2 The general system of three simultaneous linear equations

In the previous subsection we met systems of equations which could be solved by back-substitution alone. In this Section we meet systems which are not so amenable and where preliminary work must be done before back-substitution can be used.

Consider the system

We will use the solution method known as Gauss elimination , which has three stages. In the first stage the equations are written in matrix form. In the second stage the matrix equations are replaced by a system of equations having the same solution but which are in triangular form . In the final stage the new system is solved by back-substitution .

Stage 1 : Matrix Formulation

The first step is to write the equations in matrix form:

$\left[\begin{array}{ccc}\hfill 1& \hfill 3& \hfill 5\\ \hfill 2& \hfill -1& \hfill -3\\ \hfill 4& \hfill 5& \hfill -1\end{array}\right]\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \\ \hfill x_3\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 14\\ \hfill 3\\ \hfill 7\end{array}\right].$

Then, for conciseness, we combine the matrix of coefficients with the column vector of right-hand sides to produce the augmented matrix :

$\left[\begin{array}{cccc}\hfill 1& \hfill 3& \hfill 5& \hfill 14\\ \hfill 2& \hfill -1& \hfill -3& \hfill 3\\ \hfill 4& \hfill 5& \hfill -1& \hfill 7\end{array}\right]$

If the general system of equations is written $AX=B$  then the augmented matrix is written $\left[A|B\right].$

Hence the first equation

$x_1+3x_2+5x_3=14$

is replaced by the first row of the augmented matrix,

$135|14$ and so on.

Stage 1 has now been completed. We will next triangularise the matrix of coefficients by means of row operations . There are three possible row operations:

$\bullet$ interchange two rows;

$\bullet$ multiply or divide a row by a non-zero constant factor;

$\bullet$ add to, or subtract from, one row a multiple of another row.

Note that interchanging two rows of the augmented matrix is equivalent to interchanging the two corresponding equations. The shorthand notation we use is introduced by example. To interchange row 1 and row 3 we write $R1\leftrightarrow R3$ . To divide row 2 by 5 we write $R2÷5$ . To add three times row 1 to row 2, we write $R2+3R1$ . In the Task which follows you will see where these annotations are placed.

Note that these operations neither create nor destroy solutions so that at every step the system of equations has the same solution as the original system.

Stage 2 : Triangularisation

The second stage proceeds by first eliminating $x_1$ from the second and third equations using row operations.

$\left[\begin{array}{cccc}\hfill 1& \hfill 3& \hfill 5& \hfill 14\\ \hfill 2& \hfill -1& \hfill -3& \hfill 3\\ \hfill 4& \hfill 5& \hfill -1& \hfill 7\end{array}\right]\begin{array}{c}\hfill \hfill \\ \hfill R2-2\times R1\hfill \\ \hfill R3-4\times R1\hfill \end{array}\Rightarrow \left[\begin{array}{cccc}\hfill 1& \hfill 3& \hfill 5& \hfill 14\\ \hfill 0& \hfill -7& \hfill -13& \hfill -25\\ \hfill 0& \hfill -7& \hfill -21& \hfill -49\end{array}\right]$

In the above we have subtracted twice row (equation) 1 from row (equation) 2.

In full these operations would be written, respectively, as

$\left(2x_1-x_2-3x_3\right)-2\left(x_1+3x_2+5x_3\right)=3-2\times 14$ or $-7x_2-13x_3=-25$

and

$\left(4x_1+5x_2-x_3\right)-4\left(x_1+3x_2+5x_3\right)=7-4\times 14$ or $-7x_2-21x_3=-49.$

Now since all the elements in rows 2 and 3 are negative we multiply throughout by $-1$ :

$\left[\begin{array}{cccc}\hfill 1& \hfill 3& \hfill 5& \hfill 14\\ \hfill 0& \hfill -7& \hfill -13& \hfill -25\\ \hfill 0& \hfill -7& \hfill -21& \hfill -49\end{array}\right]\begin{array}{c}\hfill \hfill \\ \hfill R2\times \left(-1\right)\hfill \\ \hfill R3\times \left(-1\right)\hfill \end{array}\Rightarrow \left[\begin{array}{cccc}\hfill 1& \hfill 3& \hfill 5& \hfill 14\\ \hfill 0& \hfill 7& \hfill 13& \hfill 25\\ \hfill 0& \hfill 7& \hfill 21& \hfill 49\end{array}\right]$

Finally, we eliminate $x_3$ from the third equation by subtracting equation 2 from equation 3

i.e. $R3-R2$ :

$\left[\begin{array}{cccc}\hfill 1& \hfill 3& \hfill 5& \hfill 14\\ \hfill 0& \hfill 7& \hfill 13& \hfill 25\\ \hfill 0& \hfill 7& \hfill 21& \hfill 49\end{array}\right]\begin{array}{c}\hfill \hfill \\ \hfill \hfill \\ \hfill R3-R2\hfill \end{array}\Rightarrow \left[\begin{array}{cccc}\hfill 1& \hfill 3& \hfill 5& \hfill 14\\ \hfill 0& \hfill 7& \hfill 13& \hfill 25\\ \hfill 0& \hfill 0& \hfill 8& \hfill 24\end{array}\right]$

The system is now in triangular form.

Stage 3 : Back Substitution

Here we solve the equations from bottom to top. At each step of the back substitution process we encounter equations which only have a single unknown and so can be easily solved.

Task!

Now complete the solution to the above system by back-substitution.

Answer

Task!

Solve

Write down the augmented matrix for this system and then interchange rows 1 and 3:

Answer

Now subtract suitable multiples of row 1 from row 2 and from row 3 to eliminate the $x_1$ coefficient from rows 2 and 3:

Answer

Now divide row 2 by 5 and add a suitable multiple of the result to row 3:

Answer

Now complete the solution using back-substitution:

Answer

You should check these values in the original equations to ensure that the equations balance.

Again we emphasise that we chose a particular set of procedures in Stage 1. This was chosen mainly to keep the arithmetic simple by delaying the introduction of fractions. Sometimes we are courageous and take fewer, harder steps.

An important point to note is that when in Stage 2 we wrote $R2-2R1$ against row 2; what we meant is that row 2 is replaced by the combination (row 2)- $2\times$ (row 1).

In general, the operation

$\text{row}i-\alpha \times \text{row}j$

means replace row $i$ by the combination

$\text{row }i-\alpha \times \text{row}j.$