Q 225 - Intermediate Algebra

Question

Solve each system of equations using a matrix.

$\left\{\begin{array}{l}4x-3y+2z=0\\ -2x+3y-7z=1\\ 2x-2y+3z=6\end{array}\right.$

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,

$\left\{\begin{array}{l}4x-3y+2z=0\\ -2x+3y-7z=1\\ 2x-2y+3z=6\end{array}\right.$

Step 2. Write in augmented form.

The augmented matrix for the given system of equations is

$\left[\begin{array}{cccr}4& -3& 2& 0\\ -2& 3& -7& 1\\ 2& -2& 3& 6\end{array}\right]$

Step 3. Apply row operations.

Apply $\frac{R_1}{4}\to R_1$ and $R_2+2\times R_1\to R_2$,

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

Apply $R_3-2\times R_1\to R_3$ and $R_2\to R_2\times \frac{2}{3}$,

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

Apply $R_3+\frac{1}{2}\times R_2\to R_3$,

$\left[\begin{array}{cccr}1& -\frac{3}{4}& \frac{1}{2}& 0\\ 0& 1& -4& \frac{2}{3}\\ 0& 0& 0& \frac{19}{3}\end{array}\right]$

Apply $R_3\times \frac{3}{19}\to R_3$,

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

Now, the matrix is in row-echelon form.

Step 4. Write in system of equations.

Writing the corresponding system of equations,

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

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.