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Kaycee Roche

Answered question

2021-02-08

Use matrices to solve the given system of equations. Use back-substitution Gaussian elimination or Gauss-Jordan elimination.
${\begin{cases} x + 3 y = 0 \\ x + y + z = 1 \\ 3 x - y - z = 11 \end{cases}$

Answer & Explanation

cheekabooy

Skilled2021-02-09Added 83 answers

So,
${\begin{cases} x + 3 y = 0 \\ x + y + z = 1 \\ 3 x - y - z = 11 \end{cases}$
$A = [\begin{matrix} 1 & 3 & 0 \\ 1 & 1 & 1 \\ 3 & - 1 & - 1 \end{matrix}], B = [\begin{matrix} x \\ y \\ z \end{matrix}] and B = [\begin{matrix} 0 \\ 1 \\ 11 \end{matrix}]$
$[A | B] = [\begin{matrix} 1 & 3 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 3 & - 1 & - 1 & 11 \end{matrix}]$
$R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - 3 R_{1}$
$\Rightarrow [A | B] = [\begin{matrix} 1 & 3 & 0 & 0 \\ 0 & - 2 & 1 & 1 \\ 0 & - 10 & - 1 & 11 \end{matrix}]$
$R_{2} \to - \frac{1}{2} R_{2}$
$\Rightarrow [A | B] = [\begin{matrix} 1 & 3 & 0 & 0 \\ 0 & 1 & - \frac{1}{2} & - \frac{1}{2} \\ 0 & - 10 & - 1 & 11 \end{matrix}]$
$R_{3} \to R_{3} + 10 R_{2}$
$\Rightarrow [A | B] = [\begin{matrix} 1 & 3 & 0 & 0 \\ 0 & 1 & - \frac{1}{2} & - \frac{1}{2} \\ 0 & 0 & - 6 & 6 \end{matrix}]$
$R_{3} \to - \frac{1}{6} R_{3}$
$\Rightarrow [A | B] = [\begin{matrix} 1 & 3 & 0 & 0 \\ 0 & 1 & - \frac{1}{2} & - \frac{1}{2} \\ 0 & 0 & 1 & - 1 \end{matrix}]$
$x + 3 y = 0$
$y - \frac{1}{2} z = - \frac{1}{2}$
$z = - 1$
So,
$y - \frac{1}{2} (- 1) = - \frac{1}{2}$
$\Rightarrow y + \frac{1}{2} = - \frac{1}{2}$
$\Rightarrow y = - \frac{1}{2} - \frac{1}{2}$
$\Rightarrow y = - 1$
$x + 3 (- 1) = 0$
$\Rightarrow x - 3 = 0$
$\Rightarrow x = 3$
$x = 3, y = - 1 and z = - 1$

Jeffrey Jordon

Expert2022-01-27Added 2605 answers

Answer is given below (on video)

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