Advanced Higher Maths
Matrices

(a)  For a $2\times 2$ matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right),$

$$\begin{array}{rcl}detA& =& ad-bc\\ A^{-1}& =& \frac{1}{detA}\left(\begin{array}{cc}\phantom{-}d& -b\\ -c& \phantom{-}a\end{array}\right).\end{array}$$

So for this example:

$$\begin{array}{rcl}detA& =& (-4)(5)-(-3)(6)\\ & =& -20+18\\ & =& -2\\ A^{-1}& =& \frac{1}{-2}\left(\begin{array}{cc}\phantom{-}5& \phantom{-}3\\ -6& -4\end{array}\right)\\ & =& \frac{1}{2}\left(\begin{array}{cc}-5& -3\\ \phantom{-}6& \phantom{-}4\end{array}\right)\\ & \mathsf{\text{or}}& \left(\begin{array}{cc}-\frac{5}{2}& -\frac{3}{2}\\ \phantom{-}3& \phantom{-}2\end{array}\right)\end{array}$$

(b)  The transpose of matrix $A$ is often denoted by $A^{\mathsf{\text{T}}}$ although the SQA uses the notation $A^{\prime }.$ It is the matrix obtained by interchanging the rows and columns of $A.$

So $A^{\prime }=\left(\begin{array}{cc}-4& 6\\ -3& 5\end{array}\right).$

Singular means that the inverse matrix $P^{-1}$ does not exist. This only occurs if the determinant of $P$ equals zero.

$$\begin{array}{c}detP=0\\ -9(4)-3n=0\\ -36-3n=0\\ -3n=36\\ n=-12\end{array}$$

$$\begin{array}{rcl}B& =& 2A^{\prime }\\ \left(\begin{array}{cc}8& 2\\ q& 1\end{array}\right)& =& 2\left(\begin{array}{cc}4& 1\\ 2& p\end{array}\right)\\ \left(\begin{array}{cc}8& 2\\ q& 1\end{array}\right)& =& \left(\begin{array}{cc}8& 2\\ 4& 2p\end{array}\right)\end{array}$$

Comparing the bottom left entries, $q=4.$

Comparing the bottom right entries, $2p=1$ so $p=\frac{1}{2}.$

$A$ is orthogonal if $A{A}^{\mathsf{\text{T}}}=A^{\mathsf{\text{T}}}A=I.$ So we could multiply this matrix by its transpose and show that it equals the $2\times 2$ identity matrix.

Another way of defining an orthogonal matrix is that it is a matrix whose transpose equals its inverse. So we could find $A^{-1}$ and show that it equals $A^{\mathsf{\text{T}}}.$

The first test is usually easier, especially as it is sufficient to test either $A{A}^{\mathsf{\text{T}}}$ or $A^{\mathsf{\text{T}}}A$ (not both). So:

$$\begin{array}{rcl}A{A}^{\mathsf{\text{T}}}& =& \left(\begin{array}{cc}\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ \frac{1}{2}& \phantom{-}\frac{\sqrt{3}}{2}\end{array}\right)\left(\begin{array}{cc}\frac{\sqrt{3}}{2}& \phantom{-}\frac{1}{2}\\ -\frac{1}{2}& \phantom{-}\frac{\sqrt{3}}{2}\end{array}\right)\\ & =& \left(\begin{array}{cc}\frac{3}{4}+\frac{1}{4}& \frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\\ \frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}& \frac{1}{4}+\frac{3}{4}\end{array}\right)\\ & =& \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I\end{array}$$

Hence $A$ is orthogonal.

Let $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right).$ So $A^{\prime }=\left(\begin{array}{cc}a& c\\ b& d\end{array}\right).$

$$\begin{array}{rcl}\mathsf{\text{So}}A+A^{\prime }& =& \left(\begin{array}{cc}a& b\\ c& d\end{array}\right)+\left(\begin{array}{cc}a& c\\ b& d\end{array}\right)\\ & =& \left(\begin{array}{cc}2a& b+c\\ b+c& 2d\end{array}\right)\\ \mathsf{\text{Now,}}(A+A^{\prime }{)}^{\prime }& =& \left(\begin{array}{cc}2a& b+c\\ b+c& 2d\end{array}\right)\end{array}$$

We have shown that $A+A^{\prime }$ is self-transpose, which is the definition of symmetric.

$$\begin{array}{rcl}A-A^{\prime }& =& \left(\begin{array}{cc}a& b\\ c& d\end{array}\right)+\left(\begin{array}{cc}a& c\\ b& d\end{array}\right)\\ & =& \left(\begin{array}{cc}0& b-c\\ c-b& 0\end{array}\right)\\ \mathsf{\text{Now,}}(A-A^{\prime }{)}^{\prime }& =& \left(\begin{array}{cc}0& c-b\\ b-c& 0\end{array}\right)\\ & =& \left(\begin{array}{cc}0& -(b-c)\\ -(c-b)& 0\end{array}\right)\\ & =& -(A-A^{\prime })\end{array}$$

We have shown that $A-A^{\prime }$ equals the negative of its transpose, which is the definition of skew-symmetric.

$$\begin{array}{rcl}{A}^2& =& \left(\begin{array}{cc}3& 0\\ \lambda & -2\end{array}\right)\left(\begin{array}{cc}3& 0\\ \lambda & -2\end{array}\right)\\ & =& \left(\begin{array}{cc}9& 0\\ \lambda & 4\end{array}\right)\\ & =& p\left(\begin{array}{cc}3& 0\\ \lambda & -2\end{array}\right)+q\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\end{array}$$

Equating the lower left entries suggests that $p=1$ but we need to ensure that this applies to all the other entries, so we equate the non-zero entries (top left and bottom right).

Subtracting the two equations, we obtain $5p=5$ so $p=1,$ and substituting into either equation gives us $q=6.$

We start by obtaining an expression for $detA$ in terms of $\mu .$

$$\begin{array}{rcl}|A|& =& 2\left|\begin{array}{cc}\mu & \phantom{-}4\\ 0& -2\end{array}\right|-3\left|\begin{array}{cc}-1& \phantom{-}4\\ \phantom{-}5& -2\end{array}\right|+1\left|\begin{array}{cc}-1& \phantom{-}\mu \\ \phantom{-}5& \phantom{-}0\end{array}\right|\\ & =& 2(-2\mu )-3(2-20)+1(-5\mu )\\ & =& -4\mu +54-5\mu \\ & =& -9\mu +54\end{array}$$

Now we can solve for $\mu .$

$$\begin{array}{c}-9\mu +54=36\\ -9\mu =36-54\\ -9\mu =-18\\ \mu =2\end{array}$$

We start by obtaining an expression for $detA$ in terms of $p.$

$$\begin{array}{rcl}|A|& =& p\left|\begin{array}{cc}\phantom{-}p& \phantom{-}1\\ -1& -1\end{array}\right|-2\left|\begin{array}{cc}3& \phantom{-}1\\ 0& -1\end{array}\right|+0\left|\begin{array}{cc}3& \phantom{-}p\\ 0& -1\end{array}\right|\\ & =& p(-p+1)-2(-3-0)+0\\ & =& -p^2+p+6\end{array}$$

Matrix $A$ is singular if $|A|=0,$ so we solve:

$$\begin{array}{c}-p^2+p+6=0\\ p^2-p-6=0\\ (p-3)(p+2)=0\\ p-3=0\mathsf{\text{or}}p+2=0\\ p=3\mathsf{\text{or}}p=-2\end{array}$$

We apply elementary row operations on the augmented matrix formed by $A$ and $I_3$ until the $3\times 3$ matrix on the left becomes $I_3,$ at which point the $3\times 3$ matrix on the right will be $A^{-1}.$

$$\begin{array}{rcl}\begin{array}{c}{R}_1\\ R_2\\ R_3\end{array}& & \left(\begin{array}{cccccc}1& 2& -1& 1& 0& 0\\ -2& 0& 1& 0& 1& 0\\ 1& -1& 0& 0& 0& 1\end{array}\right)\\ \begin{array}{c}\mathsf{\text{Frozen}}\\ R_2+2R_1\\ R_1-R_3\end{array}& & \left(\begin{array}{cccccc}1& 2& -1& 1& 0& 0\\ 0& 4& -1& 2& 1& 0\\ 0& 3& -1& 1& 0& -1\end{array}\right)\\ \begin{array}{c}{R}_1-\frac{1}{2}{R}_2\\ \mathsf{\text{Frozen}}\\ 3R_2-4R_3\end{array}& & \left(\begin{array}{cccccc}1& 0& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& 4& -1& 2& 1& 0\\ 0& 0& 1& 2& 3& 4\end{array}\right)\\ \begin{array}{c}{R}_1+\frac{1}{2}{R}_3\\ R_2+R_3\\ \mathsf{\text{Frozen}}\end{array}& & \left(\begin{array}{cccccc}1& 0& 0& 1& 1& 2\\ 0& 4& 0& 4& 4& 4\\ 0& 0& 1& 2& 3& 4\end{array}\right)\\ \begin{array}{c}\mathsf{\text{Frozen}}\\ \frac{1}{4}{R}_2\\ \mathsf{\text{Frozen}}\end{array}& & \left(\begin{array}{cccccc}1& 0& 0& 1& 1& 2\\ 0& 1& 0& 1& 1& 1\\ 0& 0& 1& 2& 3& 4\end{array}\right)\end{array}$$

The left of the augmented matrix is now the $3\times 3$ identity matrix, so $A^{-1}$ is on the right.

So $A^{-1}=\left(\begin{array}{ccc}1& 1& 2\\ 1& 1& 1\\ 2& 3& 4\end{array}\right).$

You may wish to learn these transformation matrices. If not, they can be deduced easily. Click here for an explanation of a simple algorithm to let you quickly derive transformation matrices.

(a)  Reflection of a point in the $x$-axis negates its $y$-coordinate. Its $x$-coordinate is invariant (unchanged).

So $M_1=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right).$

(b)  Reflection of a point in the line $y=x$ involves the $x$- and $y$-coordinates swapping places (as you've known all the way back to Nat 5 with inverse functions). Reflection in the line $y=-x$ swaps and also negates the coordinates.

So $M_2=\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right).$

(c)  We pre-multiply by the second transformation to be applied. So:

$$\begin{array}{rcl}{M}_3& =& M_1{M}_2\\ & =& \left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\\ & =& \left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\end{array}$$

(d)  The matrix $\left(\begin{array}{cc}cos\theta & -sin\theta \\ sin\theta & \phantom{-}cos\theta \end{array}\right)$ is associated with anti-clockwise rotation through an angle $\theta$ about the origin.

$cos\theta =0⟹\theta =\frac{\pi }{2},$ which is consistent with $sin\theta =1⟹\theta =\frac{\pi }{2}.$

So $M_3$ represents an anti-clockwise rotation of $\frac{\pi }{2}$ radians about the origin.

(a)  The matrix associated with an anticlockwise rotation of $\theta$ radians about the origin is given on the formula sheet. It is:

This 1-mark question simply requires us to substitute $\theta =\frac{\pi }{3},$ as follows:

The matrix above is enough to be awarded the mark, although in preparation for part (c) below, you might prefer to evaluate each entry, using exact values as follows:

(b)  Reflection in the $x$-axis negates $y$-coordinates, so the associated matrix is as follows:

$B=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right).$

(c)  We pre-multiply by the second transformation to be applied. So:

$$\begin{array}{rcl}P& =& BA\\ & =& \left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \phantom{-}\frac{1}{2}\end{array}\right)\\ & =& \left(\begin{array}{cc}\phantom{-}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}& -\frac{1}{2}\end{array}\right)\end{array}$$

(d)  The matrix $\left(\begin{array}{cc}cos\theta & -sin\theta \\ sin\theta & \phantom{-}cos\theta \end{array}\right)$ is associated with anti-clockwise rotation through an angle $\theta$ about the origin.

Note that the top-left and bottom-right elements in this matrix are equal. They are both $cos\theta .$

In matrix $P,$ the top-left element is $\frac{1}{2}$ and the bottom-right element is $-\frac{1}{2}.$

There is no angle $\theta$ for which $cos\theta =\frac{1}{2}$ and $cos\theta =-\frac{1}{2},$ so $P$ is not associated with rotation about the origin.