Question

Solve the linear system of equations. You may use technology.

$\left|\begin{array}{c}3X+6y+14z=22\\ 7X+14y+30z=46\\ 4X+8y+7z=6\end{array}\right|$

Short answer

For the given system of equation there is unique solution. $Y=-6,y=2,z=2$.

Step-by-step solution

Augmented matrix

Solving system of equations using gauss Jordan’s elimination method.

First of all we will make the augmented matrix to the corresponding system of equations.

$\left|\begin{array}{c}3X+6y+14Z=22\\ 7X+14y+30z=46\\ 4X+8y+7Z=6\end{array}\right|$

Corresponding augmented matrix

$\left[\begin{array}{ccc}3& 6& 14\\ 7& 14& 30\\ 4& 8& 7\end{array}\overline{)\begin{array}{c}22\\ 46\\ 6\end{array}}\right]$

Solving the Augmented matrix

$$\begin{gathered} \left[\begin{array}{ccc}3& 6& 14\\ 7& 14& 30\\ 4& 8& 7\end{array}\overline{)\begin{array}{c}22\\ 46\\ 6\end{array}}\right] \\ {R}_1\leftrightarrow R_3 \\ {R}_1\to R_1-R_3 \\ \left[\begin{array}{ccc}1& 2& -7\\ 7& 14& 30\\ 3& 6& 14\end{array}\overline{)\begin{array}{c}-16\\ 46\\ 22\end{array}}\right] \end{gathered}$$

$$\begin{gathered} R_2\to R_2-7R_1 \\ {R}_3\to R_3-3R_1 \\ \left[\begin{array}{ccc}1& 2& -7\\ 0& 0& 79\\ 0& 0& 35\end{array}\overline{)\begin{array}{c}-16\\ 2\\ 2\end{array}}\right] \\ {R}_2\to \frac{1}{79}{R}_2 \\ {R}_3\to \frac{1}{35}-R_3 \\ \left[\begin{array}{ccc}1& 2& -7\\ 0& 0& 1\\ 0& 0& 1\end{array}\overline{)\begin{array}{c}-16\\ 2\\ 2\end{array}}\right] \end{gathered}$$

 Step 3:Value of variables

Corresponding system of equations will be

$\left|\begin{array}{c}X=-2y+7Z-16\\ y=2\\ Z=2\end{array}\right|$

Hence, the solution of equation will be$x=-6,y=2z=2$ .

Final answer

For the given system of equation there is unique solution $x=-6,y=2z=2$.