| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Find inverse then solve system |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard matrix techniques: computing the adjugate (cofactor matrix transposed), finding the inverse, and applying it to solve simultaneous equations. While it involves 3×3 matrices requiring careful arithmetic with nine cofactors, the method is entirely procedural with no problem-solving or insight required. It's slightly easier than average even for Further Maths due to its routine nature. |
| Spec | 4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{adj}(\mathbf{M}) = \begin{bmatrix} 11 & 1 & -7 \\ 1 & 1 & -1 \\ -7 & -1 & 5 \end{bmatrix}\) | M1, A1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\det(\mathbf{M}) = 2 \times (15-4) + 1 \times (6-5) + 3 \times (2-9)\) | M1 | AO1 |
| \(= 2\) | A1 | AO1 |
| \(\mathbf{M}^{-1} = \frac{1}{2}\begin{bmatrix} 11 & 1 & -7 \\ 1 & 1 & -1 \\ -7 & -1 & 5 \end{bmatrix}\) | B1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{2}\begin{bmatrix} 11 & 1 & -7 \\ 1 & 1 & -1 \\ -7 & -1 & 5 \end{bmatrix}\begin{bmatrix} 13 \\ 13 \\ 22 \end{bmatrix}\) | M1 | AO1 |
| \(= \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\) | A1 | AO1 |
## Part (a)(i)
$\text{adj}(\mathbf{M}) = \begin{bmatrix} 11 & 1 & -7 \\ 1 & 1 & -1 \\ -7 & -1 & 5 \end{bmatrix}$ | M1, A1 | AO1
**Guidance:** Award M1 if at least 5 correct
## Part (a)(ii)
$\det(\mathbf{M}) = 2 \times (15-4) + 1 \times (6-5) + 3 \times (2-9)$ | M1 | AO1
$= 2$ | A1 | AO1
$\mathbf{M}^{-1} = \frac{1}{2}\begin{bmatrix} 11 & 1 & -7 \\ 1 & 1 & -1 \\ -7 & -1 & 5 \end{bmatrix}$ | B1 | AO1
## Part (b)
$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{2}\begin{bmatrix} 11 & 1 & -7 \\ 1 & 1 & -1 \\ -7 & -1 & 5 \end{bmatrix}\begin{bmatrix} 13 \\ 13 \\ 22 \end{bmatrix}$ | M1 | AO1
$= \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ | A1 | AO1
**Total: [7]**
---The matrix $\mathbf{M}$ is given by
$$\mathbf{M} = \begin{pmatrix} 2 & 1 & 3 \\ 1 & 3 & 2 \\ 3 & 2 & 5 \end{pmatrix}.$$
\begin{enumerate}[label=(\alph*)]
\item Find
\begin{enumerate}[label=(\roman*)]
\item the adjugate matrix of $\mathbf{M}$,
\item hence determine the inverse matrix $\mathbf{M}^{-1}$. [5]
\end{enumerate}
\item Use your result to solve the simultaneous equations
\begin{align}
2x + y + 3z &= 13\\
x + 3y + 2z &= 13\\
3x + 2y + 5z &= 22
\end{align} [2]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 Q6 [7]}}