Question #265802

Consider the following matrix equation.

( 210121013\begin{matrix} 2 & -1 & 0 \\ 1 & 2 & -1 \\ 0& 1&3 \end{matrix} ) ( x1x2x3\begin{matrix} x1 \\ x2 \\ x3 \end{matrix} ) = ( 213\begin{matrix} 2 \\ 1 \\ 3 \end{matrix} )

Solve the equation by using Thomas algorithm.


1
Expert's answer
2021-11-15T16:11:22-0500

Solution. From the first equation of the system


2x1x2=22x_1-x_2=2

Divide through by 2 get


x10.5x2=1.x_1 -0.5x_2=1.

We write the system of equations as


(10.50121013)(x1x2x3)=(113).\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


x1+2x2x3=1x_1 +2x_2-x_3=1

Replacing x_1 -0.5*x_2=1 get

1+2.5x2x3=12.5x2x3=01 +2.5x_2-x_3=1 \to 2.5x_2-x_3=0

Divide through by 2.5 get


x225x3=0.x_2-\frac{2}{5}x_3=0.

We write the system of equations as


(10.500125013)(x1x2x3)=(103).\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


x2+3x3=3x_2+3x_3=3

Replacing x225x3=0x_2-\frac{2}{5}x_3=0 get


175x3=3\frac {17}{5} x_3=3

Divide through by 175\frac {17}{5} get


x3=1517x_3= \frac{15}{17}

We write the system of equations as


(10.500125001)(x1x2x3)=(101517).\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

x3=1517x_3= \frac{15}{17}

x2=0+25x3=25×1517=617x_2=0+\frac{2}{5}x_3 = \frac{2}{5}\times \frac{15}{17}= \frac{6}{17}

x1=12x2+1=12×617+1=317+1=2017x_1 = \frac{1}{2}x_2+1=\frac{1}{2} \times \frac{6}{17} +1=\frac{3}{17} + 1 = \frac{20}{17}

Answer. x1=2017x_1 = \frac{20}{17}; x2=617x_2= \frac{6}{17}; x3=1517x_3= \frac{15}{17}.


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