Jawaban:
Penjelasan dengan langkah-langkah:
matrix
Diket Matrix
�
=
[
2
−
1
3
0
]
A=
⎣
⎢
⎡
x
−x
−1
⎦
⎥
⎤
Jika A adalah Matrix singular ( det A = 0 ) , maka x adalah
det Matrix 3×3
→
(
.
)
+
a
b
c
det→a
(b
.c
−b
)−a
)+a
)=0
Maka
det A
1.0
1.3
1.1
detA=2(1.0−−1.3)−2(−x.0−−1.1)+x(−x.3−1.1)=0
2(3)−2(1)+x(−3x−1)=0
4
4−3x
−x=0
a = -3 , b = -1 , c = 4
akar akar x
±
x=
2a
−b±
−4ac
3.4
2(−3)
−(−1)±
(−1)
−4(−3.4)
48
6
x=−
1±
1−(−48)
49
7
1±7
untuk
−1−7
=−
−1+7
=1
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Jawaban:
Penjelasan dengan langkah-langkah:
matrix
Diket Matrix
�
=
[
2
−
�
1
2
1
3
�
−
1
0
]
A=
⎣
⎢
⎡
2
2
x
−x
1
−1
1
3
0
⎦
⎥
⎤
Jika A adalah Matrix singular ( det A = 0 ) , maka x adalah
det Matrix 3×3
[
�
1
�
1
�
1
�
2
�
2
�
2
�
3
�
3
�
3
]
�
�
�
→
�
1
(
�
2
.
�
3
−
�
3
.
�
2
)
−
�
2
(
�
1
.
�
3
−
�
3
.
�
1
)
+
�
3
(
�
1
.
�
2
−
�
2
.
�
1
)
=
0
⎣
⎢
⎡
a
1
a
2
a
3
b
1
b
2
b
3
c
1
c
2
c
3
⎦
⎥
⎤
det→a
1
(b
2
.c
3
−b
3
.c
2
)−a
2
(b
1
.c
3
−b
3
.c
1
)+a
3
(b
1
.c
2
−b
2
.c
1
)=0
Maka
det A
�
�
�
�
=
2
(
1.0
−
−
1.3
)
−
2
(
−
�
.
0
−
−
1.1
)
+
�
(
−
�
.
3
−
1.1
)
=
0
detA=2(1.0−−1.3)−2(−x.0−−1.1)+x(−x.3−1.1)=0
2
(
3
)
−
2
(
1
)
+
�
(
−
3
�
−
1
)
=
0
2(3)−2(1)+x(−3x−1)=0
4
−
3
�
2
−
�
=
0
4−3x
2
−x=0
a = -3 , b = -1 , c = 4
akar akar x
�
=
−
�
±
�
2
−
4
�
�
2
�
x=
2a
−b±
b
2
−4ac
�
=
−
(
−
1
)
±
(
−
1
)
2
−
4
(
−
3.4
)
2
(
−
3
)
x=
2(−3)
−(−1)±
(−1)
2
−4(−3.4)
�
=
−
1
±
1
−
(
−
48
)
6
x=−
6
1±
1−(−48)
�
=
−
1
±
49
6
x=−
6
1±
49
�
=
−
1
±
7
6
x=−
6
1±7
untuk
�
1
x
1
�
1
=
−
1
−
7
6
=
−
4
3
x
1
=
6
−1−7
=−
3
4
untuk
�
2
x
2
�
2
=
−
1
+
7
6
=
1
x
2
=
6
−1+7
=1