Q 222 - Intermediate Algebra

Question

Solve each system of equations using a matrix.

$\left\{\begin{array}{l}2x+3y+z=12\\ x+y+z=9\\ 3x+4y+2z=20\end{array}\right.$

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,

$\left\{\begin{array}{l}2x+3y+z=12\\ x+y+z=9\\ 3x+4y+2z=20\end{array}\right.$

Step 2. Write in augmented form.

The augmented matrix for the given system of equations is

$\left[\begin{array}{cccr}2& 3& 1& 12\\ 1& 1& 1& 9\\ 3& 4& 2& 20\end{array}\right]$

Step 3. Apply row operations.

Apply $\frac{R_1}{2}\to R_1$ and $\frac{R_3}{-1}\to R_3$,

$\left[\begin{array}{cccr}1& \frac{3}{2}& \frac{1}{2}& 6\\ 0& -\frac{1}{2}& \frac{1}{2}& 3\\ 3& 4& 2& 20\end{array}\right]$

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

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

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

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

Apply $\frac{R_3}{-1}\to R_3$,

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

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.