Example 13.1

$\begin{array}{l}\text{A}=\left(2????1\right)??\text{B}=\left(6????2\right)\hfill \\ \text{C}=\left(\begin{array}{cc}3& 7\\ 8& 3\end{array}\right)\hfill \\ \text{D}=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)\hfill \\ \text{E}=\left(\begin{array}{lll}8\hfill & 2\hfill & 1\hfill \\ 6\hfill & 1\hfill & 3\hfill \end{array}\right)\hfill \\ \text{F}=\left(\begin{array}{lll}2\hfill & 6\hfill & 3\hfill \\ 12\hfill & -2\hfill & -6\hfill \end{array}\right)\hfill \end{array}$

Find where possible: (a) A + B, (b) C + D, (c) E – F, (d) A + D.

Solution

(a) A + B = (2 1) + (6 2) = (8 3)

(b) C − D = $\left(\begin{array}{cc}3& 7\\ 8& 3\end{array}\right)+\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)=\left(\begin{array}{cc}4& 9\\ 10& 4\end{array}\right)$

(c) E + F$\begin{array}{c}=\left(\begin{array}{ccc}8& 2& 1\\ 6& 1& 3\end{array}\right)-\left(\begin{array}{ccc}2& 6& 3\\ 12& -2& -6\end{array}\right)\\ =\left(\begin{array}{ccc}6& -4& -2\\ -6& 3& 9\end{array}\right)\end{array}$

(d) A + D cannot be found because the two matrices are of different dimensions.

Example 13.2

$\text{If}?\text{A}=\left(\begin{array}{cc}2& 5\\ 6& 1\end{array}\right)$

find 2A and $\frac{1}{3}$A

Solution

$\begin{array}{c}2\text{A}=2\left(\begin{array}{cc}2& 5\\ 6& 1\end{array}\right)=\left(\begin{array}{cc}4& 10\\ 12& 2\end{array}\right)\\ \frac{1}{3}\text{A}=\frac{1}{3}\left(\begin{array}{cc}2& 5\\ 6& 1\end{array}\right)=\left(\begin{array}{cc}\frac{2}{3}& \frac{5}{3}\\ 2& \frac{1}{3}\end{array}\right)\end{array}$

Example 13.3

$\begin{array}{c}\text{A}=\left(\begin{array}{cc}1& -1\\ 3& 1\end{array}\right)\\ \text{B}=\left(\begin{array}{ccc}6& 0& -1\\ 2& 2& 3\end{array}\right)\end{array}$

Find, if possible, AB and BA

BA cannot be found because the number of columns in B is not equal to the number of rows in A.

We can justify the practical reasons for this method of matrix multiplication as in the following two examples. In the first, we return to our household shopping example.

Example 13.4

Every weekday a household orders pints of milk, loaves of bread and yoghurt from a milk lorry. The orders for the week are as follows:

Next week, the dairy introduces a special offer and reduces its prices. The prices for this week and the next are as follows:

Calculate the cost each day for this week and the next.

Solution The cost each day is made up of the number of pints of milk times the cost of a pint plus the number of loaves of bread times the cost of a loaf plus the number of cartons of yoghurt times the cost of the yoghurt. In other words, we can find the cost each day by performing matrix multiplication

$\begin{array}{c}\left(\begin{array}{ccc}3& 2& 4\\ 4& 1& 0\\ 2& 2& 4\\ 5& 1& 0\\ 1& 1& 4\end{array}\right)\left(\begin{array}{cc}0.34& 0.32\\ 0.60& 0.50\\ 0.33& 0.30\end{array}\right) \\ =\left(\begin{array}{cc}3\times \left(0.34\right)+2\times \left(0.60\right)+4\times \left(0.33\right)& 3\times \left(0.32\right)+2\times \left(0.50\right)+4\times \left(0.30\right)\\ 4\times \left(0.34\right)+1\times \left(0.60\right)+0\times \left(0.33\right)& 4\times \left(0.32\right)+1\times \left(0.50\right)+0\times \left(0.30\right)\\ 2\times \left(0.34\right)+2\times \left(0.60\right)+4\times \left(0.33\right)& 2\times \left(0.32\right)+2\times \left(0.50\right)+4\times \left(0.30\right)\\ 5\times \left(0.34\right)+1\times \left(0.60\right)+0\times \left(0.33\right)& 5\times \left(0.32\right)+1\times \left(0.50\right)+0\times \left(0.30\right)\\ 1\times \left(0.34\right)+1\times \left(0.60\right)+4\times \left(0.33\right)& 1\times \left(0.32\right)+1\times \left(0.50\right)+4\times \left(0.30\right)\end{array}\right)\\ =\left(\begin{array}{cc}3.54& 3.16\\ 1.96& 1.78\\ 3.20& 2.84\\ 2.30& 2.10\\ 2.26& 2.02\end{array}\right) \end{array}$

The rows now represent the days of the week and the columns represent this week and the next week. Hence, for instance, the cost for Thursday of next week is given by the element a₄₂=2.10.

Example 13.5

Figure 13.1 represents a communication network where the vertices a,b,f,g represent offices and vertices c,d,e represent switching centres. The numbers marked along the edges represent the number of connections between any two vertices. Calculate the number of routes from a,b to f,g.

Solution The number of routes from a to f can be calculated by taking the number via c plus the number via d plus the number via e. In each case, this is given by multiplying the number of connections along the edges connecting a to c, c to f, etc giving the number of routes from a to f as: 3 × 2 + 4 × 6 + 1 × 1.

We can see that we can get the number of routes by matrix multiplication. The network from ab to cde is represented by:

and from cde to fg by

So, the total number of routes is given by

$\begin{array}{l}\left(\begin{array}{lll}3\hfill & 4\hfill & 1\hfill \\ 2\hfill & 1\hfill & 3\hfill \end{array}\right)\left(\begin{array}{ll}2\hfill & 1\hfill \\ 6\hfill & 3\hfill \\ 1\hfill & 2\hfill \end{array}\right)\hfill \\ ??=\left(\begin{array}{ll}3\times 2+4\times 6+1\times 1\hfill & 3\times 1+4\times 3+1\times 2\hfill \\ 2\times 2+1\times 6+3\times 1\hfill & 2\times 1+3\times 1+3\times 2\hfill \end{array}\right)\hfill \end{array}$

Hence, by interpreting the rows and columns of the resulting matrix we can see that there are 31 routes from a to f, 17 from a to g, 13 from b to f and 11 from b to g.

Example 13.6

$\text{A}=\left(\begin{array}{cc}2& -1\\ 0& 1\end{array}\right),?\text{B}=\left(\begin{array}{c}3\\ 2\end{array}\right)$

Show that AI =IA =A and IB =B.

Solution

Example 13.7

Given

$\text{A}=\left(\begin{array}{cc}2& -1\\ 6& 3\end{array}\right),?\text{B}=\left(\begin{array}{lll}2\hfill & 1\hfill & 8\hfill \\ -1\hfill & 0\hfill & 1\hfill \end{array}\right)$

find A^Tand B^T

Solution The first row of A is (2–1) therefore this is the first column of A^T. The second row of A is (6 3) therefore this is the second column of A^T. This gives A^Tas follows.

${\text{A}}^{\text{T}}=\left(\begin{array}{cc}2& 6\\ -1& 3\end{array}\right)$

Similarly

${\text{B}}^{\text{T}}=\left(\begin{array}{ll}2\hfill & -1\hfill \\ 1\hfill & 0\hfill \\ 8\hfill & 1\hfill \end{array}\right)$

Example 13.8

Show that

$\text{A}=\left(\begin{array}{cc}0& 6\\ -6& 0\end{array}\right)$

is skew symmetric.

Solution

${\text{A}}^{\text{T}}=\left(\begin{array}{cc}0& -6\\ 6& 0\end{array}\right)$

Multiplying A by −1, we get

$-\text{A}=\left(\begin{array}{cc}0& -6\\ 6& 0\end{array}\right)$

We can see that A^T = −A and hence we have shown that A is skew symmetric.

Example 13.9

Show that

$\text{A}=\left(\begin{array}{cc}3& 7+\text{j}2\\ 7-\text{j}2& -2\end{array}\right)?\text{and}?\text{B}=\left(\begin{array}{cc}2& 3{\text{e}}^{-\text{j}2}\\ 3{\text{e}}^{\text{j}2}& 1\end{array}\right)$

are Hermitian.

Solution Taking the complex conjugates of each of the elements in A and B gives

${\text{A}}^{*}=\left(\begin{array}{cc}3& 7-\text{j}2\\ 7+\text{j}2& -2\end{array}\right)?\text{and}?\text{B}*=\left(\begin{array}{cc}2& 3{\text{e}}^{\text{j}2}\\ 3{\text{e}}^{-\text{j}2}& 1\end{array}\right)$

Now taking the transposes of A and B, we get

${\text{A}}^{*\text{T}}=\left(\begin{array}{cc}3& 7+\text{j}2\\ 7-\text{j}2& -2\end{array}\right)?\text{and}?{\text{B}}^{*\text{T}}=\left(\begin{array}{cc}2& 3{\text{e}}^{-\text{j}2}\\ 3{\text{e}}^{\text{j}2}& 1\end{array}\right)$

So we can see that

${\text{A}}^{*\text{T}}=\text{A}?\text{and}?{\text{B}}^{*\text{T}}=\text{B}$

showing that they are Hermitian.

In the rest of this chapter we shall assume that our matrices are real. A column vector is a matrix with only one column, for example

$\text{v}=\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)$

A row vector is a matrix with only one row, for example

$\text{v}=\left(1?2?3\right).$

Example 13.10

Show that

$\left(\begin{array}{cc}\frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& -\frac{2}{3}\end{array}\right)$

is the inverse of

$\left(\begin{array}{cc}2& 1\\ 1& -1\end{array}\right).$

Solution Multiply:

Also

Not all matrices have inverses and only square matrices can possibly have inverses. A matrix does not have an inverse if its determinant is 0.

The determinant of

$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$

is given by

$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-cb$

If the determinant of a matrix is 0 then it has no inverse and the matrix is said to be singular. If the determinant is non- zero then the inverse exists. The inverse of the 2 × 2 matrix

$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$

is

$\frac{1}{\left(ad-cb\right)}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$

That is, to find the inverse of a 2 × 2 matrix, we swap the diagonal elements, negate the off-diagonal elements, and divide the resulting matrix by the determinant.

Example 13.11

Find the determinants of the following matrices and state if the matrix has an inverse or is singular. Find the inverse in the cases where is exists and check that AA ⁻¹ =A ⁻¹ A =I

$\left(\text{a}\right)?\left(\begin{array}{cc}-1& 3\\ 2& 1\end{array}\right),?\left(\text{b}\right)\left(\begin{array}{cc}6& -2\\ -3& 1\end{array}\right),?\left(\text{c}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right).$

Solution

$\left(\text{a}\right)\left|\begin{array}{cc}-1& 3\\ 2& 1\end{array}\right|=\left(-1\right)\times 1-2\times 3=-7.$

As the determinant is not zero the matrix

$\left(\begin{array}{cc}-1& 3\\ 2& 1\end{array}\right)$

has an inverse found by swapping the diagonal elements and negating the off-diagonal elements, then dividing by the determinant. This gives

$\frac{1}{-7}\left(\begin{array}{cc}1& -3\\ -2& -1\end{array}\right)=\frac{1}{7}\left(\begin{array}{cc}-1& 3\\ 2& 1\end{array}\right).$

Check that AA ⁻¹ =I

$\begin{array}{l}\left(\begin{array}{cc}-1& 3\\ 2& 1\end{array}\right)\frac{1}{7}\left(\begin{array}{cc}-1& 3\\ 2& 1\end{array}\right)\hfill \\ =\frac{1}{7}\left(\begin{array}{ll}\left(-1\right)\left(-1\right)+\left(3\right)\left(2\right)\hfill & \left(-1\right)3+\left(3\right)\left(1\right)\hfill \\ \left(2\right)\left(-1\right)+\left(1\right)\left(2\right)\hfill & \left(2\right)\left(3\right)+\left(1\right)\left(1\right)\hfill \end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\hfill \end{array}$

and that A ⁻¹ A =I

$\begin{array}{l}\frac{1}{7}\left(\begin{array}{cc}-1& 3\\ 2& 1\end{array}\right)\left(\begin{array}{cc}-1& 3\\ 2& 1\end{array}\right)\hfill \\ ?=\frac{1}{7}\left(\begin{array}{ll}\left(-1\right)\left(-1\right)+\left(3\right)\left(2\right)\hfill & \left(-1\right)3+\left(3\right)\left(1\right)\hfill \\ \left(2\right)\left(-1\right)+\left(1\right)\left(2\right)\hfill & \left(2\right)\left(3\right)+\left(1\right)\left(1\right)\hfill \end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right).\hfill \\ \left(\text{b}\right)\left|\begin{array}{cc}6& -2\\ -3& 1\end{array}\right|=6\cdot 1-\left(-3\right)\left(-2\right)=0\hfill \end{array}$

As the determinant is zero the matrix

$\left(\begin{array}{cc}6& -2\\ -3& 1\end{array}\right)$

has no inverse. It is singular.

$\begin{array}{l}\left(\text{c}\right)?\left|\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right|\hfill \\ =\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)\frac{1}{\sqrt{2}}=1.\hfill \end{array}$

Therefore, the matrix is invertible. Its inverse is given by swapping the diagonal elements, and negating the off-diagonal elements, and then dividing by the determinant. This gives

$\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)$

Check that AA ⁻¹ =I:

${\text{AA}}^{-1}=\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$

Similarly, A⁻¹ A =I.

Example 13.12

Given that A, B, and C are matrices and AB =C where A and B are non-singular, find expressions for B and A.

Solution In this case, we are told that A and B are invertible, so they must be square and therefore C must also be square and of the same dimension. To find B we wish to ‘get rid’ of the A term on the left-hand side. We pre-multiply both sides of the equation by A⁻¹

AB =C and given A is invertible

⇔ A⁻¹AB=A⁻¹C.

Now using A⁻¹ A = I, the unit matrix, we have

IB = A^−lC.

As the unit matrix multiplied by any matrix leaves it unchanged, we have

⇔ B=A⁻¹C.

To find an expression for A, use

AB =C

given that B is invertible, we post-multiply by B⁻¹

⇔ABB⁻¹ = CB⁻¹.

Now using BB⁻¹=I, the unit matrix, we have

AI =CB⁻¹.

As the unit matrix multiplied by any matrix leaves it unchanged, we have

⇔A =CB⁻¹.

Remember that it is always important to specify whether you are pre- or post-multiplying when solving matrix equations. A term like B⁻¹ AB cannot be simplified because we cannot swap the order, as we would do with numbers.

Example 13.13

Find and draw the image of the unit square with vertices A(0,0), B(1,0), C(1,1), D(0, 1) after rotation through 30° about the origin.

Solution Rotation through 30° about the origin is found by multiplying the position vectors of the points by

$\left(\begin{array}{cc}cos\left({30}^{\circ }\right)& -sin\left({30}^{\circ }\right)\\ sin\left({30}^{\circ }\right)& cos\left({30}^{\circ }\right)\end{array}\right)\approx \left(\begin{array}{cc}0.866& -0.5\\ 0.5& 0.866\end{array}\right)$

To find the image of the unit square, we multiply the position vectors of the vertices by this matrix

$\begin{array}{c}\left(\begin{array}{cc}0.866& -0.5\\ 0.5& 0.866\end{array}\right)\left(\begin{array}{c}0\\ 0\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \left(\begin{array}{cc}0.866& -0.5\\ 0.5& 0.866\end{array}\right)\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}0.866\\ 0.5\end{array}\right)\\ \left(\begin{array}{cc}0.866& -0.5\\ 0.5& 0.866\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right)=\left(\begin{array}{c}0.366\\ 1.366\end{array}\right)\\ \left(\begin{array}{cc}0.866& -0.5\\ 0.5& 0.866\end{array}\right)\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}-0.5\\ 0.866\end{array}\right)\end{array}$

This transformation is shown in Figure 13.2.

Sometimes, it is useful to be able to rotate the axes rather than the object. For instance, the object may be held by a robotic arm and we want the arm to rotate but keep the orientation of the object the same. This is picture for the tea drinking robot in Figure 13.3.

In this case, if we rotate the axes Ox, Oy, by the position of the object remains the same but even so has new coordinates relative to the the transformed axes OX, OY. If the axes rotate through 30°, then the object moves relative to the axes by – 30°. So to rotate the axes by we multiply the position vectors of the points

$\left(\begin{array}{c}x\\ y\end{array}\right)$

by the matrix

$\left(\begin{array}{cc}cos\left(-\theta \right)& -sin\left(-\theta \right)\\ sin\left(-\theta \right)& cos\left(-\theta \right)\end{array}\right)=\left(\begin{array}{cc}cos\left(\theta \right)& sin\left(\theta \right)\\ -sin\left(\theta \right)& cos\left(\theta \right)\end{array}\right).$

Example 13.14

A unit square has vertices A(0,0), B(1,0), C (1,1), D(0, 1) relative to axes Ox, Oy. The axes are rotated through 30° to OX, OY, without moving the square. Find the coordinates of the vertices relative to the new axes OX, OY.

Solution The effect of rotating the axes through 30° is found by multiplying the position vectors of the points by

$\left(\begin{array}{cc}cos\left({30}^{\circ }\right)& sin\left({30}^{\circ }\right)\\ sin\left(-{30}^{\circ }\right)& cos\left({30}^{\circ }\right)\end{array}\right)=\left(\begin{array}{cc}0.866& 0.5\\ -0.5& 0.866\end{array}\right).$

To find the coordinates of the unit square relative to the new axes, we multiply the position vectors of the vertices by this matrix

$\begin{array}{c}\left(\begin{array}{ll}0.866\hfill & 0.5\hfill \\ -0.5\hfill & 0.866\hfill \end{array}\right)\left(\begin{array}{c}0\\ 0\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \left(\begin{array}{cc}0.866& 0.5\\ -0.5& 0.866\end{array}\right)\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}0.866\\ -0.5\end{array}\right)\\ \left(\begin{array}{cc}0.866& 0.5\\ -0.5& 0.866\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right)=\left(\begin{array}{c}1.366\\ 0.366\end{array}\right)\\ \left(\begin{array}{cc}0.866& 0.5\\ -0.5& 0.866\end{array}\right)\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}0.5\\ 0.866\end{array}\right).\end{array}$

This is shown in Figure 13.4.

Example 13.15

Find and draw the image of the unit square with vertices A(0,0), B(1,0), C(1,1), D(0,l) after translation through

$\left(\begin{array}{c}3\\ 4\end{array}\right).$

Solution Add

$\left(\begin{array}{c}3\\ 4\end{array}\right)$

to the position vectors of the vertices, that is

$\text{v}+\left(\begin{array}{c}3\\ 4\end{array}\right)$

which gives A′ as (3,4), B′ as (4,4), C′ as (4,5), and D′ as (3,5). This transformation is shown in Figure 13.6.

It is again often useful to consider what happens if the object stays where it is and the axes are translated. If the axes are translated through

$\left(\begin{array}{c}3\\ 4\end{array}\right)$

then the object appears to move relative to the axes by

$\left(\begin{array}{c}-3\\ -4\end{array}\right).$

Therefore, we subtract

$\left(\begin{array}{c}3\\ 4\end{array}\right)$

from the coordinates defining it. This is shown in Figure 13.7.

Find the coordinates of the vertices of the unit square after: (a) rotation about the origin through 50° followed by a translation of (–1, 2); (b) translation of (–1, 2) followed by rotation about the origin through 50°.

Solution (a) We can write this combined transformation as

p′ =Rp + t

where p′ is the position vector of the image point, p is the position vector of the original point, R is the matrix representing the rotation, and t is the vector representing the translation.

In this case

$\text{R}=\left(\begin{array}{cc}cos\left({50}^{\circ }\right)& -sin\left({50}^{\circ }\right)\\ sin\left({50}^{\circ }\right)& cos\left({50}^{\circ }\right)\end{array}\right)\approx \left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)$

and

$\text{t}=\left(\begin{array}{c}-1\\ 2\end{array}\right),?{\text{p}}^{\prime }=\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right),?\text{p}=\left(\begin{array}{c}x\\ y\end{array}\right)$

So we have

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}-1\\ 2\end{array}\right)$

For the coordinates of A′ substitute x =0 and y =0 giving

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)\left(\begin{array}{c}0\\ 0\end{array}\right)+\left(\begin{array}{c}-1\\ 2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)+\left(\begin{array}{c}-1\\ 2\end{array}\right)=\left(\begin{array}{c}-1\\ 2\end{array}\right)$

for B′

$\begin{array}{c}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)\left(\begin{array}{c}1\\ 0\end{array}\right)+\left(\begin{array}{c}-1\\ 2\end{array}\right)=\left(\begin{array}{c}0.643\\ 0.766\end{array}\right)+\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ =\left(\begin{array}{c}-0.357\\ 2.766\end{array}\right) \end{array}$

for C′

$\begin{array}{c}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right)+\left(\begin{array}{c}-1\\ 2\end{array}\right)=\left(\begin{array}{c}-0.123\\ 1.409\end{array}\right)+\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ =\left(\begin{array}{c}-1.123\\ 3.409\end{array}\right) \end{array}$

for D′

$\begin{array}{c}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)\left(\begin{array}{c}0\\ 1\end{array}\right)+\left(\begin{array}{c}-1\\ 2\end{array}\right)=\left(\begin{array}{c}-0.766\\ 0.643\end{array}\right)+\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ =\left(\begin{array}{c}-1.766\\ 2.643\end{array}\right) \end{array}$

The image of the unit square is pictured in Figure 13.9(b).

(b) We can write this combined transformation as

p″= R(p + t)

where p″ is the position vector of the image point, p is the position vector of the original point, R is the matrix representing the rotation, and t is the vector representing the translation. We have put the brackets in to

indicate that the translation is performed first. As before

$\text{R}=\left(\begin{array}{cc}cos\left({50}^{\circ }\right)& -sin\left({50}^{\circ }\right)\\ sin\left({50}^{\circ }\right)& cos\left({50}^{\circ }\right)\end{array}\right)\approx \left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)$

and

$\text{t}=\left(\begin{array}{c}-1\\ 2\end{array}\right),?{\text{p}}^{″}=\left(\begin{array}{c}{x}^{″}\\ y^{″}\end{array}\right),?\text{p}=\left(\begin{array}{c}x\\ y\end{array}\right)$

So we have

$\left(\begin{array}{c}{x}^{″}\\ y^{″}\end{array}\right)=\left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)\left(\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}-1\\ 2\end{array}\right)\right)$

which is the same as

$\left(\begin{array}{c}{x}^{″}\\ y^{″}\end{array}\right)=\left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)\left(\begin{array}{c}x-1\\ y+2\end{array}\right)$

For the coordinates of A″, substitute x =0 and y =0 giving

$\begin{array}{c}\left(\begin{array}{c}{x}^{″}\\ y^{″}\end{array}\right)=\left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)\left(\begin{array}{c}0-1\\ 0+2\end{array}\right)=\left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ =\left(\begin{array}{c}-2.175\\ 0.52\end{array}\right) \end{array}$

for B″

$\begin{array}{c}\left(\begin{array}{c}{x}^{″}\\ y^{″}\end{array}\right)=\left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)\left(\begin{array}{c}1-1\\ 0+2\end{array}\right)=\left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)\left(\begin{array}{c}0\\ 2\end{array}\right)\\ =\left(\begin{array}{c}-1.532\\ 1.286\end{array}\right) \end{array}$

for C″

$\begin{array}{c}\left(\begin{array}{c}{x}^{″}\\ y^{″}\end{array}\right)=\left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)\left(\begin{array}{c}1-1\\ 1+2\end{array}\right)=\left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)\left(\begin{array}{c}0\\ 3\end{array}\right)\\ =\left(\begin{array}{c}-2.298\\ 1.929\end{array}\right) \end{array}$

for D′

$\begin{array}{c}\left(\begin{array}{c}{x}^{″}\\ y^{″}\end{array}\right)=\left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)\left(\begin{array}{c}0-1\\ 1+2\end{array}\right)=\left(\begin{array}{cc}0.643& -0.766\\ 0.766& 0.643\end{array}\right)\left(\begin{array}{c}-1\\ 3\end{array}\right)\\ =\left(\begin{array}{c}-2.941\\ 1.163\end{array}\right) \end{array}$

The image of the unit square is pictured in Figure 13.9(b).

Note that the order of the transformations is important.

Sometimes, we might need to use a trick of temporarily moving the axes in order to perform certain transformations. Supposing we want to scale by 2 along the line x =y we can rotate the axes temporarily so that the new X -axis lies along the line that was previously x =y, then perform X scaling, and then rotate back again, so the axes are back in their original position. This is done in the next example.

Example 13.17

Find a matrix that performs scaling by a factor of 2 along the direction x =y and draw the image of the unit square defined by the vertices

$\text{A}\left(-\frac{1}{2},-\frac{1}{2}\right),\text{B}\left(\frac{1}{2},-\frac{1}{2}\right),\text{C}\left(\frac{1}{2},\frac{1}{2}\right),\text{D}\left(-\frac{1}{2},\frac{1}{2}\right).$

Solution First, we rotate the axes by 45°, so that the OX -axis will lie along the line that was previously x =y. This is pictured in Figure 13.10.

The matrix that transforms the coordinates so they are relative to the new axes at an angle of 45° is given by:

$\left(\begin{array}{cc}cos\left({45}^{\circ }\right)& sin\left({45}^{\circ }\right)\\ -sin\left({45}^{\circ }\right)& cos\left({45}^{\circ }\right)\end{array}\right)$

A scaling of 2 in the X -direction is then performed by multiplying by

$\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$

We then need to rotate the axes back to their original position, that is, rotate the axes by −45°, this is done by multiplying by

$\left(\begin{array}{cc}cos\left(-{45}^{\circ }\right)& sin\left(-{45}^{\circ }\right)\\ -sin\left(-{45}^{\circ }\right)& cos\left(-{45}^{\circ }\right)\end{array}\right)=\left(\begin{array}{cc}cos\left({45}^{\circ }\right)& -sin\left({45}^{\circ }\right)\\ sin\left({45}^{\circ }\right)& cos\left({45}^{\circ }\right)\end{array}\right)$

Putting the three transformation matrices together we get

$\left(\begin{array}{cc}cos\left(-{45}^{\circ }\right)& sin\left(-{45}^{\circ }\right)\\ -sin\left(-{45}^{\circ }\right)& cos\left(-{45}^{\circ }\right)\end{array}\right)\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}cos\left({45}^{\circ }\right)& -sin\left({45}^{\circ }\right)\\ sin\left({45}^{\circ }\right)& cos\left({45}^{\circ }\right)\end{array}\right)$

which gives the matrix that represents a scaling along the line x = y.

Using cos $\left({45}^{\circ }\right)=\frac{1}{\sqrt{2}}=sin\left({45}^{\circ }\right),$ we get.

$\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)$

Taking out the two factors of $\frac{1}{\sqrt{2}}$ gives

$\frac{1}{2}\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)$

Multiplying the second two matrices gives

$\frac{1}{2}\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)\left(\begin{array}{cc}2& 2\\ -1& 1\end{array}\right)$

and multiplying out the remaining two matrices gives

$\frac{1}{2}\left(\begin{array}{cc}3& 1\\ 1& 3\end{array}\right)=\left(\begin{array}{cc}\frac{3}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{3}{2}\end{array}\right)$

We can now multiply the position vectors representing the vertices of the square

$\begin{array}{c}\left(\begin{array}{cc}\frac{3}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{3}{2}\end{array}\right)\left(\begin{array}{c}-\frac{1}{2}\\ -\frac{1}{2}\end{array}\right)=\left(\begin{array}{c}-1\\ -1\end{array}\right)\\ \left(\begin{array}{cc}\frac{3}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{3}{2}\end{array}\right)\left(\begin{array}{c}\frac{1}{2}\\ -\frac{1}{2}\end{array}\right)=\left(\begin{array}{c}\frac{1}{2}\\ -\frac{1}{2}\end{array}\right)\\ \left(\begin{array}{cc}\frac{3}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{3}{2}\end{array}\right)\left(\begin{array}{c}\frac{1}{2}\\ \frac{1}{2}\end{array}\right)=\left(\begin{array}{c}1\\ 1\end{array}\right)\\ \left(\begin{array}{cc}\frac{3}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{3}{2}\end{array}\right)\left(\begin{array}{c}-\frac{1}{2}\\ \frac{1}{2}\end{array}\right)=\left(\begin{array}{c}-\frac{1}{2}\\ \frac{1}{2}\end{array}\right)\end{array}$

The transformed figure is shown in Figure 13.11. We can see that has been stretched along the x = y direction but has not been scaled along the other diagonal. The image is no longer a square but a rhombus.

Example 13.18

Find a transformation that will rotate any point p about (1,1) through an angle of 90°.

Solution To rotate about a point not at the origin, we translate the origin temporarily, rotate, and then translate the origin back again.

Rotation through 90° is performed by multiplying by

$\left(\begin{array}{cc}cos\left({90}^{\circ }\right)& -sin\left({90}^{\circ }\right)\\ sin\left({90}^{\circ }\right)& cos\left({90}^{\circ }\right)\end{array}\right)=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$

The combined transformation on a point p can be represented by

${\text{p}}^{\prime }=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\text{p}-\left(\begin{array}{c}1\\ 1\end{array}\right)\right)+\left(\begin{array}{c}1\\ 1\end{array}\right).$

Example 13.19

Using Ohm's law and Kirchoff's laws for the electrical network in Figure 13.12, show that

$\begin{array}{ccccccc}{I}_1& -& I_2& -& I_3& =& 0\\ & & 3I_2& -& 2I_3& =& 0\\ 7I_1& & & +& 2I_3& =& 8\end{array}$

Solution Kirchoff's laws for an electrical network are as follows:

Kirchoff's voltage law (KVL): The sum of all the voltage drops around any closed loop is zero. This can also be expressed as: the voltage impressed on a closed loop is equal to the sum of the voltage drops in the rest of the loop.

Kirchoff's current law (KCL): At any point of a circuit, the sum of the in-flowing currents is equal to the sum of the out-flowing currents.

By Ohm's law we know the voltage drop across a resistor is given by V =IR, where R is the resistance of the resistor. Two loops have been identified in Figure 13.12 and by using KVL and Ohm's law in loop 1 we get

$3I_2-2I_3=0.$