Answer:
First, we write the augmented matrix.
⎡
⎢
⎣
1
−
1
1
2
3
−
1
3
−
2
−
9
|
8
−
2
9
⎤
⎥
⎦
[1−1123−13−2−9 | 8−29]
Next, we perform row operations to obtain row-echelon form.
−
2
R
1
+
R
2
=
R
2
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
3
−
2
−
9
|
8
−
18
9
⎤
⎥
⎦
−
3
R
1
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
0
1
−
12
|
8
−
18
−
15
⎤
⎥
⎦
−2R1+R2=R2→[1−1105−33−2−9|8−189]−3R1+R3=R3→[1−1105−301−12|8−18−15]
The easiest way to obtain a 1 in row 2 of column 1 is to interchange
\displaystyle {R}_{2}
R 2
and
\displaystyle {R}_{3}
R 3
.
Interchange
R
2
and
R
3
→
⎡
⎢
⎣
1
−
1
1
8
0
1
−
12
−
15
0
5
−
3
−
18
⎤
⎥
⎦
InterchangeR2andR3→[1−11801−12−1505−3−18]
Then
−
5
R
2
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
57
|
8
−
15
57
⎤
⎥
⎦
−
1
57
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
1
|
8
−
15
1
⎤
⎥
⎦
−5R2+R3=R3→[1−1101−120057|8−1557]−157R3=R3→[1−1101−12001|8−151]
The last matrix represents the equivalent system.
x
−
y
+
z
=
8
y
−
12
z
=
−
15
z
=
1
x−y+z=8 y−12z=−15 z=1
Using back-substitution, we obtain the solution as
\displaystyle \left(4,-3,1\right)
(4,−3,1)