| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Adjugate matrix calculation |
| Difficulty | Standard +0.3 This is a straightforward Further Maths linear algebra question requiring standard techniques: finding when det(A)=0 for singularity, computing the adjugate matrix (cofactor matrix transposed), and finding the inverse. All steps are routine calculations with no novel insight required, making it slightly easier than average even for Further Maths. |
| Spec | 4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(\det A = 4(\lambda \times \lambda) - (8 \times 8)\) | M1 | oe |
| \(\det A = 4\lambda^2 - 64\) | A1 | |
| Singular when \(\det \mathbf{A} = 0\) |
| Answer | Marks |
|---|---|
| \(4\lambda^2 - 64 = 0\) | M1 |
| \(\lambda^2 = 16\) | |
| \(\lambda = \pm 4\) | A1 |
| so there are two values where A is singular |
| Answer | Marks | Guidance |
|---|---|---|
| \(4\lambda^2 - 64 = 0\) | (M1) | |
| Discriminant = \(0^2 - (4 \times 4 \times -64) = 1024\) | (A1) | Must reference >0 |
| As \(1024 > 0\) there are two roots of the equation | ||
| so there are two values where A is singular |
| Answer | Marks | Guidance |
|---|---|---|
| Cofactor matrix: \(\begin{pmatrix}9 & -8 & -12\\-24 & 12 & 32\\-16 & 8 & 12\end{pmatrix}\) | ||
| Adjugate matrix = \(\begin{pmatrix}9 & -24 & -16\\-8 & 12 & 8\\-12 & 32 & 12\end{pmatrix}\) | B3 | All correct. B2 for 7 or 8 correct. B1 for 5 or 6 correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(\det A = (4 \times 3^2) - 64 = -28\) | B1 | FT their (a) |
| \(\therefore A^{-1} = \frac{1}{-28}\begin{pmatrix}9 & -24 & -16\\-8 & 12 & 8\\-12 & 32 & 12\end{pmatrix}\) | B1 | FT their adjugate. Mark final answer |
**Part a)**
$\det A = 4(\lambda \times \lambda) - (8 \times 8)$ | M1 | oe
$\det A = 4\lambda^2 - 64$ | A1 |
Singular when $\det \mathbf{A} = 0$ | |
METHOD 1:
$4\lambda^2 - 64 = 0$ | M1 |
$\lambda^2 = 16$ | |
$\lambda = \pm 4$ | A1 |
so there are two values where **A** is singular | |
METHOD 2:
$4\lambda^2 - 64 = 0$ | (M1) |
Discriminant = $0^2 - (4 \times 4 \times -64) = 1024$ | (A1) | Must reference >0
As $1024 > 0$ there are two roots of the equation | |
so there are two values where **A** is singular | |
**Part b) i)**
Cofactor matrix: $\begin{pmatrix}9 & -8 & -12\\-24 & 12 & 32\\-16 & 8 & 12\end{pmatrix}$ | |
Adjugate matrix = $\begin{pmatrix}9 & -24 & -16\\-8 & 12 & 8\\-12 & 32 & 12\end{pmatrix}$ | B3 | All correct. B2 for 7 or 8 correct. B1 for 5 or 6 correct
**Part b) ii)**
$\det A = (4 \times 3^2) - 64 = -28$ | B1 | FT their (a)
$\therefore A^{-1} = \frac{1}{-28}\begin{pmatrix}9 & -24 & -16\\-8 & 12 & 8\\-12 & 32 & 12\end{pmatrix}$ | B1 | FT their adjugate. Mark final answer
---The matrix $\mathbf{A}$ is defined by
$$\mathbf{A} = \begin{pmatrix} 4 & 8 & 0 \\ 0 & \lambda & -2 \\ 4 & 0 & \lambda \end{pmatrix}.$$
\begin{enumerate}[label=(\alph*)]
\item Show that there are two values of $\lambda$ for which $\mathbf{A}$ is singular. [4]
\item Given that $\lambda = 3$,
\begin{enumerate}[label=(\roman*)]
\item determine the adjugate matrix of $\mathbf{A}$,
\item determine the inverse matrix $\mathbf{A}^{-1}$. [5]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2022 Q10 [9]}}