| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Matrix inverse calculation |
| Difficulty | Standard +0.3 This is a straightforward Further Maths Core Pure question requiring calculation of a 3×3 determinant (showing it's never zero) and finding the inverse using cofactors. While it involves algebraic manipulation with parameter a, the techniques are standard and methodical with no novel problem-solving required. |
| Spec | 4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices4.03n Inverse 2x2 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(det(\mathbf{M}) = a(6) - 2(4) - 3(2a-12)\) | M1 | Finds the determinant of matrix M. Must be seen in part (a). Allow one slip if no method shown. |
| \(det(\mathbf{M}) = 28 \neq 0\) therefore, non-singular for all values of \(a\) | A1 | Correct value for determinant, states doesn't equal 0 (accept > 0) and draws conclusion that matrix is non-singular. Must have a conclusion. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Matrix of minors: \(\begin{pmatrix} 6 & 4 & 2a-12 \\ 4+3a & 2a+12 & a^2-8 \\ 9 & 6 & 3a-4 \end{pmatrix}\) | M1 | Finds the matrix of minors, at least 5 correct values. |
| Matrix of cofactors transposed: \(\begin{pmatrix} 6 & -4-3a & 9 \\ -4 & 2a+12 & -6 \\ 2a-12 & 8-a^2 & 3a-4 \end{pmatrix}\) | M1 | Finds matrix of cofactors and transposes (in either order). Allow minor slips if process is clearly correct. |
| \(\frac{1}{28}\begin{pmatrix} 6 & -4-3a & 9 \\ -4 & 2a+12 & -6 \\ 2a-12 & 8-a^2 & 3a-4 \end{pmatrix}\) | M1, A1 | M1: Completes process, divides by determinant. A1: Correct matrix. |
## Question 5:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $det(\mathbf{M}) = a(6) - 2(4) - 3(2a-12)$ | M1 | Finds the determinant of matrix **M**. Must be seen in part (a). Allow one slip if no method shown. |
| $det(\mathbf{M}) = 28 \neq 0$ therefore, non-singular for all values of $a$ | A1 | Correct value for determinant, states doesn't equal 0 (accept > 0) and draws conclusion that matrix is non-singular. Must have a conclusion. |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Matrix of minors: $\begin{pmatrix} 6 & 4 & 2a-12 \\ 4+3a & 2a+12 & a^2-8 \\ 9 & 6 & 3a-4 \end{pmatrix}$ | M1 | Finds the matrix of minors, at least 5 correct values. |
| Matrix of cofactors transposed: $\begin{pmatrix} 6 & -4-3a & 9 \\ -4 & 2a+12 & -6 \\ 2a-12 & 8-a^2 & 3a-4 \end{pmatrix}$ | M1 | Finds matrix of cofactors and transposes (in either order). Allow minor slips if process is clearly correct. |
| $\frac{1}{28}\begin{pmatrix} 6 & -4-3a & 9 \\ -4 & 2a+12 & -6 \\ 2a-12 & 8-a^2 & 3a-4 \end{pmatrix}$ | M1, A1 | M1: Completes process, divides by determinant. A1: Correct matrix. |
---5.
$$\mathbf { M } = \left( \begin{array} { r r r }
a & 2 & - 3 \\
2 & 3 & 0 \\
4 & a & 2
\end{array} \right) \quad \text { where } a \text { is a constant }$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathbf { M }$ is non-singular for all values of $a$.
\item Determine, in terms of $a , \mathbf { M } ^ { - 1 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP1 2022 Q5 [6]}}