Question

Solve each system of equations using a matrix.

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

Short answer

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

Step-by-step solution

Step 1. Given information.

Consider the given system of equations,

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

Step 2. Write in augmented form.

The augmented matrix for the given system of equations is

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

Step 3. Apply row operations.

Apply$\frac{R_1}{2}\to R_1$,

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

Apply $R_3-4\times R_1\to R_3$,

$\left[\begin{array}{rrrr}1& \frac{5}{2}& 0& 2\\ 0& 3& -1& 3\\ 0& -10& 3& -11\end{array}\right]$

Apply $\frac{R_2}{3}\to R_2$,

$\left[\begin{array}{rrrr}1& \frac{5}{2}& 0& 2\\ 0& 1& -\frac{1}{3}& 1\\ 0& -10& 3& -11\end{array}\right]$

Step 4. Again apply row operations.

Apply $R_3+10\times R_2\to R_3$,

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

Apply $-3\times R_3\to R_3$,

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

Now, the matrix is in row-echelon form.

Step 5. Write in system of equations. 

Writing the corresponding system of equations,

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

Substitute $z=3$in equation (ii),

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

Substitute $y=2$in equation (i),

localid="1644429836175" $$\begin{gathered} x+\frac{5}{2}\times 2=2 \\ x=2-5 \\ x=-3 \end{gathered}$$

Step 6. Check the answers.

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

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

This is true.

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

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

This is also true.