Advanced Higher Maths
Matrices

(a)  Recall the definitions of determinant and inverse for \(2\!\times\!2\) matrices:

If \(A=\left(\begin{array}{@{\,}rr@{\,}} a & b\\ c & d \end{array}\right)\,\) then \(\text{det}\,A=ad\!-\!bc\,\) and \(A^{-1}=\large\frac{1}{\text{det}\,A}\normalsize\left(\begin{array}{@{\,}rr@{\,}} d & -b\\ -c & a \end{array}\right)\small.\)

So for this example:

$$ \begin{eqnarray} \text{det}\,A &=& (-4)(5)-(-3)(6) \\[6pt] &=& -\!20+18\\[6pt] &=& -\!2\\[12pt] A^{-1} &=& \small\frac{1}{-2}\normalsize\left(\begin{array}{@{\,}rr@{\,}} 5 & 3\\ -6 & -4 \end{array}\right) \\[6pt] &=& \small\frac{1}{2}\normalsize\left(\begin{array}{@{\,}rr@{\,}} -5 & -3\\ 6 & 4 \end{array}\right) \\[6pt] &\small\textsf{or}& \left(\begin{array}{@{\,}rr@{\,}} -\frac{5}{2} & -\frac{3}{2}\\ 3\phantom{.} & 2\phantom{.} \end{array}\right) \\[6pt] \end{eqnarray} $$

(b)  \(A'\) is the transpose of \(A\small,\) which is the matrix obtained by interchanging the rows and columns of \(A\small.\)

So \(A'=\left(\begin{array}{@{\,}rr@{\,}} \,-4 & 6\,\cr \,-3 & 5\, \end{array}\right)\small.\)

Historical note: Although \(\raise 0.3pt{A^T}\) is the globally standard notation, \(\raise 0.3pt{A'}\) is still used in Scotland... read more  

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

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

So \(P\) is singular if \(n=-12\small.\)

We simply calculate \(2A'\) and compare its entries with \(B\small.\)

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

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

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

You are expected to know that \(A\) is orthogonal if \(A\tiny\,\normalsize A'=A'A=I\small.\)

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 \(A'\small.\)

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

$$ \begin{eqnarray} A\tiny\,\normalsize A' &=& \left(\begin{array}{@{\,}rr@{\,}} \large\frac35 & -\large\frac45\\ \large\frac45 & \large\frac35 \end{array}\right)\left(\begin{array}{@{\,}rr@{\,}} \large\frac35 & \large\frac45\cr -\large\frac45 & \large\frac35 \end{array}\right) \\[12pt] &=& \left(\begin{array}{@{\,}rr@{\,}} \large\frac{9}{15}\normalsize\!+\!\large\frac{16}{25} & \large\frac{12}{25}\normalsize\!-\!\large\frac{12}{25}\cr \large\frac{12}{25}\normalsize\!-\!\large\frac{12}{25} & \large\frac{16}{15}\normalsize\!+\!\large\frac{9}{25} \end{array}\right) \\[12pt] &=& \:\left(\begin{array}{@{\,}rr@{\,}} 1 & 0\\ 0 & 1 \end{array}\right) \end{eqnarray} $$

We have shown that \(A\tiny\,\normalsize A'=I\small,\) so \(A\) is orthogonal.

Note: There are infinitely many \(2\!\times\!2\) orthogonal matrices with rational entries. They are very closely related to the Pythagorean triples: \((3,\,4,\,5)\small,\) \((5,\,12,\,13)\small,\) \((8,\,15,\,17)\small,\) \((7,\,24,\,25)\small,\) \((20,\,21,\,29)\small,\) \((12,\,35,\,37)\) etc.

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

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

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

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

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

Note: The rather naïve proofs given above could easily be extended to square matrices of any dimension, by using two of the properties of transpose listed above, which you are expected to know. Specifically:

$$ \begin{eqnarray} (A + A’)’ &=& A’+(A’)’ \\[6pt] &=& A’ + A\\[6pt] &=& A + A’ \\[18pt] (A - A’)’ &=& A’-(A’)’ \\[6pt] &=& A’ - A\\[6pt] &=& -\!(A - A’) \\[6pt] \end{eqnarray} $$

We start by squaring \(A\) and comparing its entries to \(pA\!+\!qI\small.\)

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

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

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

So \(p\!=\!1\) and \(q\!=\!6\small.\) Therefore \(A^2=A+6I\small.\)

We start by obtaining an expression for \(\text{det}\,A\) 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 start by obtaining an expression for \(\text{det}\,A\) in terms of \(p\small.\)

$$ \begin{eqnarray} \vert\,A\,\vert &=& p\,\begin{vmatrix} \,\phantom{-}p &\!\!\!\!\phantom{-}1\,\cr \,-1&\!\!\!\!-1 \end{vmatrix} -2\begin{vmatrix} \,3 &\!\!\!\!\phantom{-}1\,\cr \,0 &\!\!\!\!-1\, \end{vmatrix}+0\begin{vmatrix} \,3 &\!\!\!\!\phantom{-}p\,\cr \,0 &\!\!\!\!-1\, \end{vmatrix} \\[6pt] &=& p(-p+1)-2(-3-0)+0 \\[6pt] &=& -\!p^2+p+6 \\[6pt] \end{eqnarray} $$

Matrix \(A\) is singular if \(\text{det}\,A=0\small,\) so we solve:

$$ \begin{gather} -p^2+p+6=0 \\[6pt] p^2-p-6=0 \\[6pt] (p-3)(p+2)=0 \\[6pt] p-3=0\:\:\small\textsf{or}\normalsize\ \:p+2=0 \\[6pt] p=3\:\:\small\textsf{or}\normalsize\ \:p=-2 \\[6pt] \end{gather} $$

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

Note 1: There are other methods of finding the inverse of a \(3\!\times\!3\) matrix, but EROs are usually more efficient.

Note 2: EROs could also be used for the inverse of a \(2\!\times\!2\) matrix, but the method in example 1 is quicker.

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 \(\left(\begin{array}{@{\,}rr@{\,}} \text{cos}\,\theta & -\text{sin}\,\theta\\ \text{sin}\,\theta & \text{cos}\,\theta\, \end{array}\right)\) is associated with anti-clockwise rotation through an angle \(\theta\) about the origin.

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

So \(M_3\) represents an anti-clockwise rotation of \(\large\frac{\pi}{2}\) 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 \(\theta = \large\frac{\pi}{3}\small,\) 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 \(\raise 0.3pt{x}\)-axis negates \(y\)-coordinates, so the associated matrix is as follows:

$$\raise 0.2pt{B=\begin{pmatrix} \,1 & 0\cr \,0 & -1 \end{pmatrix}}$$

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

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

(d)  The matrix \(\left(\begin{array}{@{\,}rr@{\,}} \text{cos}\,\theta & -\text{sin}\,\theta\\ \text{sin}\,\theta & \text{cos}\,\theta\, \end{array}\right)\) 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 \(cos\,\theta\small.\)

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

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

(a)  This is a simple arithmetic operation:

$$ \begin{eqnarray} 3A+2B &=& 3\begin{pmatrix} \,-3 &\!\!\!\phantom{-}2\,\cr \,\phantom{-}0 &\!\!\!\phantom{-}1\, \end{pmatrix}+2\begin{pmatrix} \,2 &\!2\,\cr \,5 &\!\lambda\, \end{pmatrix}\\[9pt] &=& \begin{pmatrix} \,-9 &\!\!\!\phantom{-}6\,\cr \,\phantom{-}0 &\!\!\!\phantom{-}3\, \end{pmatrix}+\begin{pmatrix} \,4 &\!4\,\cr \,10 &\!2\lambda\, \end{pmatrix}\\[9pt] &=& \begin{pmatrix} \,-5 & 10\cr \,10 & 3\!+\!2\lambda \end{pmatrix}\\[9pt] \end{eqnarray} $$

(b) (i)  Creating \(A'\) by swapping the rows and columns of \(A\):

$$ \begin{eqnarray} A'B &=& \begin{pmatrix} \,-3 &\!\!\!\phantom{-}0\,\cr \,\phantom{-}2 &\!\!\!\phantom{-}1\, \end{pmatrix}\begin{pmatrix} \,2 &\!2\,\cr \,5 &\!\lambda\, \end{pmatrix}\\[9pt] &=& \begin{pmatrix} \,-6+0 & \!-6+0\lambda\,\cr \,\phantom{-}4+5 &\!\phantom{-}4+\lambda\, \end{pmatrix}\\[9pt] &=& \begin{pmatrix} \,-6 & \!\!-6\,\cr \,\phantom{-}9 & \!\!\phantom{-}4+\lambda\, \end{pmatrix}\\[9pt] \end{eqnarray} $$

(b) (ii)  We just calculate the \(2\!\times\!2\) determinant in the usual way:

$$ \begin{eqnarray} \text{det}\,(A'B)&=&(-6)(4+\lambda)\!-\!(-6)(9) \\[6pt] &=& -\!24-6\lambda+54\\[6pt] &=& 30-6\lambda\\[8pt] \end{eqnarray} $$

(b) (iii)  A matrix is singular if its determinant equals zero:

$$ \begin{gather} 30-6\lambda = 0\\[6pt] 30 = 6\lambda\\[6pt] \lambda = 5\\[6pt] \end{gather} $$