| Exam Board | WJEC |
|---|---|
| Module | Further Unit 1 (Further Unit 1) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Solving matrix equations for unknown matrix |
| Difficulty | Moderate -0.8 This is a straightforward two-part question requiring routine application of the 2×2 matrix inverse formula and matrix multiplication. Finding the inverse of a 2×2 matrix is a standard procedure (determinant and formula), and solving AX=B by computing X=A⁻¹B is direct application with no problem-solving insight needed. Easier than average for Further Maths. |
| Spec | 4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix |
2. The matrices $\mathbf { A }$ and $\mathbf { B }$ are such that $\mathbf { A } = \left[ \begin{array} { c c } 2 & - 1 \\ 4 & - 7 \end{array} \right]$ and $\mathbf { B } = \left[ \begin{array} { c c c } 2 & 0 & 9 \\ 4 & - 20 & 13 \end{array} \right]$.
\begin{enumerate}[label=(\alph*)]
\item Find the inverse of $\mathbf { A }$.
\item Hence, find the matrix $\mathbf { X }$, where $\mathbf { A X } = \mathbf { B }$.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 1 2023 Q2 [5]}}