1 Basic concepts

1.1 Determinants

A square matrix possesses an associated determinant. Unlike a matrix, which is an array of numbers, a determinant has a single value.

A two by two matrix $C=\left[\begin{array}{cc}{c}_{11}\hfill & c_{12}\hfill \\ c_{21}\hfill & c_{22}\hfill \end{array}\right]$ has an associated determinant

$det\; <!--nolimits\; -->\left(C\right)=\left|\begin{array}{cc}{c}_{11}\hfill & c_{12}\hfill \\ c_{21}\hfill & c_{22}\hfill \end{array}\right|=c_{11}{c}_{22}-c_{21}{c}_{12}$

(Note square or round brackets denote a matrix, straight vertical lines denote a determinant.)

A three by three matrix has an associated determinant

$det\; <!--nolimits\; -->\left(C\right)=\left|\begin{array}{ccc}\hfill c_{11}\hfill & \hfill c_{12}\hfill & \hfill c_{13}\hfill \\ \hfill c_{21}\hfill & \hfill c_{22}\hfill & \hfill c_{23}\hfill \\ \hfill c_{31}\hfill & \hfill c_{32}\hfill & \hfill c_{33}\hfill \end{array}\right|$

Among other ways this determinant can be evaluated by an “expansion about the top row”:

$det\; <!--nolimits\; -->\left(C\right)=c_{11}\left|\begin{array}{cc}{c}_{22}\hfill & c_{23}\hfill \\ c_{32}\hfill & c_{33}\hfill \end{array}\right|-c_{12}\left|\begin{array}{cc}{c}_{21}\hfill & c_{23}\hfill \\ c_{31}\hfill & c_{33}\hfill \end{array}\right|+c_{13}\left|\begin{array}{cc}{c}_{21}\hfill & c_{22}\hfill \\ c_{31}\hfill & c_{32}\hfill \end{array}\right|$

Note the minus sign in the second term.

Task!

Evaluate the determinants

$det\; <!--nolimits\; -->\left(A\right)=\left|\begin{array}{cc}\hfill 4\hfill & \hfill 6\hfill \\ \hfill 3\hfill & \hfill 1\hfill \end{array}\right|det\; <!--nolimits\; -->\left(B\right)=\left|\begin{array}{cc}\hfill 4\hfill & \hfill 8\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right|det\; <!--nolimits\; -->\left(C\right)=\left|\begin{array}{ccc}\hfill 6& \hfill 5& \hfill 4\hfill \\ \hfill 2& \hfill -1& \hfill 7\hfill \\ \hfill -3& \hfill 2& \hfill 0\hfill \end{array}\right|$

Answer

1.2 Systems of linear equations

We first recall some basic results in linear (matrix) algebra. Consider a system of $n$ equations in $n$ unknowns $x_1,x_2,\dots ,x_n$ :

$\begin{array}{ccccccccc}\hfill c_{11}{x}_1\hfill & \hfill +\hfill & \hfill c_{12}{x}_2\hfill & \hfill +\hfill & \hfill \dots \hfill & \hfill +\hfill & \hfill c_{1n}{x}_n\hfill & \hfill =\hfill & \hfill k_1\hfill \\ \hfill c_{21}{x}_1\hfill & \hfill +\hfill & \hfill c_{22}{x}_2\hfill & \hfill +\hfill & \hfill \dots \hfill & \hfill +\hfill & \hfill c_{2n}{x}_n\hfill & \hfill =\hfill & \hfill k_2\hfill \\ \hfill ⋮\hfill & \hfill +\hfill & \hfill ⋮\hfill & \hfill +\hfill & \hfill \dots \hfill & \hfill +\hfill & \hfill ⋮\hfill & \hfill =\hfill & \hfill ⋮\hfill \\ \hfill c_{n1}{x}_1\hfill & \hfill +\hfill & \hfill c_{n2}{x}_2\hfill & \hfill +\hfill & \hfill \dots \hfill & \hfill +\hfill & \hfill c_{nn}{x}_n\hfill & \hfill =\hfill & \hfill k_n\hfill \end{array}$

We can write such a system in matrix form:

$\left[\begin{array}{cccc}\hfill c_{11}\hfill & \hfill c_{12}\hfill & \hfill \dots \hfill & \hfill c_{1n}\hfill \\ \hfill c_{21}\hfill & \hfill c_{22}\hfill & \hfill \dots \hfill & \hfill c_{2n}\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill \dots \hfill & \hfill ⋮\hfill \\ \hfill c_{n1}\hfill & \hfill c_{n2}\hfill & \hfill \dots \hfill & \hfill c_{nn}\hfill \end{array}\right]\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \\ \hfill ⋮\hfill \\ \hfill x_n\hfill \end{array}\right]=\left[\begin{array}{c}\hfill k_1\hfill \\ \hfill k_2\hfill \\ \hfill ⋮\hfill \\ \hfill k_n\hfill \end{array}\right],\text{or equivalently}CX=K.$

We see that $C$ is an $n\times n$ matrix (called the coefficient matrix), $X={\left\{x_1,x_2,\dots ,x_n\right\}}^T$ is the $n\times 1$ column vector of unknowns and $K={\left\{k_1,k_2,\dots ,k_n\right\}}^T$ is an $n\times 1$ column vector of given constants.

The zero matrix will be denoted by $\underset{̲}{O}$ .

If $K\ne \underset{̲}{O}$ the system is called inhomogeneous ; if $K=\underset{̲}{O}$ the system is called homogeneous .

1.3 Basic results in linear algebra

Consider the system of equations $CX=K$ .

We are concerned with the nature of the solutions (if any) of this system. We shall see that this system only exhibits three solution types:

$\bullet$   The system is consistent and has a unique solution for $X$

$\bullet$   The system is consistent and has an infinite number of solutions for $X$

$\bullet$   The system is inconsistent and has no solution for $X$

consider:

$det\; <!--nolimits\; -->\left(C\right)\ne 0$ or $det\; <!--nolimits\; -->\left(C\right)=0$

Case 1 :  $det\; <!--nolimits\; -->\left(C\right)\ne 0$

In this case $C^{-1}$ exists and the unique solution to $CX=K$ is

$X=C^{-1}K$

Case 2 :  $det\; <!--nolimits\; -->\left(C\right)=0$

In this case $C^{-1}$ does not exist.

1. If $K\ne \underset{̲}{O}$ the system $CA=K$ has no solutions .
2. If $K=\underset{̲}{O}$ the system $CX=\underset{̲}{O}$ has an infinite number of solutions.

We note that a homogeneous system

$CX=\underset{̲}{O}$

has a unique solution $X=\underset{̲}{O}$ if $det\; <!--nolimits\; -->\left(C\right)\ne 0$ (this is called the trivial solution ) or an infinite number of solutions if $det\; <!--nolimits\; -->\left(C\right)=0$ .

Example 1

(Case 1) Solve the inhomogeneous system of equations

$x_1+x_2=1$ $2x_1+x_2=2$

which can be expressed as $CX=K$ where

$C=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right]X=\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \end{array}\right]K=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \end{array}\right]$

Solution

Here $det\; <!--nolimits\; -->\left(C\right)=-1\ne 0$ .

The system of equations has the unique solution : $X=\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right].$

Example 2

(Case 2a) Examine the following inhomogeneous system for solutions

$x_1+2x_2=1$

$3x_1+6x_2=0$

Solution

Here $det\; <!--nolimits\; -->\left(C\right)=\left|\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill 6\hfill \end{array}\right|=0.$   In this case there are no solutions.

To see this we see the first equation of the system states $x_1+2x_2=1$ whereas the second equation (after dividing through by 3) states $x_1+2x_2=0$ , a contradiction.

Example 3

(Case 2b) Solve the homogeneous system

$x_1+x_2=0$

$2x_1+2x_2=0$

Solution

Here $det\; <!--nolimits\; -->\left(C\right)=\left|\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 2\hfill \end{array}\right|=0.$ The solutions are any pairs of numbers $\left\{x_1,x_2\right\}$ such that $x_1=-x_2$ , i.e. $X=\left[\begin{array}{c}\hfill \alpha \\ \hfill -\alpha \end{array}\right]\text{ where}\alpha \text{ is arbitrary}.$

There are an infinite number of solutions .

1.4 A simple eigenvalue problem

We shall be interested in simultaneous equations of the form:

$AX=\lambda X,$

where $A$ is an $n\times n$ matrix, $X$ is an $n\times 1$ column vector and $\lambda$ is a scalar (a constant) and, in the first instance, we examine some simple examples to gain experience of solving problems of this type.

Example 4

Consider the following system with $n=2$ :

so that

$A=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 2\hfill \end{array}\right]\text{ and}X=\left[\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right].$

It appears that there are three unknowns $x,y,\lambda$ . The obvious questions to ask are: can we find $x,y$ ? what is $\lambda$ ?

Solution

To solve this problem we firstly re-arrange the equations (take all unknowns onto one side);

$\left(2-\lambda \right)x+3y=0$ (1)

$3x+\left(2-\lambda \right)y=0$ (2)

Therefore, from equation (2):

$x=-\frac{\left(2-\lambda \right)}{3}y.$ (3)

Then when we substitute this into (1)

$-\frac{{\left(2-\lambda \right)}^2}{3}y+3y=0$ which simplifies to $\left[-{\left(2-\lambda \right)}^2+9\right]y=0.$

We conclude that either $y=0$ or $9={\left(2-\lambda \right)}^2$ . There are thus two cases to consider:

Case 1

If $y=0$ then $x=0$ (from (3)) and we get the trivial solution . (We could have guessed this solution at the outset.)

Case 2

$9={\left(2-\lambda \right)}^2$

which gives, on taking square roots:

$\pm 3=2-\lambda \text{ giving}\lambda =2\pm 3$ so $\lambda =5\text{ or}\lambda =-1.$

Now, from equation (3), if $\lambda =5$ then $x=+y$ and if $\lambda =-1$ then $x=-y$ .

We have now completed the analysis. We have found values for $\lambda$ but we also see that we cannot obtain unique values for $x$ and $y$ : all we can find is the ratio between these quantities. This behaviour is typical, as we shall now see, of an eigenvalue problem.