Question

# Solve the system of equations (Use matrices.):x-2y+z = 16, 2x-y-z = 14, 3x+5y-4z =-10

Matrices

Solve the system of equations (Use matrices.):
$$x-2y+z = 16$$,
$$2x-y-z = 14$$,
$$3x+5y-4z =-10$$

2021-02-03

Step 1
Given,
$$x-2y+z = 16,$$
$$2x-y-z = 14,$$
$$3x+5y-4z = -10$$
Step 2
Consider the Augmented matrix is of the form $$AX=B$$
$$\begin{bmatrix}1 & -2&1 \\2 & -1&-1\\3&5&-4 \end{bmatrix}\begin{bmatrix}x \\y\\z \end{bmatrix}=\begin{bmatrix}16 \\14\\-10 \end{bmatrix}$$
Here,
$$A=\begin{bmatrix}1 & -2&1 \\2 & -1&-1\\3&5&-4 \end{bmatrix},X=\begin{bmatrix}x \\y\\z \end{bmatrix}\text{ and }B=\begin{bmatrix}16 \\14\\-10 \end{bmatrix}$$
Use Gauss Elimination method,
Consider,
$$[A/B]=\begin{bmatrix}1 & -2&1&16 \\2 & -1&-1&14\\3&5&-4&-10 \end{bmatrix}$$
$$R_2 \rightarrow R_2-2R_1$$
$$\sim \begin{bmatrix}1 & -2&1&16 \\0 & 3&-3&-18\\3&5&-4&-10 \end{bmatrix}$$
$$R_3 \rightarrow R_3-3R_1$$
$$\sim \begin{bmatrix}1 & -2&1&16 \\0 & 3&-3&-18\\0&11&-7&-58 \end{bmatrix}$$
$$R_2 \rightarrow \frac{1}{3}R_2$$
$$\sim \begin{bmatrix}1 & -2&1&16 \\0 & 1&-1&-6\\0&11&-7&-58 \end{bmatrix}$$
$$R_3 \rightarrow R_3-11R_2$$
$$\sim \begin{bmatrix}1 & -2&1&16 \\0 & 1&-1&-6\\0&0&4&8 \end{bmatrix}$$
$$R_3 \rightarrow \frac{1}{4}R_3$$
$$\sim \begin{bmatrix}1 & -2&1&16 \\0 & 1&-1&-6\\0&0&1&2 \end{bmatrix}$$
Step 3
The above matrix is in the row echelon form
By back substitution we get,
$$z=2 \dots(i)$$
$$y-z=−6 \dots (ii)$$
$$x-2y+z=16 \dots (iii)$$
Substitute the value of z in (ii) we get,
$$y-2=-6$$
$$y=-4$$
Substitute the value of y and z in (iii) we get,
$$x-2(-4)+2=16$$
$$x+8+2=16$$
$$x+10=16$$
$$x=16-10$$
$$x=6$$
Therefore the solution set is $$(x,y,z)=(6,-4,2)$$