Q 220 - Intermediate Algebra

Question

Solve each system of equations using a matrix.

$\left\{\begin{array}{l}2y+3z=-1\\ 5x+3y=-6\\ 7x+z=1\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of linear equation is $\left(0,-2,1\right)$.

Step 1. Given information.

Consider the given system of equations,

$\left\{\begin{array}{l}2y+3z=-1\\ 5x+3y=-6\\ 7x+z=1\end{array}\right.$

Step 2. Write in augmented form.

The augmented matrix for the given system of equations is

$\left[\begin{array}{cccr}0& 2& 3& -1\\ 5& 3& 0& -6\\ 7& 0& 1& 1\end{array}\right]$

Step 3. Apply row operations.

Interchanging rows $R_1 \mathrm{and} R_2$,

$\left[\begin{array}{cccr}5& 3& 0& -6\\ 0& 2& 3& -1\\ 7& 0& 1& 1\end{array}\right]$

Apply $\frac{R_1}{5}\to R_1$ and $R_3-7\times R_1\to R_3$,

$\left[\begin{array}{cccr}1& \frac{3}{5}& 0& -\frac{6}{5}\\ 0& 2& 3& -1\\ 0& -\frac{21}{5}& 1& \frac{47}{5}\end{array}\right]$

Apply $\frac{R_2}{2}\to R_2$ and $R_3+\frac{21}{5}\times R_2\to R_3$,

$\left[\begin{array}{cccr}1& \frac{3}{5}& 0& -\frac{6}{5}\\ 0& 1& \frac{3}{2}& -\frac{1}{2}\\ 0& 0& \frac{73}{10}& \frac{73}{10}\end{array}\right]$

Apply $R_3\times \frac{10}{73}\to R_3$,

$\left[\begin{array}{cccr}1& \frac{3}{5}& 0& -\frac{6}{5}\\ 0& 1& \frac{3}{2}& -\frac{1}{2}\\ 0& 0& 1& 1\end{array}\right]$

Now, the matrix is in row-echelon form.

Step 4. Write in system of equations.

Writing the corresponding system of equations,

$$\begin{gathered} x+\frac{3}{5}y=-\frac{6}{5} ...... \left(i\right) \\ y+\frac{3}{2}z=-\frac{1}{2} ...... \left(ii\right) \\ z=1 ...... \left(iii\right) \end{gathered}$$

Substitute $z=1$ in equation (ii),

$$\begin{gathered} y+\frac{3}{2}\times 1=-\frac{1}{2} \\ y=-\frac{3}{2}-\frac{1}{2} \\ y=-\frac{4}{2} \\ y=-2 \end{gathered}$$

Substitute $y=-2$ in equation (i),

$$\begin{gathered} x+\frac{3}{5}\times \left(-2\right)=-\frac{6}{5} \\ x-\frac{6}{5}=-\frac{6}{5} \\ x=0 \end{gathered}$$

Step 5. Check the answers.

Substitute the values $x,y$ in equation (i),

$$\begin{gathered} 0+\frac{3}{5}\left(-2\right)=-\frac{6}{5} \\ 0-\frac{6}{5}=-\frac{6}{5} \\ -\frac{6}{5}=-\frac{6}{5} \end{gathered}$$

This is true.

Substitute the values $y,z$ in equation (ii),

$$\begin{gathered} -2+\frac{3}{2}\left(1\right)=-\frac{1}{2} \\ -2+\frac{3}{2}=-\frac{1}{2} \\ -\frac{1}{2}=-\frac{1}{2} \end{gathered}$$

This is also true.