Q 221 - Intermediate Algebra

Question

Solve each system of equations using a matrix.

$\left\{\begin{array}{l}3x-z=-3\\ 5y+2z=-6\\ 4x+3y=-8\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of linear equation is $\left(-2,0,-3\right)$.

Step 1. Given information.

Consider the given system of equations,

$\left\{\begin{array}{l}3x-z=-3\\ 5y+2z=-6\\ 4x+3y=-8\end{array}\right.$

Step 2. Write in augmented form.

The augmented matrix for the given system of equations is

$\left[\begin{array}{cccr}3& 0& -1& -3\\ 0& 5& 2& -6\\ 4& 3& 0& -8\end{array}\right]$

Step 3. Apply row operations.

Apply $\frac{R_1}{3}\to R_1$ and $R_3-4\times R_1\to R_3$,

$\left[\begin{array}{cccr}1& 0& -\frac{1}{3}& -1\\ 0& 5& 2& -6\\ 0& 3& \frac{4}{3}& -4\end{array}\right]$

Apply $\frac{R_2}{5}\to R_2$ and $R_3-3\times R_2\to R_3$,

$\left[\begin{array}{cccr}1& 0& -\frac{1}{3}& -1\\ 0& 1& \frac{2}{5}& -\frac{6}{5}\\ 0& 0& \frac{2}{15}& -\frac{2}{5}\end{array}\right]$

Apply $R_3\times \frac{15}{2}\to R_3$,

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

Now, the matrix is in row-echelon form.

Step 4. Write in system of equations.

Writing the corresponding system of equations,

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

Substitute $z=-3$ in equation (i),

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

Substitute $z=-3$ in equation (ii),

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

Step 5. Check the answers.

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

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

This is true.

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

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

This is also true.