Solution.
2 x + 7 y + z = 14 2x + 7y + z = 14 2 x + 7 y + z = 14 ;
x + 3 y − z = 2 x + 3y - z = 2 x + 3 y − z = 2 ;
x + 7 y + 12 z = 45 ; x + 7y + 12z = 45; x + 7 y + 12 z = 45 ;
[ 2 7 1 14 1 3 − 1 2 1 7 12 45 ] \begin{bmatrix}
2 & 7&1&14 \\
1 & 3&-1&2\\
1&7&12&45
\end{bmatrix} ⎣ ⎡ 2 1 1 7 3 7 1 − 1 12 14 2 45 ⎦ ⎤
Find the pivot in the 1st column and swap the 2nd and the 1st rows
[ 1 3 − 1 2 2 7 1 14 1 7 12 45 ] \begin{bmatrix}
1& 3&-1&2 \\
2 & 7&1&14\\
1&7&12&45
\end{bmatrix} ⎣ ⎡ 1 2 1 3 7 7 − 1 1 12 2 14 45 ⎦ ⎤
Eliminate the 1st column
[ 1 3 − 1 2 0 1 3 10 0 4 13 43 ] \begin{bmatrix}
1& 3&-1&2 \\
0 & 1&3&10\\
0&4&13&43
\end{bmatrix} ⎣ ⎡ 1 0 0 3 1 4 − 1 3 13 2 10 43 ⎦ ⎤
Find the pivot in the 2nd column in the 2nd row
[ 1 3 − 1 2 0 1 3 10 0 4 13 43 ] \begin{bmatrix}
1& 3&-1&2 \\
0 & 1&3&10\\
0&4&13&43
\end{bmatrix} ⎣ ⎡ 1 0 0 3 1 4 − 1 3 13 2 10 43 ⎦ ⎤
Eliminate the 2nd column
[ 1 0 − 10 − 28 0 1 3 10 0 0 1 3 ] \begin{bmatrix}
1& 0&-10&-2 8\\
0 & 1&3&10\\
0&0&1&3
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 0 − 10 3 1 − 28 10 3 ⎦ ⎤
Find the pivot in the 3rd column in the 3rd row
[ 1 0 − 10 − 28 0 1 3 10 0 0 1 3 ] \begin{bmatrix}
1& 0&-10&-2 8\\
0 & 1&3&10\\
0&0&1&3
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 0 − 10 3 1 − 28 10 3 ⎦ ⎤
Eliminate the 3rd column
[ 1 0 0 2 0 1 0 1 0 0 1 3 ] \begin{bmatrix}
1& 0&0&2 \\
0 & 1&0&1\\
0&0&1&3
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 0 0 0 1 2 1 3 ⎦ ⎤
Solution set:
x = 2 ; y = 1 ; z = 3. x = 2;
y = 1;
z = 3. x = 2 ; y = 1 ; z = 3.
Answer: x = 2 ; y = 1 ; z = 3. x = 2;
y = 1;
z = 3. x = 2 ; y = 1 ; z = 3.
Comments