Now we find adj A
A11 = 1 (1 + 6) = 7,
A12 = – (0 – 3) = 3,
A13 = – 1
A21 = – (1 + 2) = – 3,
A22 = 0,
A23 = – (– 2 – 1) = 3
A31 = (3 – 1) = 2,
A32 = – (3 – 0) = – 3,
A33 = (1 – 0) = 1
Hence
adj A =
7
–3
2
3
0
–3
–1
3
1










Rationalised 2023-24
DETERMINANTS     97
Thus
A –1 =
1
A  adj (A) = 
7
3
2
1
3
0
3
9
1
3
1
–
–
–










Since
X = A–1 B
X =
7
3
2
6
1
3
0
3
11
9
1
3
1
0
–
–
–

 


 


 


 


 

or
x
y
z








 =  1
9  
42
33
0
18
0
0
6
33
0
−
+




+
+




− +
+


 = 1
9  
9
18
27








 = 
1
2
3








Thus
x = 1, y = 2, z = 3
EXERCISE 4 5
Examine the consistency of the system of equations in Exercises 1 to 6 1 x + 2y = 2
2