The given system of equation is

2

x

+

3

y

+

10

z

= 4 ,

4

x

−

6

y

+

5

z

= 1 ,

6

x

+

9

y

−

20

z

= 2
The given system of equation can be written as
[

2	3	10
4	−6	5
6	9	−20

][

1 x

1 y

1 z

] = [

4

1

2

]
or AX B,Where A = [

2	3	10
4	−6	5
6	9	−20

] , X = [

1 x

1 y

1 z

] and B = [

4

1

2

]
Now, |A| = [

2	3	10
4	−6	5
6	9	−20

]
= 2 (120 - 45) - 3 (- 80 - 30) + 10 (36 + 36)
= 1200 ≠ 0
Hence, the unique solution of the system of equation is given by X = A−1B
Now, the cofactors of A are computed as :
C11 = (−1)2 (120 - 45) = 75, C12 = (−1)3 (- 80 - 30) = 110, C13 = (−1)4 (36 + 36) = 72
C21 = (−1)3 (- 60 - 90) = 150, C22 = (−1)4 (- 40 - 60) = - 100, C23 = (−1)5 (18 - 18) = 0
C31 = (−1)4 (15 + 60) = 75, C12 = (−1)5 (10 - 40) = 30, C33 = (−1)6 (- 12 - 12) = - 24
∴ Adj A = [

75	110	72
150	−100	0
75	30	−24

]T = [

75	150	75
110	−100	30
72	0	−24

]
⇒ S−1 =

AdjA

|A|

=

1

1200

[

75	150	75
110	−100	30
72	0	−24

]
X = A−1B
=
=

1

1200

[

400+150+150

440−100+60

288+0−48

] =

1

1200

[

600

400

240

]
X = [

600 1200

400 1200

240 1200

] = [

1 2

1 3

1 5

] ⇒ [

1 x

1 y

1 z

] = [

1 2

1 3

1 5

]
⇒

1

x

=

1

2

,

1

y

=

1

3

and

1

z

=

1

5

⇒ x = 2 , y = 3 and z = 5
Thus, solution of given system of equation is given by x = 2, y = 3 and z = 5.
OR
The given matrix is A = [

1	3	−2
−3	0	−1
2	1	0

]
We have AA−1 = I
Thus, A = IA
Or, [

1	3	−2
−3	0	−1
2	1	0

] = [

1	0	0
0	1	0
0	0	1

] A
Applying R2 → R2+3R1 and R3 → R3−2R1
[

1	3	−2
0	9	−7
0	−5	4

] = [

1	0	0
3	1	0
−2	0	1

] A
Now,applying R2 →

1

9

R2
[

1	3	−2
0	1	− 7 9
0	−5	4

] = [

1	0	0
1 3	1 9	0
−2	0	1

] A
Applying R1 → R1−3R2 and R3 → R3+5R2
[

1	0	1 3
0	1	− 7 9
0	0	1 9

] = [

0	− 1 3	0
1 3	1 9	0
− 1 3	5 9	1

] A
Applying R3 → 9R3
[

1	0	1 3
0	1	− 7 9
0	0	1

] = [

0	− 1 3	0
1 3	1 9	0
−3	5	9

] A
Applying R1 → R1−

1

3

R3 and R2 → R2+

7

9

R3
[

1	0	0
0	1	0
0	0	1

] = [

1	−2	−3
−2	4	7
−3	5	9

] A ⇒ I = [

1	−2	−3
−2	4	7
−3	5	9

] A
∴ A−1 = [

1	−2	−3
−2	4	7
−3	5	9

]
Hence, inverse of the matrix A is [

1	−2	−3
−2	4	7
−3	5	9

]