Let we solve the sysyem by using Gauss method:
( 1 1 − 1 = 4 2 0 1 = 7 3 − 2 0 = 5 ) \begin{pmatrix}
1 & 1 &-1&=&4\\
2 & 0 & 1&=&7\\
3& -2& 0&=&5
\end{pmatrix} ⎝ ⎛ 1 2 3 1 0 − 2 − 1 1 0 = = = 4 7 5 ⎠ ⎞
1) Add 2 equation to the1 equation
( 3 1 0 = 11 2 0 1 = 7 3 − 2 0 = 5 ) \begin{pmatrix}
3 & 1 &0&=&11\\
2 & 0 & 1&=&7\\
3& -2& 0&=&5
\end{pmatrix} ⎝ ⎛ 3 2 3 1 0 − 2 0 1 0 = = = 11 7 5 ⎠ ⎞
2) We add 1 equation multiplied by 2 to to the 3 equation:
( 3 1 0 = 4 2 0 1 = 7 9 0 0 = 27 ) \begin{pmatrix}
3 & 1 &0&=&4\\
2 & 0 & 1&=&7\\
9& 0& 0&=&27
\end{pmatrix} ⎝ ⎛ 3 2 9 1 0 0 0 1 0 = = = 4 7 27 ⎠ ⎞
3) Divide 3 eqution by 9
( 3 1 0 = 11 2 0 1 = 7 1 0 0 = 27 9 = 3 ) \begin{pmatrix}
3 & 1 &0&=&11\\
2 & 0 & 1&=&7\\
1& 0& 0&=&\frac{27}{9}=3
\end{pmatrix} ⎝ ⎛ 3 2 1 1 0 0 0 1 0 = = = 11 7 9 27 = 3 ⎠ ⎞
4) Substract from 1 equation 3 equation multiplied by 3 and fron 2 equation 3 equation multiplied by 2
( 0 1 0 = 11 − 3 ⋅ 3 = 2 0 0 1 = 7 − 2 ⋅ 3 = 1 1 0 0 = 3 ) \begin{pmatrix}
0 & 1 &0&=&11-3\cdot 3=2\\
0 & 0 & 1&=&7-2\cdot 3=1\\
1& 0& 0&=&3
\end{pmatrix} ⎝ ⎛ 0 0 1 1 0 0 0 1 0 = = = 11 − 3 ⋅ 3 = 2 7 − 2 ⋅ 3 = 1 3 ⎠ ⎞
Matrix to the left is unit now then problem is solved.
Answer:
X = 3 Y = 2 Z = 1 X=3\\
Y=2\\
Z=1 X = 3 Y = 2 Z = 1
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