| Exam Board | WJEC |
|---|---|
| Module | Further Unit 1 (Further Unit 1) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Solving matrix equations for unknown matrix |
| Difficulty | Moderate -0.5 This is a straightforward matrix equation requiring finding A^{-1} and computing X = A^{-1}B. While it involves multiple steps (finding determinant, inverse, then multiplication), these are routine procedures for Further Maths students with no conceptual challenges or novel problem-solving required. Slightly easier than average due to its mechanical nature. |
| Spec | 4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(X = A^{-1}B\), \(\det A = 14\), \(A^{-1} = \frac{1}{14}\begin{pmatrix}0 & -7\\2 & 3\end{pmatrix}\), \(X = \frac{1}{14}\begin{pmatrix}0 & -7\\2 & 3\end{pmatrix}\begin{pmatrix}5 & 1\\0 & 4\end{pmatrix}\), \(X = \begin{pmatrix}0 & -2\\5/7 & 1\end{pmatrix}\) | M1, B1, M1A1, m1, A1 | M1 valid attempt to find \(A^{-1}\) dependent on 1st M1; cao |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(X = \begin{pmatrix}a & b\\c & d\end{pmatrix}\). Then \(\begin{pmatrix}3 & 7\\-2 & 0\end{pmatrix}\begin{pmatrix}a & b\\c & d\end{pmatrix} = \begin{pmatrix}5 & 1\\0 & 4\end{pmatrix}\) leads to \(3a + 7c = 5\) and \(-2a + 0c = 0\), \(3b + 7d = 1\) and \(-2b + 0d = 4\). Solving: \(a = 0, b = -2, c = 5/7, d = 1\) OR \(a = 0, c = 5/7, b = -2, d = 1\) | (M1), (m1), (A1), (m1), (A1), (A1) | Use of system; cao; cao |
**Method 1:**
$X = A^{-1}B$, $\det A = 14$, $A^{-1} = \frac{1}{14}\begin{pmatrix}0 & -7\\2 & 3\end{pmatrix}$, $X = \frac{1}{14}\begin{pmatrix}0 & -7\\2 & 3\end{pmatrix}\begin{pmatrix}5 & 1\\0 & 4\end{pmatrix}$, $X = \begin{pmatrix}0 & -2\\5/7 & 1\end{pmatrix}$ | M1, B1, M1A1, m1, A1 | M1 valid attempt to find $A^{-1}$ dependent on 1st M1; cao
**Method 2:**
Let $X = \begin{pmatrix}a & b\\c & d\end{pmatrix}$. Then $\begin{pmatrix}3 & 7\\-2 & 0\end{pmatrix}\begin{pmatrix}a & b\\c & d\end{pmatrix} = \begin{pmatrix}5 & 1\\0 & 4\end{pmatrix}$ leads to $3a + 7c = 5$ and $-2a + 0c = 0$, $3b + 7d = 1$ and $-2b + 0d = 4$. Solving: $a = 0, b = -2, c = 5/7, d = 1$ OR $a = 0, c = 5/7, b = -2, d = 1$ | (M1), (m1), (A1), (m1), (A1), (A1) | Use of system; cao; cao
---\begin{enumerate}
\item The matrices $\mathbf { A }$ and $\mathbf { B }$ are given by $\mathbf { A } = \left( \begin{array} { r r } 3 & 7 \\ - 2 & 0 \end{array} \right)$, $\mathbf { B } = \left( \begin{array} { l l } 5 & 1 \\ 0 & 4 \end{array} \right)$.
\end{enumerate}
The matrix $\mathbf { X }$ is such that $\mathbf { A X } = \mathbf { B }$. Showing all your working, find the matrix $\mathbf { X }$.\\
\hfill \mbox{\textit{WJEC Further Unit 1 2019 Q1 [6]}}