Question

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

Matrices
ANSWERED
asked 2021-02-02

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

Answers (1)

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 \((x,y,z)=(6,-4,2)\)

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