Q. 4.85 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The system of linear equations doesn't have any 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 $R_3-R_1\to R_3$,

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

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

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

Apply $R_3\times \frac{3}{16}\to R_3$,

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

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.