Question

Consider the following matrix equation.

( $\begin{matrix} 2 & -1 & 0 \\ 1 & 2 & -1 \\ 0& 1&3 \end{matrix}$ ) ( $\begin{matrix} x1 \\ x2 \\ x3 \end{matrix}$ ) = ( $\begin{matrix} 2 \\ 1 \\ 3 \end{matrix}$ )

Solve the equation by using Thomas algorithm.

Answer 1

Solution. From the first equation of the system

$$2x_1-x_2=2$$

Divide through by 2 get

$$x_1 -0.5x_2=1.$$

We write the system of equations as

$$\begin{pmatrix} 1 & -0.5 & 0 \\ 1 & 2 & -1 \\ 0 & 1 & 3 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 3 \end{pmatrix} .$$

From the second equation of the system

$$x_1 +2x_2-x_3=1$$

Replacing x_1 -0.5*x_2=1 get

$$1 +2.5x_2-x_3=1 \to 2.5x_2-x_3=0$$

Divide through by 2.5 get

$$x_2-\frac{2}{5}x_3=0.$$

We write the system of equations as

$$\begin{pmatrix} 1 & -0.5 & 0 \\ 0 & 1 & -\frac{2}{5} \\ 0 & 1 & 3 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 3 \end{pmatrix} .$$

From the third equation of the system

$$x_2+3x_3=3$$

Replacing $x_2-\frac{2}{5}x_3=0$ get

$$\frac {17}{5} x_3=3$$

Divide through by $\frac {17}{5}$ get

$$x_3= \frac{15}{17}$$

We write the system of equations as

$$\begin{pmatrix} 1 & -0.5 & 0 \\ 0 & 1 & -\frac{2}{5} \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ \frac {15}{17} \end{pmatrix} .$$

Let us find the roots of the equations of the system

$$x_3= \frac{15}{17}$$

$$x_2=0+\frac{2}{5}x_3 = \frac{2}{5}\times \frac{15}{17}= \frac{6}{17}$$

$$x_1 = \frac{1}{2}x_2+1=\frac{1}{2} \times \frac{6}{17} +1=\frac{3}{17} + 1 = \frac{20}{17}$$

Answer. $x_1 = \frac{20}{17}$; $x_2= \frac{6}{17}$; $x_3= \frac{15}{17}$.