Q. 4.84 - Intermediate Algebra

Question

Solve the system of equations using a matrix:$\left\{\begin{array}{l}-3x+y+z=-4\\ -x+2y-2z=1\\ 2x-y-z=-1\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of linear equation is $\left(5,7,4\right)$.

Step 1. Given information.

Consider the given system of equations,

$\left\{\begin{array}{l}-3x+y+z=-4\\ -x+2y-2z=1\\ 2x-y-z=-1\end{array}\right.$

Step 2. Write in augmented form.

The augmented matrix for the given system of equations is

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

Step 3. Apply row operations.

Apply $\frac{R_1}{-3}\to R_1$ and $R_2+R_1\to R_2$,

$\left[\begin{array}{cccr}1& -\frac{1}{3}& -\frac{1}{3}& \frac{4}{3}\\ 0& \frac{5}{3}& -\frac{7}{3}& \frac{7}{3}\\ 2& -1& -1& -1\end{array}\right]$

Apply $R_3-2\times R_1\to R_3$ and $R_2\times \frac{3}{5}\to R_2$,

$\left[\begin{array}{cccr}1& -\frac{1}{3}& -\frac{1}{3}& \frac{4}{3}\\ 0& 1& -\frac{7}{5}& \frac{7}{5}\\ 0& -\frac{1}{3}& -\frac{1}{3}& -\frac{11}{3}\end{array}\right]$

Apply $R_3+\frac{1}{3}\times R_2\to R_3$,

$\left[\begin{array}{cccr}1& -\frac{1}{3}& -\frac{1}{3}& \frac{4}{3}\\ 0& 1& -\frac{7}{5}& \frac{7}{5}\\ 0& 0& -\frac{4}{5}& -\frac{16}{5}\end{array}\right]$

Apply $R_3\times \left(-\frac{5}{4}\right)\to R_3$,

$\left[\begin{array}{cccr}1& -\frac{1}{3}& -\frac{1}{3}& \frac{4}{3}\\ 0& 1& -\frac{7}{5}& \frac{7}{5}\\ 0& 0& 1& 4\end{array}\right]$

Now, the matrix is in row-echelon form.

Step 4. Write in system of equations.

Writing the corresponding system of equations,
$$\begin{gathered} x-\frac{1}{3}y-\frac{1}{3}z=\frac{4}{3} ...... \left(i\right) \\ y-\frac{7}{5}z=\frac{7}{5} ...... \left(ii\right) \\ z=4 ...... \left(iii\right) \end{gathered}$$

Substitute the value of $y$ in equation (ii),

$$\begin{gathered} y-\frac{7}{5}\times \left(4\right)=\frac{7}{5} \\ y=\frac{28}{5}+\frac{7}{5} \\ y=\frac{35}{5} \\ y=7 \end{gathered}$$

Substitute the value of $y,z$ in equation (i),

$$\begin{gathered} x-\frac{1}{3}\times \left(7\right)-\frac{1}{3}\times \left(4\right)=\frac{4}{3} \\ x-\frac{7}{3}-\frac{4}{3}=\frac{4}{3} \\ x=\frac{4}{3}+\frac{11}{3} \\ x=5 \end{gathered}$$

Step 5. Check the answers.

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

$$\begin{gathered} 5-\frac{1}{3}\left(7\right)-\frac{1}{3}\left(4\right)=\frac{4}{3} \\ 5-\frac{7}{3}-\frac{4}{3}=\frac{4}{3} \\ \frac{15-7-4}{3}=\frac{4}{3} \\ \frac{4}{3}=\frac{4}{3} \end{gathered}$$

This is true.

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

$$\begin{gathered} 7-\frac{7}{5}\left(4\right)=\frac{7}{5} \\ 7-\frac{28}{5}=\frac{7}{5} \\ \frac{7}{5}=\frac{7}{5} \end{gathered}$$

This is also true.