| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Determinant calculation and singularity |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question on matrix determinants. Part (a) requires computing a 3×3 determinant using cofactor expansion—a standard technique. Part (b) asks to show the quadratic determinant is always non-zero by completing the square or using the discriminant, which is routine algebra. While it's Further Maths content, both parts follow well-practiced procedures with no novel insight required, making it slightly easier than average. |
| Spec | 4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices |
| Answer | Marks | Guidance |
|---|---|---|
| a) | \(\det A = \lambda(2\lambda - 16) + 1(-\lambda + 24) + 14(-2 + 6)\) | M1 |
| \(\det A = \lambda(2\lambda - 16) - 1(-\lambda + 24) + 14(-2 + 6)\) | A1 | |
| \(\det A = 2\lambda^2 - 15\lambda + 32\) | A1 | Fully simplified |
| (3) | ||
| b) | \(A\) is singular if \(\det A = 0\) | B1 |
| FT from (a) | ||
| METHOD 1: | ||
| \(2\lambda^2 - 15\lambda + 32 = 2\left(\lambda - \frac{15}{4}\right)^2 + \frac{31}{8}\) | M1, A1 | |
| \(2\left(\lambda - \frac{15}{4}\right)^2 + \frac{31}{8} > 0\) for all values of \(\lambda\) | B1 | Accept \(2\left(\lambda - \frac{15}{4}\right)^2 + \frac{31}{8} = 0\) for real values of \(\lambda\) |
| Therefore \(A\) is non-singular. | ||
| METHOD 2: | ||
| When \(2\lambda^2 - 15\lambda + 32 = 0\) | (M1), (A1) | |
| \(b^2 - 4ac = 225 - 256 = -31\) | (B1) | |
| As \(-31 < 0\) there are no real roots. | ||
| Therefore \(A\) is non-singular. | ||
| METHOD 3: | ||
| When \(2\lambda^2 - 15\lambda + 32 = 0\) | (M1) | |
| \(\lambda = \frac{15 \pm \sqrt{225 - 256}}{4} = \frac{15 \pm \sqrt{-31}}{4} = \frac{15}{4} \pm \frac{\sqrt{31}}{4}i\) | (A1) | |
| Therefore, there are no real roots. | (B1) | |
| Therefore \(A\) is non-singular. | ||
| (4) | ||
| [7] |
a) | $\det A = \lambda(2\lambda - 16) + 1(-\lambda + 24) + 14(-2 + 6)$ | M1 | Or equivalent rows/columns |
| $\det A = \lambda(2\lambda - 16) - 1(-\lambda + 24) + 14(-2 + 6)$ | A1 | |
| $\det A = 2\lambda^2 - 15\lambda + 32$ | A1 | Fully simplified |
| | (3) | |
b) | $A$ is singular if $\det A = 0$ | B1 | si |
| | | FT from (a) |
| **METHOD 1:** | | |
| $2\lambda^2 - 15\lambda + 32 = 2\left(\lambda - \frac{15}{4}\right)^2 + \frac{31}{8}$ | M1, A1 | |
| $2\left(\lambda - \frac{15}{4}\right)^2 + \frac{31}{8} > 0$ for all values of $\lambda$ | B1 | Accept $2\left(\lambda - \frac{15}{4}\right)^2 + \frac{31}{8} = 0$ for real values of $\lambda$ |
| Therefore $A$ is non-singular. | | |
| **METHOD 2:** | | |
| When $2\lambda^2 - 15\lambda + 32 = 0$ | (M1), (A1) | |
| $b^2 - 4ac = 225 - 256 = -31$ | (B1) | |
| As $-31 < 0$ there are no real roots. | | |
| Therefore $A$ is non-singular. | | |
| **METHOD 3:** | | |
| When $2\lambda^2 - 15\lambda + 32 = 0$ | (M1) | |
| $\lambda = \frac{15 \pm \sqrt{225 - 256}}{4} = \frac{15 \pm \sqrt{-31}}{4} = \frac{15}{4} \pm \frac{\sqrt{31}}{4}i$ | (A1) | |
| Therefore, there are no real roots. | (B1) | |
| Therefore $A$ is non-singular. | | |
| | (4) | |
| | [7] | |The matrix $\mathbf{A}$ is given by $\mathbf{A} = \begin{pmatrix} \lambda & 1 & 14 \\ -1 & 2 & 8 \\ -3 & 2 & \lambda \end{pmatrix}$, where $\lambda$ is a real constant.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the determinant of $\mathbf{A}$ in terms of $\lambda$. Give your answer in the form $a\lambda^2 + b\lambda + c$, where $a$, $b$, $c$ are integers whose values are to be determined. [3]
\item Show that $\mathbf{A}$ is non-singular for all values of $\lambda$. [4]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2023 Q2 [7]}}