(a)  For a \(2\!\times\!2\) matrix \(A=\begin{pmatrix} a &\!\!b\,\cr c &\!d\, \end{pmatrix}\small,\)

$$ \begin{eqnarray} det\,A&=&ad\!-\!bc\\[6pt] A^{-1}&=&\small\frac{1}{det\,A}\normalsize\begin{pmatrix} \phantom{-}d &\!\!\!\!-\!b\,\cr -c &\!\!\!\!\phantom{-}a\, \end{pmatrix}\small. \end{eqnarray} $$

So for this example:

$$ \begin{eqnarray} det\,A&=&(-4)(5)\!-\!(-3)(6) \\[6pt] &=& -20+18\\[6pt] &=& -2\\[8pt] A^{-1}&=&\small\frac{1}{-2}\normalsize\begin{pmatrix} \phantom{-}5 &\!\!\!\!\phantom{-}3\,\cr -6 &\!\!\!\!-\!4\, \end{pmatrix} \\[6pt] &=&\small\frac{1}{2}\normalsize\begin{pmatrix} -5 &\!\!\!\!-\!3\,\cr \phantom{-}6 &\!\!\!\!\phantom{-}4\, \end{pmatrix} \\[6pt] &\small\textsf{or}&\begin{pmatrix} -\frac{5}{2} &\!\!\!\!-\!\frac{3}{2}\,\cr \phantom{-}3 &\!\!\!\!\phantom{-}2\, \end{pmatrix} \\[6pt] \end{eqnarray} $$

(b)  The transpose of matrix \(A\) is often denoted by \(A^{\textsf{T}}\) although the SQA uses the notation \(A'\small.\) It is the matrix obtained by interchanging the rows and columns of \(A\small.\)

So \(A'=\begin{pmatrix} -4 & 6\,\cr -3 & 5\, \end{pmatrix}\small.\)

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

$$ \begin{gather} det\,P = 0\\[6pt] -9(4)-3n = 0\\[6pt] -36-3n = 0\\[6pt] -3n = 36\\[6pt] n = -\!12 \end{gather} $$

$$ \begin{eqnarray} B&=&2A' \\[6pt] \begin{pmatrix} \,8 &\!2\,\cr \,q &\!1\, \end{pmatrix}&=& 2\begin{pmatrix} \,4 &\!1\,\cr \,2 &\!p\, \end{pmatrix}\\[6pt] \begin{pmatrix} \,8 &\!2\,\cr \,q &\!1\, \end{pmatrix}&=& \begin{pmatrix} \,8 &\!2\,\cr \,4 &\!2p\, \end{pmatrix}\\[6pt] \end{eqnarray} $$

Comparing the bottom left entries, \(q\!=\!4\small.\)

Comparing the bottom right entries, \(2p\!=\!1\) so \(p\!=\!\frac{1}{2}\small.\)

\(A\) is orthogonal if \(A\tiny\,\normalsize A^{\textsf{T}}=A^{\textsf{T}}A=I\small.\) 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 \(A^{\textsf{T}}\small.\)

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

$$ \begin{eqnarray} A\tiny\,\normalsize A^{\textsf{T}}&=&\begin{pmatrix} \frac{\sqrt{3}}{2} &\!\!\!\!-\!\frac{1}{2}\cr \frac{1}{2} &\!\!\!\!\!\phantom{-}\!\frac{\sqrt{3}}{2} \end{pmatrix}\begin{pmatrix} \frac{\sqrt{3}}{2} &\!\!\!\!\phantom{-}\!\frac{1}{2}\cr -\frac{1}{2} &\!\!\!\!\!\phantom{-}\!\frac{\sqrt{3}}{2} \end{pmatrix} \\[6pt] &=&\begin{pmatrix} \frac{3}{4}\!+\!\frac{1}{4}&\frac{\sqrt{3}}{4}\!-\!\frac{\sqrt{3}}{4}\cr \frac{\sqrt{3}}{4}\!-\!\frac{\sqrt{3}}{4}&\frac{1}{4}\!+\!\frac{3}{4} \end{pmatrix} \\[6pt] &=&\begin{pmatrix} 1&0\cr 0&1 \end{pmatrix} = I \end{eqnarray} $$

Hence \(A\) is orthogonal.

Let \(A=\begin{pmatrix} a&b\cr c&d \end{pmatrix}\small.\) So \(A'=\begin{pmatrix} a&c\cr b&d \end{pmatrix}\small.\)

$$ \begin{eqnarray} \small\textsf{So}\normalsize\:A\!+\!A'&=& \begin{pmatrix} a&b\cr c&d \end{pmatrix}+\begin{pmatrix} a&c\cr b&d \end{pmatrix}\\[6pt] &=& \begin{pmatrix} 2a& b+c \cr b+c& 2d \end{pmatrix}\\[6pt] \small\textsf{Now,}\normalsize\:(A\!+\!A')'&=& \begin{pmatrix} 2a& b+c \cr b+c& 2d \end{pmatrix} \end{eqnarray} $$

We have shown that \(\raise 0.2pt{A\!+\!A'}\) is self-transpose, which is the definition of symmetric.

$$ \begin{eqnarray} A\!-\!A'&=& \begin{pmatrix} a&b\cr c&d \end{pmatrix}+\begin{pmatrix} a&c\cr b&d \end{pmatrix}\\[6pt] &=& \begin{pmatrix} 0& b-c \cr c-b& 0 \end{pmatrix}\\[6pt] \small\textsf{Now,}\normalsize\:(A\!-\!A')'&=& \begin{pmatrix} 0& c-b \cr b-c& 0 \end{pmatrix}\\[6pt] &=& \begin{pmatrix} 0& -(b-c) \cr -(c-b)& 0 \end{pmatrix}\\[6pt] &=& -(A\!-\!A') \end{eqnarray} $$

We have shown that \(\raise 0.2pt{A\!-\!A'}\) equals the negative of its transpose, which is the definition of skew-symmetric.

$$ \begin{eqnarray} A^2 &=& \begin{pmatrix} 3 & \!0\cr \lambda & \!\!\!-\!2 \end{pmatrix}\begin{pmatrix} 3 & \!0\cr \lambda & \!\!\!-\!2 \end{pmatrix} \\[6pt] &=&\begin{pmatrix} 9 & \!0\cr \lambda & \!4 \end{pmatrix} \\[6pt] &=&p\begin{pmatrix} 3 & \!0\cr \lambda & \!\!\!-\!2 \end{pmatrix}+q\begin{pmatrix} 1 & \!0\cr 0 & \!1 \end{pmatrix} \end{eqnarray} $$

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).

$$\eqalign{ 3p+q &=9\cr -2p+q&=4\cr }$$

Subtracting the two equations, we obtain \(5p\!=\!5\) so \(p\!=\!1\small,\) and substituting into either equation gives us \(q\!=\!6\small.\)

We start by obtaining an expression for \(det\,A\) in terms of \(\mu\small.\)

$$ \begin{eqnarray} \vert A\vert &=& 2\,\begin{vmatrix} \mu &\!\!\!\!\phantom{-}4\,\cr 0 &\!\!\!\!-\!\!2\, \end{vmatrix} -3\,\begin{vmatrix} -1 &\!\!\!\!\phantom{-}4\,\cr \phantom{-}5 &\!\!\!\!-\!\!2\, \end{vmatrix}+1\,\begin{vmatrix} -1 &\!\!\!\!\phantom{-}\mu\,\cr \phantom{-}5 &\!\!\!\!\phantom{-}0\, \end{vmatrix} \\[6pt] &=& 2(-2\mu)-3(2-20)+1(-5\mu) \\[6pt] &=& -4\mu+54-5\mu \\[6pt] &=& -9\mu+54 \\[12pt] \end{eqnarray} $$

Now we can solve for \(\mu\small.\)

$$ \begin{gather} -9\mu+54=36 \\[6pt] -9\mu=36-54 \\[6pt] -9\mu=-18 \\[6pt] \mu=2 \\[6pt] \end{gather} $$

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 \(I_3\small,\) at which point the \(3\!\times\!3\) matrix on the right will be \(A^{-1}\small.\)

$$ \begin{eqnarray} \begin{array}{r} \small R_1\normalsize\cr \small R_2\normalsize\cr \small R_3\normalsize \end{array} && \left( \begin{array}{ccc:ccc} 1 & 2 & -1 & 1 & 0 & 0\cr -2 & 0 & 1 & 0 & 1 & 0\cr 1 & -1 & 0 & 0 & 0 & 1 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small R_2\!+\!2R_1\normalsize \cr \small R_1\!-\!R_3\normalsize \end{array} && \left( \begin{array}{ccc:ccc} 1 & 2 & -1 & 1 & 0 & 0\cr 0 & 4 & -1 & 2 & 1 & 0\cr 0 & 3 & -1 & 1 & 0 & -1 \end{array} \right) \\[10pt] \begin{array}{r} \small R_1\!-\!\frac{1}{2}R_2\normalsize\cr \small\textsf{Frozen}\normalsize\cr \small 3R_2\!-\!4R_3\normalsize \end{array} && \left( \begin{array}{ccc:ccc} 1 & 0 & -\frac12 & 0 & -\frac12 & 0\cr 0 & 4 & -1 & 2 & 1 & 0\cr 0 & 0 & 1 & 2 & 3 & 4 \end{array} \right) \\[10pt] \begin{array}{r} \small R_1\!+\!\frac{1}{2}R_3\normalsize\cr \small R_2\!+\!R_3\normalsize\cr \small\textsf{Frozen}\normalsize \end{array} && \left( \begin{array}{ccc:ccc} 1 & 0 & 0 & 1 & 1 & 2\cr 0 & 4 & 0 & 4 & 4 & 4\cr 0 & 0 & 1 & 2 & 3 & 4 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small \frac{1}{4}R_2\normalsize\cr \small\textsf{Frozen}\normalsize \end{array} && \left( \begin{array}{ccc:ccc} 1 & 0 & 0 & 1 & 1 & 2\cr 0 & 1 & 0 & 1 & 1 & 1\cr 0 & 0 & 1 & 2 & 3 & 4 \end{array} \right) \\[10pt] \end{eqnarray} $$

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

So \(A^{-1}= \begin{pmatrix} 1 & 1 & 2\cr 1 & 1 & 1\cr 2 & 3 & 4 \end{pmatrix}\small.\)

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 \(\raise 0.2pt{x}\)-axis negates its \(\raise 0.2pt{y}\)-coordinate. Its \(\raise 0.2pt{x}\)-coordinate is invariant (unchanged).

So \(\raise 0.2pt{M_1=\begin{pmatrix} 1 & 0\cr 0 & -1 \end{pmatrix}\small.}\)

(b)  Reflection of a point in the line \(\raise 0.3pt{y\!=\!x}\) involves the \(\raise 0.2pt{x}\)- and \(y\)-coordinates swapping places (as you've known all the way back to Nat 5 with inverse functions). Reflection in the line \(\raise 0.3pt{y\!=\!\!-\!x}\) swaps and also negates the coordinates.

So \(\raise 0.2pt{M_2=\begin{pmatrix} 0 & -1\cr -1 & 0 \end{pmatrix}\small.}\)

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

$$ \begin{eqnarray} M_3 &=& M_{1}\,M_2\\[6pt] &=& \begin{pmatrix} 1 & 0\cr 0 & -1 \end{pmatrix}\,\begin{pmatrix} 0 & -1\cr -1 & 0 \end{pmatrix}\\[6pt] &=& \begin{pmatrix} 0 & -1\cr 1 & 0 \end{pmatrix}\\[6pt] \end{eqnarray} $$

(d)  The matrix \(\begin{pmatrix} cos\,\theta & -sin\,\theta\,\cr sin\,\theta & \phantom{-}cos\,\theta\, \end{pmatrix}\) is associated with anti-clockwise rotation through an angle \(\theta\) about the origin.

\(\raise 0.2pt{cos\,\theta\!=\!0 \implies \theta\!=\!\frac{\pi}{2}\small},\) which is consistent with \(\raise 0.2pt{sin\,\theta\!=\!1 \implies \theta\!=\!\frac{\pi}{2}\small.}\)

So \(M_3\) represents an anti-clockwise rotation of \(\raise 0.2pt{\frac{\pi}{2}}\) 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:

$$\begin{pmatrix} cos\,\theta & -sin\,\theta\,\cr sin\,\theta & \phantom{-}cos\,\theta\, \end{pmatrix}$$

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

$$A=\begin{pmatrix} cos\,\frac{\pi}{3} & -sin\,\frac{\pi}{3}\,\cr sin\,\frac{\pi}{3} & \phantom{-}cos\,\frac{\pi}{3}\, \end{pmatrix}$$

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:

$$A=\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2}\,\cr \frac{\sqrt{3}}{2} & \phantom{-}\frac{1}{2}\, \end{pmatrix}$$

(b)  Reflection in the \(\raise 0.3pt{x}\)-axis negates \(y\)-coordinates, so the associated matrix is as follows:

\(\raise 0.2pt{B=\begin{pmatrix} 1 & 0\cr 0 & -1 \end{pmatrix}\small.}\)

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

$$ \begin{eqnarray} P &=& B\,A\\[6pt] &=& \begin{pmatrix} 1 & 0\cr 0 & -1 \end{pmatrix}\,\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2}\,\cr \frac{\sqrt{3}}{2} & \phantom{-}\frac{1}{2}\, \end{pmatrix}\\[6pt] &=& \begin{pmatrix} \phantom{-}\frac{1}{2} & -\frac{\sqrt{3}}{2}\,\cr -\frac{\sqrt{3}}{2} & -\frac{1}{2}\, \end{pmatrix} \end{eqnarray} $$

(d)  The matrix \(\begin{pmatrix} cos\,\theta & -sin\,\theta\,\cr sin\,\theta & \phantom{-}cos\,\theta\, \end{pmatrix}\) 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 \(cos\,\theta\small.\)

In matrix \(P\small,\) the top-left element is \(\frac12\) and the bottom-right element is \(-\frac12\small.\)

There is no angle \(\theta\) for which \(cos\,\theta=\frac12\) and \(cos\,\theta=-\frac12\small,\) so \(P\) is not associated with rotation about the origin.