Solve-
Answers
First, we write the augmented matrix.
⎡
⎢
⎣
1
−
1
1
2
3
−
1
3
−
2
−
9
|
8
−
2
9
⎤
⎥
⎦
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
⎤
⎥
⎦
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
⎤
⎥
⎦
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
⎤
⎥
⎦
The last matrix represents the equivalent system.
x
−
y
+
z
=
8
y
−
12
z
=
−
15
z
=
1
Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).
Step-by-step explanation:
a)
−
30
×
1
10
−30 ×
10
1
= -3
(b)
50
×
−
1
5
50×
5
−1
= -10
(c)
−
36
×
−
1
9
−36 ×
9
−1
= 4
(d)
(
−
49
)
×
1
49
(−49)×
49
1
= -1
(e)
13
×
[
(
−
2
)
+
1
]
13×[(−2) + 1]
13
×
−
1
1
13×
1
−1
= -13
(f)
0
×
−
1
12
0×
12
−1
= 0
(g)
(
−
31
)
÷
[
(
−
30
)
−
1
]
(−31)÷[(−30)−1]
(
−
31
)
÷
(
−
31
)
(−31)÷(−31)
−
31
×
−
1
31
−31 ×
31
−1
= 1
(h)
[
(
−
36
)
÷
12
]
÷
3
[(−36)÷12]÷3
[
(
−
36
)
×
1
12
]
×
1
3
[(−36)×
12
1
]×
3
1
−
36
12
×
1
3
12
−36
×
3
1
−
3
×
1
3
−3×
3
1
= 1
(i)
[
−
6
+
5
]
÷
[
−
2
+
1
]
[−6+5]÷[−2+1]
−
1
×
−
1
1
−1×
1
−1
= 1