Q 218 - Intermediate Algebra

Question

Solve each system of equations using a matrix.

$\left\{\begin{array}{l}x+2z=0\\ 4y+3z=-2\\ 2x-5y=3\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of linear equation is $\left(4,1,-2\right)$.

Step 1. Given information.

Consider the given system of equations,

$\left\{\begin{array}{l}x+2z=0\\ 4y+3z=-2\\ 2x-5y=3\end{array}\right.$

Step 2. Write in augmented form.

The augmented matrix for the given system of equations is 

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

Step 3. Apply row operations.

Apply $R_3-2\times R_1\to R_3$,

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

Apply $\frac{R_2}{4}\to R_2$,

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

Apply $R_3+5\times R_2\to R_3$,

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

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

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

Now, the matrix is in row-echelon form.

Step 4. Write in system of equations.

Writing the corresponding system of equations,

$$\begin{gathered} x+2z=0 ...... \left(i\right) \\ y+\frac{3}{4}z=-\frac{1}{2} ...... \left(ii\right) \\ z=-2 ...... \left(iii\right) \end{gathered}$$

Substitute $z=-2$ in equation (i),

$$\begin{gathered} x+2\left(-2\right)=0 \\ x-4=0 \\ x=4 \end{gathered}$$

Substitute $z=-2$ in equation (ii),

$$\begin{gathered} y+\frac{3}{4}\left(-2\right)=-\frac{1}{2} \\ y-\frac{3}{2}=-\frac{1}{2} \\ y=\frac{3}{2}-\frac{1}{2} \\ y=\frac{2}{2} \\ y=1 \end{gathered}$$

Step 5. Check the answers.

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

$$\begin{gathered} 4+2\left(-2\right)=0 \\ 4-4=0 \\ 0=0 \end{gathered}$$

This is true.

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

$$\begin{gathered} 1+\frac{3}{4}\left(-2\right)=-\frac{1}{2} \\ 1-\frac{3}{2}=-\frac{1}{2} \\ -\frac{1}{2}=-\frac{1}{2} \end{gathered}$$

This is also true.