Question

Solve each system of equations using a matrix:$\left\{\begin{array}{l}x-2y+2z=1\\ -2x+y-z=2\\ x-y+z=5\end{array}\right.$

Short answer

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

Step-by-step solution

Step 1. Given information.

Consider the given system of equations,

$\left\{\begin{array}{l}x-2y+2z=1\\ -2x+y-z=2\\ x-y+z=5\end{array}\right.$

Step 2. Write in augmented form.

The augmented matrix for the given system of equations is

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

Step 3. Apply row operations.

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

localid="1644521947631" $\left[\begin{array}{cccr}1& -2& 2& 1\\ 0& -3& 3& 4\\ 0& 1& -1& 4\end{array}\right]$

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

localid="1644522015605" $\left[\begin{array}{cccr}1& -2& 2& 1\\ 0& 1& -1& -\frac{4}{3}\\ 0& 0& 0& \frac{16}{3}\end{array}\right]$

Apply localid="1644522038148" $R_3\times \frac{3}{16}\to R_3$,

localid="1644522048942" $\left[\begin{array}{cccr}1& -2& 2& 1\\ 0& 1& -1& -\frac{4}{3}\\ 0& 0& 0& 1\end{array}\right]$

Now, the matrix is in row-echelon form.

Step 4. Write in system of equations.

Writing the corresponding system of equations,

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

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.