Q 223 - Intermediate Algebra

Question

Solve each system of equations using a matrix.

$\left\{\begin{array}{l}x+2y+6z=5\\ -x+y-2z=3\\ x-4y-2z=1\end{array}\right.$

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,

$\left\{\begin{array}{l}x+2y+6z=5\\ -x+y-2z=3\\ x-4y-2z=1\end{array}\right.$

Step 2. Write in augmented form.

The augmented matrix for the given system of equations is

$\left[\begin{array}{cccr}1& 2& 6& 5\\ -1& 1& -2& 3\\ 1& -4& -2& 1\end{array}\right]$

Step 3. Apply row operations.

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

$\left[\begin{array}{cccr}1& 2& 6& 5\\ 0& 3& 4& 8\\ 0& -6& -8& -4\end{array}\right]$

Apply $\frac{R_2}{3}\to R_2$ and $R_3+6\times R_2\to R_3$,

$\left[\begin{array}{cccr}1& 2& 6& 5\\ 0& 1& \frac{4}{3}& \frac{8}{3}\\ 0& 0& 0& 12\end{array}\right]$

Apply $\frac{R_3}{12}\to R_3$,

$\left[\begin{array}{cccr}1& 2& 6& 5\\ 0& 1& \frac{4}{3}& \frac{8}{3}\\ 0& 0& 0& 1\end{array}\right]$

Now, the matrix is in row-echelon form.

Step 4. Write in system of equations.

Writing the corresponding system of equations,

$$\begin{gathered} x+\frac{3}{2}y+\frac{1}{2}z=6 ...... \left(i\right) \\ y-z=-6 ...... \left(ii\right) \\ 0=1 ...... \left(iii\right) \end{gathered}$$

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.