Question

Solve the system of equation by Gauss elimination method:

x + y + z = 7

3x +3y + 4z =24

2x + y + 3z = 16​

Answer 1

Given:

x + y + z = 7

3x + 3y + 4z = 24

2x + y + 3z = 16

To find:

Solve the system of equation by Gauss elimination method.

Solution:

x + y + z = 7

3x + 3y + 4z = 24

2x + y + 3z = 16

$\left[\begin{array}{ccc}1&1&1 \ | \ 7 \\3&3&4 \ \ | \ 24\\2&1&3 \ \ | \ 16\end{array}\right]$

R₂ - 3R₁ → R₂ ; R₃ - 2R₁ → R₃

$\left[\begin{array}{ccc}1&1&1 \ \ | \ 7 \\0&0&1 \ \ | \ 3\\0&-1&1 \ \ | \ 2\end{array}\right]$

R₂ ↔ R₃

$\left[\begin{array}{ccc}1&1&1 \ \ | \ 7 \\0&-1&1 \ \ | \ 2\\0&0&1 \ \ | \ 3\end{array}\right]$

R₂ / -1 → R₂

$\left[\begin{array}{ccc}1&1&1 \ | \ 7 \\0&1&-1 \ \ | \ -2\\0&0&1 \ | \ 3\end{array}\right]$

R₁ - R₂ → R₁

$\left[\begin{array}{ccc}1&0&2 \ | \ 9 \\0&1&-1 \ \ | \ -2\\0&0&1 \ | \ 3\end{array}\right]$

R₁ - 2R₃ → R₁ ; R₂ + R₃ → R₂

$\left[\begin{array}{ccc}1&0&0 \ | \ 3 \\0&1&0 \ | \ 1 \\0&0&1 \ | \ 3\end{array}\right]$

x = 3

y = 1

z = 3