Q. 383 - Intermediate Algebra

Question

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

$\left\{\begin{array}{l}2x-y+3z=-3\\ -x+2y-z=10\\ x+y+z=5\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the given system of equations is: 

$$\begin{gathered} x=-2 \\ y=5 \\ z=2 \end{gathered}$$

Step 1. Given information

The given system of equations is: 

$\left\{\begin{array}{l}2x-y+3z=-3\\ -x+2y-z=10\\ x+y+z=5\end{array}\right.$

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

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

$\left[\begin{array}{rrrr}2& -1& 3& -3\\ -1& 2& -1& 10\\ 1& 1& 1& 5\end{array}\right]$

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

Interchanging the row $1$ and $3$, we get:

$\left[\begin{array}{rrrr}2& -1& 3& -3\\ -1& 2& -1& 10\\ 1& 1& 1& 5\end{array}\right]\stackrel{R_1\leftrightarrow R_3}{⟶}\left[\begin{array}{rrrr}1& 1& 1& 5\\ -1& 2& -1& 10\\ 2& -1& 3& -3\end{array}\right]$ Add row $1$ and row $2$, we get:

$\left[\begin{array}{rrrr}1& 1& 1& 5\\ -1& 2& -1& 10\\ 2& -1& 3& -3\end{array}\right]\stackrel{R_1+R_2}{⟶}\left[\begin{array}{rrrr}1& 1& 1& 5\\ 0& 3& 0& 15\\ 2& -1& 3& -3\end{array}\right]$

Multiply row $1$ by $-2$ and then add to row $3$, we get:

$\left[\begin{array}{rrrr}1& 1& 1& 5\\ 0& 3& 0& 15\\ 2& -1& 3& -3\end{array}\right]\stackrel{-2R_1+R_3}{\to }\left[\begin{array}{rrrr}1& 1& 1& 5\\ 0& 3& 0& 15\\ 2& -3& 1& -13\end{array}\right]$

Dividing row $2$ by $3$, we get:

$\left[\begin{array}{rrrr}1& 1& 1& 5\\ 0& 3& 0& 15\\ 2& -3& 1& -13\end{array}\right]\stackrel{\frac{R_2}{3}}{\to }\left[\begin{array}{rrrr}1& 1& 1& 5\\ 0& 1& 0& 5\\ 0& -3& \text{'}1& -13\end{array}\right]$

Step 4. Compare the equation by echelon form.

The echelon form of the matrix is: 

$\left[\begin{array}{rrrr}1& 1& 1& 5\\ 0& 1& 0& 5\\ 0& -3& 1& -13\end{array}\right]$

Equation from the matrix is given as:

$$\begin{gathered} x+y+z=5 \\ y=5 \\ -3y+z=-13 \end{gathered}$$

Step 5. Find the value of x   &   z

Substituting the value $y=5$ in the equation $-3y+z=-13$:

$$\begin{gathered} -3\left(5\right)+z=-13 \\ -15+z=-13 \\ z=-13+15 \\ z=2 \end{gathered}$$

Put $$\begin{gathered} y=5 \\ z=2 \end{gathered}$$ in the equation

$x+y+z=5$:

$$\begin{gathered} x+5+2=5 \\ x+7=5 \\ x=5-7 \\ x=-2 \end{gathered}$$

Thus, the solution to the given system of equations is :

$$\begin{gathered} x=-2 \\ y=5 \\ z=2 \end{gathered}$$

Step 6. Writing solution in ordered traid

The solution of the system in the ordered triad is $(-2,5,2)$

Step 7. Check the solution.

Substitute the ordered triad $\left(-2,5,2\right)$ in the equation $2x - y + 3z = -3$:

$$\begin{gathered} 2x - y + 3z = -3 \\ 2\left(-2\right)-5+3\left(2\right)=-3 \\ -3=-3 \end{gathered}$$

This is true

Substituting the ordered triad $\left(-2,5,2\right)$ in the equation $-x+2y-z$:

 $$\begin{gathered} -x + 2y - z = 10 \\ -(-2)+2\left(5\right)-2=10 \\ 10=10 \end{gathered}$$

Substituting the ordered triad $\left(-2,5,2\right)$ in the equation $x + y + z = 5$:

$$\begin{gathered} x + y + z = 5 \\ -2+5+2=5 \\ 5=5 \end{gathered}$$

This is also true.

Thus the ordered triad $\left(-2,5,2\right)$ is the solution of the system of equations.