13.
2 3 10
+-+ = 4
х y у
4 6
5
+ - = 1
х у
6 9
20
= 2
х у
Z
+
Z
+
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12th
Maths
Determinants
Applications of Matrices and Determinants
Solve by matrix inversion m...
Maths
Solve by matrix inversion method
2/x+3/y+10/z=4
1/x-6/y+5/z=1
6/x+9/y-20/z=2
Hard
Let
x
1
=u,
y
1
=v,
z
1
=w
Now the given system of linear equation may be written as:
2u+3v+10w=4,4u−6v+5w=1and6u+9v−20w=2
Above system of equation can be written in matrix form as:
AX=B
X=A−1B
A=
∣
∣
∣
∣
∣
∣
∣
∣
2
4
6
3
−6
9
10
5
−20
∣
∣
∣
∣
∣
∣
∣
∣
,X=
∣
∣
∣
∣
∣
∣
∣
∣
u
v
w
∣
∣
∣
∣
∣
∣
∣
∣
,B=
∣
∣
∣
∣
∣
∣
∣
∣
4
1
2
∣
∣
∣
∣
∣
∣
∣
∣
∣A∣=
∣
∣
∣
∣
∣
∣
∣
∣
2
4
6
3
−6
9
10
5
−20
∣
∣
∣
∣
∣
∣
∣
∣
=2(120−45)−3(−80−30)+10(36+36)=150+330+720=1200
=0
ForadjA:A
11
=120−45=75
A
12
=−(−60−90)=150
A
13
=15+60=75
A
12
=−(−80−30)=110
A
22
=−40−60=−100
A
32
=−(10−40)=30
A
13
=36+36=72
A
23
=−(18−18)=0
A
33
=−12−12=−24
adj.A=
⎣
⎢
⎢
⎡
75
150
75
110
−100
30
72
0
−24
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
75
110
72
150
−100
0
75
30
−24
⎦
⎥
⎥
⎤
A
−1
=
∣A∣
1
.adj.A=
1200
1
⎣
⎢
⎢
⎡
75
110
72
150
−100
0
75
30
−24
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
u
v
w
⎦
⎥
⎥
⎤
=
1200
1
⎣
⎢
⎢
⎡
75
110
72
150
−100
0
75
30
−24
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
4
1
2
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
u
v
w
⎦
⎥
⎥
⎤
=
1200
1
⎣
⎢
⎢
⎡
300+150+300
440−100+60
288+0−48
⎦
⎥
⎥
⎤
\\
⎣
⎢
⎢
⎡
u
v
w
⎦
⎥
⎥
⎤
=
1200
1
⎣
⎢
⎢
⎡
600
400
240
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
u
v
w
⎦
⎥
⎥
⎤
=
1200
1
⎣
⎢
⎢
⎡
2
1
3
1
5
1
⎦
⎥
⎥
⎤
x=2,y=3,w=5