Solve the system of equation by Gauss elimination method: x + y + z = 2 2x - y + 3z =16 x + y - z = -3
Answers
Answer:
x=8/3
y=-19/6
z=5/2
Explanation:
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Answer:
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
\begin{gathered}\left[\begin{array}{ccc}1&1&1 \ | \ 7 \\3&3&4 \ \ | \ 24\\2&1&3 \ \ | \ 16\end{array}\right]\end{gathered}
⎣
⎢
⎡
1
3
2
1
3
1
1 ∣ 7
4 ∣ 24
3 ∣ 16
⎦
⎥
⎤
R₂ - 3R₁ → R₂ ; R₃ - 2R₁ → R₃
\begin{gathered}\left[\begin{array}{ccc}1&1&1 \ \ | \ 7 \\0&0&1 \ \ | \ 3\\0&-1&1 \ \ | \ 2\end{array}\right]\end{gathered}
⎣
⎢
⎡
1
0
0
1
0
−1
1 ∣ 7
1 ∣ 3
1 ∣ 2
⎦
⎥
⎤
R₂ ↔ R₃
\begin{gathered}\left[\begin{array}{ccc}1&1&1 \ \ | \ 7 \\0&-1&1 \ \ | \ 2\\0&0&1 \ \ | \ 3\end{array}\right]\end{gathered}
⎣
⎢
⎡
1
0
0
1
−1
0
1 ∣ 7
1 ∣ 2
1 ∣ 3
⎦
⎥
⎤
R₂ / -1 → R₂
\begin{gathered}\left[\begin{array}{ccc}1&1&1 \ | \ 7 \\0&1&-1 \ \ | \ -2\\0&0&1 \ | \ 3\end{array}\right]\end{gathered}
⎣
⎢
⎡
1
0
0
1
1
0
1 ∣ 7
−1 ∣ −2
1 ∣ 3
⎦
⎥
⎤
R₁ - R₂ → R₁
\begin{gathered}\left[\begin{array}{ccc}1&0&2 \ | \ 9 \\0&1&-1 \ \ | \ -2\\0&0&1 \ | \ 3\end{array}\right]\end{gathered}
⎣
⎢
⎡
1
0
0
0
1
0
2 ∣ 9
−1 ∣ −2
1 ∣ 3
⎦
⎥
⎤
R₁ - 2R₃ → R₁ ; R₂ + R₃ → R₂
\begin{gathered}\left[\begin{array}{ccc}1&0&0 \ | \ 3 \\0&1&0 \ | \ 1 \\0&0&1 \ | \ 3\end{array}\right]\end{gathered}
⎣
⎢
⎡
1
0
0
0
1
0
0 ∣ 3
0 ∣ 1
1 ∣ 3
⎦
⎥
⎤
x = 3
y = 1
z = 3