Q. 4.86 - Intermediate Algebra

Question

Solve each system of equations using a matrix: $\left\{\begin{array}{l}3x+4y-3z=-2\\ 2x+3y-z=-12\\ x+y-2z=6\end{array}\right.$

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,

$\left\{\begin{array}{l}3x+4y-3z=-2\\ 2x+3y-z=-12\\ x+y-2z=6\end{array}\right.$

Step 2. Write in augmented form.

The augmented matrix for the given system of equations is

$\left[\begin{array}{cccr}3& 4& -3& -2\\ 2& 3& -1& -12\\ 1& 1& -2& 6\end{array}\right]$

Step 3. Apply row operations.

Apply $\frac{R_1}{3}\to R_1$ and $R_2-2\times R_1\to R_2$,

$\left[\begin{array}{cccr}1& \frac{4}{3}& -1& -\frac{2}{3}\\ 0& \frac{1}{3}& 1& -\frac{32}{3}\\ 1& 1& -2& 6\end{array}\right]$

Apply $R_3-R_1\to R_3$ and $R_2\times 3\to R_2$,

$\left[\begin{array}{cccr}1& \frac{4}{3}& -1& -\frac{2}{3}\\ 0& 1& 3& -32\\ 0& -\frac{1}{3}& -1& \frac{20}{3}\end{array}\right]$

Apply $R_3+\frac{1}{3}\times R_2\to R_3$,

$\left[\begin{array}{cccr}1& \frac{4}{3}& -1& -\frac{2}{3}\\ 0& 1& 3& -32\\ 0& 0& 0& -4\end{array}\right]$

Apply $\frac{R_3}{-4}\to R_3$,

$\left[\begin{array}{cccr}1& \frac{4}{3}& -1& -\frac{2}{3}\\ 0& 1& 3& -32\\ 0& 0& 0& 1\end{array}\right]$

Now, the matrix is in row-echelon form.

Step 4. Write in system of equations.

Writing the corresponding system of equations,

$$\begin{gathered} x+\frac{4}{3}y-z=-\frac{2}{3} ...... \left(i\right) \\ y+3z=-32 ...... \left(ii\right) \\ 0=1 ...... \left(iii\right) \end{gathered}$$

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.