1 Solving two equations in two unknowns

If we have one linear equation

$ax=b$

in which the unknown is $x$ and $a$ and $b$ are constants then there are just three possibilities:

• $a\ne 0$ then $x=\frac{b}{a}=a^{-1}b$ . In this case the equation $ax=b$ has a unique solution for $x$ .
• $a=0$ , $b=0$ then the equation $ax=b$ becomes $0=0$ and any value of $x$ will do. In this case the equation $ax=b$ has infinitely many solutions .
• $a=0$ and $b\ne 0$ then $ax=b$ becomes $0=b$ which is a contradiction. In this case the equation $ax=b$ has no solution for $x$ .

What happens if we have more than one equation and more than one unknown? We shall find that the solutions to such systems can be characterised in a manner similar to that occurring for a single equation; that is, a system may have a unique solution, an infinity of solutions or no solution at all.

In this Section we examine a method, known as Cramer’s rule and employing determinants, for solving systems of linear equations.

Consider the equations

$ax+by=e$ (1)

$cx+dy=f$ (2)

where $a,b,c,d,e,f$ are given numbers. The variables $x$ and $y$ are unknowns we wish to find. The pairs of values of $x$ and $y$ which simultaneously satisfy both equations are called solutions. Simple algebra will eliminate the variable $y$ between these equations. We multiply equation (1) by $d$ , equation (2) by $b$ and subtract:

$\text{first,}\left(1\right)\times dadx+bdy=ed$

$\text{then},\left(2\right)\times bbcx+bdy=bf$

(we multiplied in this way to make the coefficients of $y$ equal.)

Now subtract to obtain

$\left(ad-bc\right)x=ed-bf$ (3)

Task!

Starting with equations (1) and (2) above, eliminate $x$ .

Answer

If we multiply this last equation in the Task above by $-1$ we obtain

$\left(ad-bc\right)y=af-ec$ (4)

Dividing equations (3) and (4) by $ad-bc$ we obtain the solutions

$x=\frac{ed-bf}{ad-bc},y=\frac{af-ec}{ad-bc}$ (5)

There is of course one proviso: if $ad-bc=0$ then neither $x$ nor $y$ has a defined value.

If we choose to express these solutions in terms of determinants we have the formulation for the solution of simultaneous equations known as Cramer’s rule .

If we define $\Delta$ as the determinant $\left|\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right|$ and provided $\Delta \ne 0$ then the unique solution of the equations

$ax+by=e$

$cx+dy=f$

is by (5) given by

$x=\frac{{\Delta }_x}{\Delta },y=\frac{{\Delta }_y}{\Delta }\text{where}{\Delta }_x=\left|\begin{array}{cc}\hfill e\hfill & \hfill b\hfill \\ \hfill f\hfill & \hfill d\hfill \end{array}\right|,{\Delta }_y=\left|\begin{array}{cc}\hfill a\hfill & \hfill e\hfill \\ \hfill c\hfill & \hfill f\hfill \end{array}\right|$

Now $\Delta$ is the determinant of coefficients on the left-hand sides of the equations. In the expression ${\Delta }_x$ the coefficients of $x$ (i.e. $\left(\begin{array}{c}\hfill a\hfill \\ \hfill c\hfill \end{array}\right)$ which is column 1 of $\Delta$ ) are replaced by the terms on the right-hand sides of the equations (i.e. by $\left(\begin{array}{c}\hfill e\hfill \\ \hfill f\hfill \end{array}\right)$ ). Similarly in ${\Delta }_y$ the coefficients of $y$ (column 2 of $\Delta$ ) are replaced by the terms on the right-hand sides of the equations.

Task!

Use Cramer’s rule to solve the simultaneous equations

Answer

Task!

Use Cramer’s rule to solve the equations

Answer

Notation

For ease of generalisation to larger systems we write the two-equation system in a different notation:

Here the unknowns are $x_1$ and $x_2$ , the right-hand sides are $b_1$ and $b_2$ and the coefficients are $a_{ij}$ where, for example, $a_{21}$ is the coefficient of $x_1$ in equation two. In general, $a_{ij}$ is the coefficient of $x_j$ in equation $i$ .

Cramer’s rule can then be stated as follows:

If $\left|\begin{array}{cc}\hfill a_{11}\hfill & \hfill a_{12}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill \end{array}\right|\ne 0$ , then the equations

have solution

$x_1=\frac{\left|\begin{array}{cc}\hfill b_1\hfill & \hfill a_{12}\hfill \\ \hfill b_2\hfill & \hfill a_{22}\hfill \end{array}\right|}{\left|\begin{array}{cc}\hfill a_{11}\hfill & \hfill a_{12}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill \end{array}\right|},x_2=\frac{\left|\begin{array}{cc}\hfill a_{11}\hfill & \hfill b_1\hfill \\ \hfill a_{21}\hfill & \hfill b_2\hfill \end{array}\right|}{\left|\begin{array}{cc}\hfill a_{11}\hfill & \hfill a_{12}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill \end{array}\right|}.$