A2 June 2021 Paper 1 Q12
12 The matrix \(\mathbf{A} = \begin{bmatrix} 1 & 5 & 3 \\ 4 & -2 & p \\ 8 & 5 & -11 \end{bmatrix}\), where \(p\) is a constant.
(a) Given that \(\mathbf{A}\) is a non-singular matrix, find \(\mathbf{A}^{-1}\) in terms of \(p\).
State any restrictions on the value of \(p\). [6 marks]
(b) The equations below represent three planes.\[\begin{aligned} x + 5y + 3z &= 5 \\ 4x - 2y + pz &= 24 \\ 8x + 5y - 11z &= -30 \end{aligned}\]
(i) Find, in terms of \(p\), the coordinates of the point of intersection of the three planes. [4 marks]
(ii) In the case where \(p = 2\), show that the planes are mutually perpendicular. [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Calculates \(|\mathbf{A}|\) using the determinants of three 2x2 matrices | M1 | 1.1a |
| Obtains correct \(|\mathbf{A}|\) | A1 | 1.1b |
| Obtains matrix of minors/cofactors with four elements correct | M1 | 1.1a |
| Obtains fully correct matrix of cofactors | A1 | 1.1b |
| Transposes their matrix of cofactors (with at most one further error) & divides by their determinant | A1F | 1.1b |
| Obtains fully correct answer including \(p \neq -10\) | R1 | 2.1 |
| (6) |
Typical solution
\[|\mathbf{A}| = 1\begin{vmatrix} -2 & p \\ 5 & -11 \end{vmatrix} - 5\begin{vmatrix} 4 & p \\ 8 & -11 \end{vmatrix} + 3\begin{vmatrix} 4 & -2 \\ 8 & 5 \end{vmatrix}\]\[|\mathbf{A}| = (22 - 5p) - 5(-44 - 8p) + 3(20 + 16)\]\[|\mathbf{A}| = 350 + 35p = 35(10 + p)\]Cofactors:
\[\begin{bmatrix} 22 - 5p & 44 + 8p & 36 \\ 70 & -35 & 35 \\ 5p + 6 & 12 - p & -22 \end{bmatrix}\]\[\mathbf{A}^{-1} = \frac{1}{350 + 35p}\begin{bmatrix} 22 - 5p & 70 & 5p + 6 \\ 44 + 8p & -35 & 12 - p \\ 36 & 35 & -22 \end{bmatrix}\]\[p \neq -10\]| Scheme | Marks | AO |
|---|---|---|
| (i) Uses their \(\mathbf{A}^{-1}\) to form a product to find the coordinates of the point of intersection. Must include \(\begin{bmatrix}5 \\ 24 \\ -30\end{bmatrix}\) or Eliminates one variable to form two simultaneous equations in two variables | M1 | 3.1a |
| One component correct from their \(\mathbf{A}^{-1}\), can be unsimplified or Obtains one correct value for \(x\), \(y\) or \(z\), can be unsimplified | A1F | 1.1b |
| Two components correct from their \(\mathbf{A}^{-1}\), can be unsimplified or Obtains a second correct value for \(x\), \(y\) or \(z\), can be unsimplified | A1F | 1.1b |
| All three correct, like terms collected, but can be unsimplified Condone any form of the answer | R1 | 2.1 |
| (4) | ||
| (ii) Obtains the scalar product of normal vectors of two planes | M1 | 3.1a |
| Calculates that the scalar product is 0 and interprets this as meaning the two planes are perpendicular. | A1 | 3.2a |
| Obtains vector product of the same two normal vectors or Obtains the scalar products of the other two pairs of normal vector combinations | M1 | 3.1a |
| Clearly shows the vector product is a multiple of third plane’s normal vector and interprets this as meaning that the three planes are mutually perpendicular or States that all three scalar products are zero and interprets this as meaning that the three planes are mutually perpendicular | R1 | 3.2a |
| (4) | ||
| (14 marks) |
Typical solution
(i)
\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{35(10 + p)}\begin{bmatrix} 22 - 5p & 70 & 5p + 6 \\ 44 + 8p & -35 & 12 - p \\ 36 & 35 & -22 \end{bmatrix}\begin{bmatrix}5 \\ 24 \\ -30\end{bmatrix}\]\[= \frac{1}{35(10 + p)}\begin{bmatrix}110 - 25p + 1680 - 150p - 180 \\ 220 + 40p - 840 - 360 + 30p \\ 180 + 840 + 660\end{bmatrix}\]\[= \frac{1}{35(10 + p)}\begin{bmatrix}1610 - 175p \\ -980 + 70p \\ 1680\end{bmatrix}\]\[= \frac{1}{10 + p}\begin{bmatrix}46 - 5p \\ -28 + 2p \\ 48\end{bmatrix}\]\[x = \frac{46 - 5p}{10 + p}; \quad y = \frac{-28 + 2p}{10 + p}; \quad z = \frac{48}{10 + p}\]Point of intersection is:
\[\left(\frac{46 - 5p}{10 + p}, \frac{-28 + 2p}{10 + p}, \frac{48}{10 + p}\right)\](ii)
\[\begin{aligned} x + 5y + 3z &= 5 \\ 4x - 2y + 2z &= 24 \\ 8x + 5y - 11z &= -30 \end{aligned}\]\[\begin{bmatrix}1 \\ 5 \\ 3\end{bmatrix} \cdot \begin{bmatrix}4 \\ -2 \\ 2\end{bmatrix} = 4 - 10 + 6 = 0\]So the planes represented by the first two equations are perpendicular
\[\begin{bmatrix}1 \\ 5 \\ 3\end{bmatrix} \times \begin{bmatrix}4 \\ -2 \\ 2\end{bmatrix} = \begin{bmatrix}16 \\ 10 \\ -22\end{bmatrix} = 2\begin{bmatrix}8 \\ 5 \\ -11\end{bmatrix}\]As the vector perpendicular to the first two planes is a multiple of the normal vector of the third plane, the three planes are mutually perpendicular