Question

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

$\left|\begin{array}{c}3x+5y+3z=25\\ 7X+9y+19z=65\\ 4X+5y+11z=5\end{array}\right|$

Short answer

For the given system of equation there is unique solution. $X=4,y=2.z=1$.

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.

$\left|\begin{array}{c}3X+5y+3z=25\\ 7X+9y+19z=65\\ -4X+5y+11z=5\end{array}\right|$

Corresponding augmented matrix

$\left[\begin{array}{ccc}3& 5& 3\\ 7& 9& 19\\ -4& 5& 11\end{array}\overline{)\begin{array}{c}25\\ 65\\ 5\end{array}}\right]$

Solving the Augmented matrix

$\left[\begin{array}{ccc}3& 5& 3\\ 7& 9& 19\\ -4& 5& 11\end{array}\overline{)\begin{array}{c}25\\ 65\\ 5\end{array}}\right]$

$$\begin{gathered} R_1\leftrightarrow R_3 \\ {R}_1\to \left(-1\right)R_1-R_3 \\ \left[\begin{array}{ccc}1& -10& -14\\ 7& 9& 19\\ 3& 5& 3\end{array}\overline{)\begin{array}{c}30\\ 65\\ 25\end{array}}\right] \end{gathered}$$

$$\begin{gathered} R_2\to R_2-7R_1 \\ {R}_3\to R_3-3R_1 \\ \left[\begin{array}{ccc}1& -10& -14\\ 0& 79& 117\\ 0& 35& 45\end{array}\overline{)\begin{array}{c}-30\\ 275\\ 115\end{array}}\right] \\ {R}_2\to \frac{1}{79}{R}_2 \\ {R}_3\to \frac{1}{5}{R}_3 \\ \left[\begin{array}{ccc}1& -10& -14\\ 0& 1& \frac{117}{79}\\ 0& 7& 9\end{array}\overline{)\begin{array}{c}-30\\ 79\\ 23\end{array}}\right] \\ {R}_1\to R_1+10R_2 \\ {R}_3\to R_3-7R_2 \\ \left[\begin{array}{ccc}1& 0& \frac{64}{79}\\ 0& 1& \frac{117}{79}\\ 0& 0& \frac{-108}{79}\end{array}\overline{)\begin{array}{c}\frac{380}{79}\\ \frac{275}{79}\\ \frac{-108}{79}\end{array}}\right] \end{gathered}$$

$$\begin{gathered} R_3\to \frac{-79}{108}{R}_3 \\ {R}_1\to R_1-\frac{64}{79}{R}_2 \\ {R}_2\to R_2-\frac{275}{79}{R}_2 \\ \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\overline{)\begin{array}{c}4\\ 2\\ 1\end{array}}\right] \end{gathered}$$

 Step 3: Value of variables

Corresponding system of equations will be

$\left|\begin{array}{c}x=4\\ y=2\\ z=1\end{array}\right|$

Hence, the solution of equation will be $x=4,y=2,z=1$

Final answer

For the given system of equation there is unique solution $x=4,y=2,z=1$.