Question

Solve the system of equations using a matrix.

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

Short answer

The solution of system of equation is $(5,7,4)$.

Step-by-step solution

Step 1. Given information.

Given linear equations:

$\begin{array}{l}-3x+y+z=-4\\ -x+2y-2z=1\\ 2x-y-z=-1\end{array}$

Step 2. Write the concept.

Let's write the equation in augmented form.

The augmented matrix is as follows:

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

Step 3. Calculation

Solve the augmented matrix

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

Let us interchange row 1 and row 2.

role="math" localid="1644679774023" $$\begin{gathered} \left[\begin{array}{cccc}-3& 1& 1& -4\\ -1& 2& -2& 1\\ 2& -1& -1& -1\end{array}\right]\stackrel{R_1\leftrightarrow R_2}{\to }\left[\begin{array}{cccc}-1& 2& -2& 1\\ -3& 1& 1& -4\\ 2& -1& -1& -1\end{array}\right] \\ \stackrel{-3R_1+R_2}{\to }\left[\begin{array}{cccc}-1& 2& -2& 1\\ 0& -5& 7& -7\\ 2& -1& -1& -1\end{array}\right] \\ \stackrel{2R_1+R_3}{\to }\left[\begin{array}{cccc}-1& 2& -2& 1\\ 0& -5& 7& -7\\ 0& -4& -5& 1\end{array}\right] \\ \stackrel{R_3\to 5R_2+7R_3}{\to }\left[\begin{array}{cccc}-1& 2& -2& 1\\ 0& -5& 7& -7\\ 0& -4& 0& -28\end{array}\right] \\ \stackrel{R_3\to \frac{1}{4}{R}_3}{\to }\left[\begin{array}{cccc}-1& 2& -2& 1\\ 0& -5& 7& -7\\ 0& 1& 0& 7\end{array}\right] \end{gathered}$$

Step 4. Echelon form

Matrix in echelon form is

$\left[\begin{array}{cccc}-1& 2& -2& 1\\ 0& -5& 7& -7\\ 0& 1& 0& 7\end{array}\right]$

Therefore the corresponding system of equation is

$$\begin{gathered} -x+2y-2z=1 \\ -5y+7z=-7 \\ y=7 \end{gathered}$$

Step 5. Substitution method for remaining values.

Put $y=7$in the equation

$$\begin{gathered} -5y+7z=-7 \\ -5\left(7\right)+7z=-7 \\ -35+7z=-7 \\ 7z=-7+35 \\ 7z=28 \\ z=4 \end{gathered}$$

Put $y=7,z=4$in the equation

$$\begin{gathered} -3x+y+z=-4 \\ -3x+7+4=-4 \\ -3x+11=-4 \\ -3x=-15 \\ x=5 \end{gathered}$$