Question

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

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

Short answer

The solution of the system of equations is$(-2,5,2)$

Step-by-step solution

Step 1. Given

A system of linear equations is given as-

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

Step 2. Concept Used

First, we will convert the given system of linear equations into its argument matrix. After that we will apply row transformation accordingly.

And we will to make zeroes in the matrix at least two in one row and one in other row through which we can find the value of one of the variable. And then we will find the value of other variable by using value of first one.

Step 3. Calculation

A system of linear equations is given as-

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

Correspond argument matrix is

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

Apply row transformations as-

$$\begin{gathered} \left[\begin{array}{rrrr}2& -1& 3& -3\\ -1& 2& -1& 10\\ 1& 1& 1& 5\end{array}\right]\stackrel{R_1\leftrightarrow R_3}{⟶}\left[\begin{array}{rrrr}1& 1& 1& 5\\ -1& 2& -1& 10\\ 2& -1& 3& -3\end{array}\right] \\ \stackrel{-2R_1+{\mathit{R}}_3}{⟶}\left[\begin{array}{rrrr}1& 1& 1& 5\\ 0& 3& 0& 15\\ 2& -1& 3& -3\end{array}\right] \\ \stackrel{\frac{1}{3}{R}_2}{\to }\left[\begin{array}{rrrr}1& 1& 1& 5\\ 0& 1& 0& 5\\ 0& -3& 1& -13\end{array}\right] \end{gathered}$$

Step 4. Further Calculation

Accordingly, the equations are-

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

Use substitution,

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

For the value of $x$,

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

Step 5. Verification

Verify the equation,

$$\begin{gathered} 2x-y+3z=-3 \\ 2(-2)-5+3\left(2\right)=-3 \\ -4-5+6=-3 \\ -3=-3 \end{gathered}$$

LHS is equal to RHS