# Chapter 8 - Section 8.1 - Matrix Solutions to Linear Systems - Exercise Set - Page 893: 28

The solution of the given linear system of equations is $\left\{ \left( 0,2,2 \right) \right\}$.

#### Work Step by Step

A system of linear equations can be solved with the help of matrices. One of the methods (Gaussian elimination) involves the augmented matrix. The augment matrix is, $\left[ \begin{matrix} 3 & 1 & -1 & 0 \\ 1 & 1 & 2 & 6 \\ 2 & 2 & 3 & 10 \\ \end{matrix} \right]$ The Gaussian elimination method is used to solve the system of linear equations by turning the augmented matrix into row echelon form. Perform the operations as follows: 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 value of $y,z$ in (I) to get, \begin{align} & x-2-8=-10 \\ & x=0 \end{align} Hence, the solution of the given linear system of equations is $\left\{ \left( 0,2,2 \right) \right\}$.

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