A2 June 2020 Paper 2 Q8
8
(a) Factorise\[\begin{vmatrix} 2a + b + x & x + b & x^2 + b^2 \\ 0 & a & -a^2 \\ a + b & b & b^2 \end{vmatrix}\]
as fully as possible. [6 marks]
(b) The matrix \(\mathbf{M}\) is defined by\[\mathbf{M} = \begin{bmatrix} 13 + x & x + 3 & x^2 + 9 \\ 0 & 5 & -25 \\ 8 & 3 & 9 \end{bmatrix}\]
Under the transformation represented by \(\mathbf{M}\), a solid of volume \(0.625\,\mathrm{m}^3\) becomes a solid of volume \(300\,\mathrm{m}^3\)
Use your answer to part (a) to find the possible values of \(x\). [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Shows understanding that value of the determinant is unchanged when a column or row operation is used. | M1 | 1.1a |
| Demonstrates understanding that a factor can be extracted from the determinant. | M1 | 1.1a |
| Expands the determinant | M1 | 1.1a |
| Correctly extracts one factor | A1 | 1.1b |
| Correctly finds two factors | A1 | 1.1b |
| Obtains the determinant in fully factorised form \(-a(a + b)(a + x)(x - b)\) OE Correct answer seen: 6 marks | A1 | 1.1b |
Typical solution
\[a\begin{vmatrix} 2 & x + b & x^2 + b^2 \\ -1 & a & -a^2 \\ 1 & b & b^2 \end{vmatrix}\]\[a\begin{vmatrix} 0 & x - b & x^2 - b^2 \\ -1 & a & -a^2 \\ 1 & b & b^2 \end{vmatrix}\]\[a(x - b)\begin{vmatrix} 0 & 1 & x + b \\ -1 & a & -a^2 \\ 1 & b & b^2 \end{vmatrix}\]\[a(x - b)(a + b)\begin{vmatrix} 0 & 1 & x + b \\ 0 & 1 & b - a \\ 1 & b & b^2 \end{vmatrix}\]\[a(x - b)(a + b)\{(b - a) - (x + b)\}\]\[-a(x - b)(a + b)(x + a)\]| Scheme | Marks | AO |
|---|---|---|
| Forms an expression for the volume scale factor \(= \dfrac{300}{0.625}\) | M1 | 3.1a |
| Forms an equation for their volume scale factor \(= \det \mathbf{M}\) | M1 | 1.1a |
| Deduces all four correct values of \(x\): \(-3\), \(1\), \(-1 \pm 2\sqrt{7}\) | A1 | 2.2a |
| (9 marks) |