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shadsiei

Answered question

2021-02-02

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

Answer & Explanation

wheezym

Skilled2021-02-03Added 103 answers

Step 1
Given,
$x - 2 y + z = 16,$
$2 x - y - z = 14,$
$3 x + 5 y - 4 z = - 10$
Step 2
Consider the Augmented matrix is of the form $A X = B$
$[\begin{matrix} 1 & - 2 & 1 \\ 2 & - 1 & - 1 \\ 3 & 5 & - 4 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 16 \\ 14 \\ - 10 \end{matrix}]$
Here,
$A = [\begin{matrix} 1 & - 2 & 1 \\ 2 & - 1 & - 1 \\ 3 & 5 & - 4 \end{matrix}], X = [\begin{matrix} x \\ y \\ z \end{matrix}] and B = [\begin{matrix} 16 \\ 14 \\ - 10 \end{matrix}]$
Use Gauss Elimination method,
Consider,
$[A / B] = [\begin{matrix} 1 & - 2 & 1 & 16 \\ 2 & - 1 & - 1 & 14 \\ 3 & 5 & - 4 & - 10 \end{matrix}]$
$R_{2} \to R_{2} - 2 R_{1}$
$\sim [\begin{matrix} 1 & - 2 & 1 & 16 \\ 0 & 3 & - 3 & - 18 \\ 3 & 5 & - 4 & - 10 \end{matrix}]$
$R_{3} \to R_{3} - 3 R_{1}$
$\sim [\begin{matrix} 1 & - 2 & 1 & 16 \\ 0 & 3 & - 3 & - 18 \\ 0 & 11 & - 7 & - 58 \end{matrix}]$
$R_{2} \to \frac{1}{3} R_{2}$
$\sim [\begin{matrix} 1 & - 2 & 1 & 16 \\ 0 & 1 & - 1 & - 6 \\ 0 & 11 & - 7 & - 58 \end{matrix}]$
$R_{3} \to R_{3} - 11 R_{2}$
$\sim [\begin{matrix} 1 & - 2 & 1 & 16 \\ 0 & 1 & - 1 & - 6 \\ 0 & 0 & 4 & 8 \end{matrix}]$
$R_{3} \to \frac{1}{4} R_{3}$
$\sim [\begin{matrix} 1 & - 2 & 1 & 16 \\ 0 & 1 & - 1 & - 6 \\ 0 & 0 & 1 & 2 \end{matrix}]$
Step 3
The above matrix is in the row echelon form
By back substitution we get,
$z = 2 \dots (i)$
$y - z = - 6 \dots (i i)$
$x - 2 y + z = 16 \dots (i i i)$
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 $(x, y, z) = (6, - 4, 2)$

Jeffrey Jordon

Expert2022-01-29Added 2605 answers

Answer is given below (on video)

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