WJEC Further Unit 1 2018 June — Question 1 6 marks

Exam BoardWJEC
ModuleFurther Unit 1 (Further Unit 1)
Year2018
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMatrices
TypeSolving matrix equations for unknown matrix
DifficultyModerate -0.8 This is a straightforward Further Maths question testing basic matrix operations: recognizing singular matrices via determinant (1 mark recall), computing a 2×2 inverse using the standard formula (routine procedure), and solving a matrix equation by pre-multiplying (direct application). All parts are standard textbook exercises requiring no problem-solving insight, though it's slightly above trivial due to the multi-step nature and being Further Maths content.
Spec4.03l Singular/non-singular matrices4.03n Inverse 2x2 matrix4.03r Solve simultaneous equations: using inverse matrix

The matrices \(\mathbf{A}\) and \(\mathbf{B}\) are such that \(\mathbf{A} = \begin{bmatrix} 4 & 2 \\ -1 & -3 \end{bmatrix}\) and \(\mathbf{B} = \begin{bmatrix} 4 & 2 \\ 2 & 1 \end{bmatrix}\).
  1. Explain why \(\mathbf{B}\) has no inverse. [1]
    1. Find the inverse of \(\mathbf{A}\). [3]
    2. Hence, find the matrix \(\mathbf{X}\), where \(\mathbf{AX} = \begin{bmatrix} -4 \\ 1 \end{bmatrix}\) [2]

Part a)
AnswerMarks
Answer: \(\det B = 0\)B1
Part b)
AnswerMarks Guidance
Answer: (i) \(\det A = -10\)B1
Answer: \(A^{-1} = -\frac{1}{10}\begin{pmatrix}-3 & -2\\1 & 4\end{pmatrix}\)M1 A1
Answer: (ii) \(X = -\frac{1}{10}\begin{pmatrix}-3 & -2\\1 & 4\end{pmatrix}\begin{pmatrix}-4\\1\end{pmatrix}\)M1
Answer: \(x = \begin{pmatrix}-1\\0\end{pmatrix}\)A1 cao
**Part a)**
Answer: $\det B = 0$ | B1 | 

**Part b)**
Answer: (i) $\det A = -10$ | B1 |
Answer: $A^{-1} = -\frac{1}{10}\begin{pmatrix}-3 & -2\\1 & 4\end{pmatrix}$ | M1 A1 |
Answer: (ii) $X = -\frac{1}{10}\begin{pmatrix}-3 & -2\\1 & 4\end{pmatrix}\begin{pmatrix}-4\\1\end{pmatrix}$ | M1 |
Answer: $x = \begin{pmatrix}-1\\0\end{pmatrix}$ | A1 | cao
The matrices $\mathbf{A}$ and $\mathbf{B}$ are such that $\mathbf{A} = \begin{bmatrix} 4 & 2 \\ -1 & -3 \end{bmatrix}$ and $\mathbf{B} = \begin{bmatrix} 4 & 2 \\ 2 & 1 \end{bmatrix}$.

\begin{enumerate}[label=(\alph*)]
\item Explain why $\mathbf{B}$ has no inverse. [1]

\item \begin{enumerate}[label=(\roman*)]
\item Find the inverse of $\mathbf{A}$. [3]
\item Hence, find the matrix $\mathbf{X}$, where $\mathbf{AX} = \begin{bmatrix} -4 \\ 1 \end{bmatrix}$ [2]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{WJEC Further Unit 1 2018 Q1 [6]}}