1 Solving three equations in three unknowns

The easiest set of three simultaneous linear equations to solve is of the following type:

$3x_1=6$ ,

$2x_2=5$ ,

$4x_3=7$

which obviously has solution ${\left[x_1,x_2,x_3\right]}^T={\left[2,\frac{5}{2},\frac{7}{4}\right]}^T$ or $x_1=2,x_2=\frac{5}{2},x_3=\frac{7}{4}$ .

In matrix form $AX=B$ the equations are

$\left[\begin{array}{ccc}\hfill 3\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 4\hfill \end{array}\right]\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \\ \hfill x_3\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 6\hfill \\ \hfill 5\hfill \\ \hfill 7\hfill \end{array}\right]$

where the matrix of coefficients, $A$ , is clearly diagonal.

Task!

Solve the equations

$\left[\begin{array}{ccc}\hfill 2& \hfill 0& \hfill 0\\ \hfill 0& \hfill -1& \hfill 0\\ \hfill 0& \hfill 0& \hfill 3\end{array}\right]\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \\ \hfill x_3\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 8\\ \hfill 2\\ \hfill -6\end{array}\right].$

Answer

The next easiest system of equations to solve is of the following kind:

The last equation can be solved immediately to give $x_3=2$ .

Substituting this value of $x_3$ into the second equation gives $2x_2+2=12$ from which $2x_2=10$ so that $x_2=5$

Substituting these values of $x_2$ and $x_3$ into the first equation gives $3x_1+5-2=0$ from which $3x_1=-3$ so that $x_1=-1$

Hence the solution is ${\left[x_1,x_2,x_3\right]}^T={\left[-1,5,2\right]}^T.$

This process of solution is called back-substitution .

In matrix form the system of equations is

$\left[\begin{array}{ccc}\hfill 3& \hfill 1& \hfill -1\\ \hfill 0& \hfill 2& \hfill 1\\ \hfill 0& \hfill 0& \hfill 3\end{array}\right]\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \\ \hfill x_3\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 0\\ \hfill 12\\ \hfill 6\end{array}\right].$

The matrix of coefficients is said to be upper triangular because all elements below the leading diagonal are zero. Any system of equations in which the coefficient matrix is triangular (whether upper or lower) will be particularly easy to solve.

Task!

Solve the following system of equations by back-substitution.

$\left[\begin{array}{ccc}\hfill 2& \hfill -1& \hfill 3\\ \hfill 0& \hfill 3& \hfill -1\\ \hfill 0& \hfill 0& \hfill 2\end{array}\right]\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \\ \hfill x_3\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 7\hfill \\ \hfill 5\hfill \\ \hfill 2\hfill \end{array}\right].$

Write the equations in expanded form:

Answer

Answer

Using this value for $x_3$ , obtain $x_2$ and $x_1$ :

Answer

Although we have worked so far with integers this will not always be the case and fractions will enter the solution process. We must then take care and it is always wise to check that the equations balance using the calculated solution.