Chapter 8 - Section 8.1 - Matrix Solutions to Linear Systems - Exercise Set - Page 895: 54

It can be solved with the help of matrices by carrying out various row operations on the augmented matrix.

It can be solved with the help of matrices by carrying out various row operations on the augmented matrix. In the Gaussian Elimination method, first we write the augmented matrix from the provided system of equations. Then we reduce the matrix into row echelon form by using the row operations. The system of equations that is obtained from this method has equations in more than one variable and the solution can be obtained by back substitution. Example: Consider a system of linear equations, $\left\{ \begin{align} & 3x+y-z=0 \\ & x+y+2z=6 \\ & 2x+2y+3z=10 \end{align} \right.$ The augmented matrix for the above system of equations is written as: $\left[ \begin{matrix} 3 & 1 & -1 & 0 \\ 1 & 1 & 2 & 6 \\ 2 & 2 & 3 & 10 \\ \end{matrix} \right]$ Convert the augmented matrix into row-echelon form. Replace ${{R}_{1}}\,\text{by}\,{{R}_{1}}-{{R}_{3}}$ to get $\left[ \begin{matrix} 1 & -1 & -4 & -10 \\ 1 & 1 & 2 & 6 \\ 2 & 2 & 3 & 10 \\ \end{matrix} \right]$ Replace ${{R}_{2}}\,\text{by}\,{{R}_{2}}-{{R}_{1}}$ to get $\left[ \begin{matrix} 1 & -1 & -4 & -10 \\ 0 & 2 & 6 & 16 \\ 2 & 2 & 3 & 10 \\ \end{matrix} \right]$ Replace ${{R}_{2}}\,\text{by}\,\frac{1}{2}{{R}_{2}}$ to get $\left[ \begin{matrix} 1 & -1 & -4 & -10 \\ 0 & 1 & 3 & 8 \\ 2 & 2 & 3 & 10 \\ \end{matrix} \right]$ Replace ${{R}_{3}}\,\text{by}\,{{R}_{3}}-2{{R}_{1}}$ to get $\left[ \begin{matrix} 1 & -1 & -4 & -10 \\ 0 & 1 & 3 & 8 \\ 0 & 4 & 11 & 30 \\ \end{matrix} \right]$ Replace ${{R}_{3}}\,\text{by}\,{{R}_{3}}-4{{R}_{2}}$ to get $\left[ \begin{matrix} 1 & -1 & -4 & -10 \\ 0 & 1 & 3 & 8 \\ 0 & 0 & -1 & -2 \\ \end{matrix} \right]$ Replace ${{R}_{3}}\,\text{by}\,-{{R}_{3}}$ to get $\left[ \begin{matrix} 1 & -1 & -4 & -10 \\ 0 & 1 & 3 & 8 \\ 0 & 0 & 1 & 2 \\ \end{matrix} \right]$ The system of equations for the above augmented matrix is given by $x-y-4z=-10$ …… (I) $y+3z=8$ …… (II) $z=2$ …… (III) Substitute the value of $z$ in equation (II) to get $\begin{align} & y+6=8 \\ & y=2 \end{align}$ Substitute the value of $y,z$ in (I) to get $\begin{align} & x-2-8=-10 \\ & x=0 \end{align}$ Thus, the solution is $\left( 0,2,2 \right)$.