\mathbf{M}^{-1}\begin{bmatrix} -3 \\ -6 \\ 13 \end{bmatrix}\) Condone \(\mathbf{M}^{-1}\) in terms of \(c\) | M1 | 1.1a |
| (iii) Completes a reasoned argument to obtain the correct solution. Do not accept \(\mathbf{r} = \begin{bmatrix} -1 \\ 3 \\ 2 \end{bmatrix}\) But do accept \(\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -1 \\ 3 \\ 2 \end{bmatrix}\) NMS or any method NOT using \(\mathbf{M}^{-1}\mathbf{v}\) scores 0 marks | R1 | 2.1 |
| (3) | ||
| (12 marks) |
Typical solution
(i)
\[\begin{aligned} |\mathbf{M}| &= 2(-c + 4) + 1(-c + 2) + 1(-2 + 1) \\ &= -3c + 9 \end{aligned}\]\[\therefore c \neq 3\](ii)
\[\text{minors } \begin{bmatrix} -c + 4 & -c + 2 & -1 \\ -c - 2 & 2c - 1 & 5 \\ 3 & -3 & -3 \end{bmatrix}\]\[\text{Cofactors } \begin{bmatrix} -c + 4 & c - 2 & -1 \\ c + 2 & 2c - 1 & -5 \\ 3 & 3 & -3 \end{bmatrix}\]\[\mathbf{M}^{-1} = \frac{1}{-3c + 9}\begin{bmatrix} -c + 4 & c + 2 & 3 \\ c - 2 & 2c - 1 & 3 \\ -1 & -5 & -3 \end{bmatrix}\](iii)
\[\begin{aligned} \begin{bmatrix} x \\ y \\ z \end{bmatrix} &= \frac{1}{-3}\begin{bmatrix} 0 & 6 & 3 \\ 2 & 7 & 3 \\ -1 & -5 & -3 \end{bmatrix}\begin{bmatrix} -3 \\ -6 \\ 13 \end{bmatrix} \\ &= \frac{1}{-3}\begin{bmatrix} 3 \\ -9 \\ -6 \end{bmatrix} \end{aligned}\]\[x = -1,\ y = 3,\ z = 2\]