CAIE A-Level Further Math AS 1.4 Matrices Questions

$\left|\begin{array}{ll}1 2 5\end{array}\right|-k\left|\begin{array}{ll}2 3 5\end{array}\right|+3\left|\begin{array}{ll}2 3 2\end{array}\right|=0 \Rightarrow-1-k+3=0 \Rightarrow k=2$
Sets determinant of A equal to zero.

$\left(\begin{array}{ccc}-2 -1 1 1 3\end{array}\right)\left(\begin{array}{ccc}1 2 2 1 3 2 5\end{array}\right)\left(\begin{array}{cc}0 -2 -1 0 0\end{array}\right)=\left(\begin{array}{ccc}-2 -1 1 1 3\end{array}\right)\left(\begin{array}{cc}-2 -1 -1 -2 0\end{array}\right)$
Multiplying two matrices correctly, correct
dimensions.

$\left(\begin{array}{cc}3 -7 -9 3\end{array}\right)$
Completing matrix multiplication, AG.

5

$\left(\begin{array}{cc}3 -7 -9 3\end{array}\right)\binom{x}{y}=\binom{3 x-7 y}{-9 x+3 y}$
Transforms $\binom{x}{y}$ to $\binom{X}{Y}$.

-9 x+3 m x=m(3 x-7 m x)
Uses y=m x and Y=m X.

$-9+3 m=3 m-7 m^{2} \Rightarrow 7 m^{2}=9$$y=\frac{3}{\sqrt{7}} x$ and $y=-\frac{3}{\sqrt{7}} x$
5

$\mathbf{D}=\left(\begin{array}{cc}\alpha 0 \alpha\end{array}\right)$$\mathbf{E}=\left(\begin{array}{ll}\beta 0 1\end{array}\right)$$\mathbf{F}=\left(\begin{array}{ll}0 1 0\end{array}\right)$$\left(\begin{array}{cc}3 -7 -9 3\end{array}\right)=\left(\begin{array}{ll}\alpha 0 \alpha\end{array}\right)-9\left(\begin{array}{ll}0 \beta 1 0\end{array}\right)$
Setting up simultaneous equations using their D and
E.

$\mathbf{D}=\left(\begin{array}{ll}3 0 3\end{array}\right) \quad \mathbf{E}=\left(\begin{array}{ll}\frac{7}{9} 0 1\end{array}\right)$
Condone $\alpha=3, \beta=\frac{7}{9}$ if it is clear that they refer to
the correct matrices.

5

Rotation [anticlockwise]
about the origin through angle $2 \theta$.
Shear [in the x-direction] followed by a rotation [anticlockwise about the
origin through angle $2 \theta$ ].
3

$\left(\begin{array}{cc}1 k 0 1\end{array}\right)^{-1}=\left(\begin{array}{cc}1 -k 0 1\end{array}\right),\left(\begin{array}{cc}\cos 2 \theta -\sin 2 \theta \sin 2 \theta \cos 2 \theta\end{array}\right)^{-1}=\left(\begin{array}{cc}\cos 2 \theta \sin 2 \theta -\sin 2 \theta \cos 2 \theta\end{array}\right)$$\mathbf{M}^{-1}=\left(\begin{array}{cc}1 -k 0 1\end{array}\right)\left(\begin{array}{cc}\cos 2 \theta \sin 2 \theta -\sin 2 \theta \cos 2 \theta\end{array}\right)$
Correct order

2

$\mathbf{M}-\mathbf{I}=\left(\begin{array}{cc}\cos 2 \theta-1 k \cos 2 \theta-\sin 2 \theta \sin 2 \theta k \sin 2 \theta+\cos 2 \theta-1\end{array}\right)$$(\cos 2 \theta-1)(k \sin 2 \theta+\cos 2 \theta-1)-k \sin 2 \theta \cos 2 \theta+\sin ^{2} 2 \theta[=0]$
Evaluates $\operatorname{det}(\mathbf{M}-\mathbf{I})$$2-2 \cos 2 \theta-k \sin 2 \theta=0$
Brackets removed correctly and =0

$4 \sin ^{2} \theta=2 k \sin \theta \cos \theta$
Uses $1-\cos 2 \theta=2 \sin ^{2} \theta$ and $\sin 2 \theta=2 \sin \theta \cos \theta$ or all
necessary double angle formulae.

$\tan \theta=\frac{1}{2} k$
5

$\mathbf{M}=\left(\begin{array}{cc}-\frac{1}{2} -\frac{3}{2} \sqrt{3} \frac{1}{2} \sqrt{3} \frac{5}{2}\end{array}\right)$.
$\left(\begin{array}{cc}-\frac{1}{2} -\frac{3}{2} \sqrt{3} \frac{1}{2} \sqrt{3} \frac{5}{2}\end{array}\right)\binom{x}{y}=\binom{-\frac{1}{2} x-\frac{3}{2} \sqrt{3} y}{\frac{1}{2} \sqrt{3} x+\frac{5}{2} y}$

B1FT

Transforms $\binom{x}{y}$ to $\binom{X}{Y}$$$\begin{aligned} -\frac{1}{2} x-\frac{3}{2} \sqrt{3} y=x\left[\Rightarrow-\frac{3}{2} x-\frac{3}{2} \sqrt{3} y=0 \Rightarrow x+\sqrt{3} y=0\right] \text { and } \frac{1}{2} \sqrt{3} x+\frac{5}{2} y=y\left[\Rightarrow \frac{1}{2} \sqrt{3} x+\frac{3}{2} y=0 \Rightarrow \sqrt{3} x+3 y=0\right] \end{aligned}$$
Sets $\binom{X}{Y}=\binom{x}{y}$$\sqrt{3} x+3 y=0$
AG.

4