Question

Solve each system of equations using a matrix.

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

Short answer

The system of linear equations doesn't have any solution.

Step-by-step solution

Step 1. Given information.

Step 1. Given information.

Consider the given system of equations,

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

Step 2. Write in augmented form.

The augmented matrix for the given system of equations is

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

Step 3. Apply row operations.

Apply $R_2-R_1\to R_2$and $R_3-2\times R_1\to R_3$,

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

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

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

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

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

Now, the matrix is in row-echelon form.

Step 4. Write in system of equations.

Writing the corresponding system of equations,

$$\begin{gathered} x+2y-3z=-1......\left(i\right) \\ y-\frac{4}{5}z=-\frac{2}{5}......\left(ii\right) \\ 0=1......\left(iii\right) \end{gathered}$$

As equation (iii) is a false statement.

Therefore, it is not possible to solve and is an inconsistent system.

Hence, the system of linear equations has no solution.