Question

In the following exercise, solve each system of equations using a matrix.

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

Short answer

The given system of equations has no solution.

Step-by-step solution

Step 1. Given information  

The given system of equations is:

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

Step 2. Convert the system of equations into an augmented matrix.  

The augmented matrix of the given system of equations is given as:

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

In the augmented matrix, the first equation gives the first row and the second equation gives the second row. The vertical line replaces the equal signs.

Step 3. Convert the augmented matrix into echelon form   

Perform row operation $R_2\to R_1-R_2$

role="math" localid="1644579738873" $\left[\begin{array}{rrrr}1& 2& -3& -1\\ 1& -3& 1& 1\\ 2& -1& -2& 2\end{array}\right]\stackrel{R_2\to R_1-R_2}{⟶}\left[\begin{array}{rrrr}1& 2& -3& -1\\ 0& 5& -4& -2\\ 2& -1& -2& 2\end{array}\right]$

Performing row operation $R_3\to 2R_1-R_3$:

$\left[\begin{array}{rrrr}1& 2& -3& -1\\ 0& 5& -4& -2\\ 2& -1& -2& 2\end{array}\right]\stackrel{R_3\to 2R_1-R_3}{\to }\left[\begin{array}{lllr}1& 2& -3& -1\\ 0& 5& -4& -2\\ 0& 5& -4& -4\end{array}\right]$

Perform row operation $R_3\to R_2-R_3$

$\left[\begin{array}{lllr}1& 2& -3& -1\\ 0& 5& -4& -2\\ 0& 5& -4& -4\end{array}\right]\stackrel{R_1\to R_2-R_3}{\to }\left[\begin{array}{rrrr}1& 2& -3& -1\\ 0& 5& -4& -2\\ 0& 0& 0& 2\end{array}\right]$

Step 4. Compare the equation by echelon form.  

The echelon form of the matrix is:

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

The system of equation by the above matrix is:

$$\begin{gathered} x+2y-3z=-1 \\ 5y-4z=-2 \\ 0=2 \end{gathered}$$

But we know that $0\ne 2$

So,

This system of equation has no solution.