x
y
 
 
   = −
−



= −




1
11
33
11
3
1
Hence
x = 3, y = – 1
Example 17 Solve the following system of equations by matrix method 3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4
Solution The system of equations can be written in the form AX = B, where
3
2
3
8
A
2
1
1 , X
and B
1
4
3
2
4
x
y
z
−


 
 


 
 
=
−
=
=


 
 


 
 
−


 
 
We see that
A  = 3 (2 – 3) + 2(4 + 4) + 3 (– 6 – 4) = – 17 ≠ 0
Hence, A is nonsingular and so its inverse exists Now
A11 = –1,
A12 = – 8,
A13 = –10
A21 = –5,
A22 = – 6,
A23 = 1
A31 = –1,
A32 = 9,
A33 = 7
Rationalised 2023-24
 96
MATHEMATICS
Therefore
A–1 =
1
5
1
1
8
6
9
17
10
1
7
−
−
−




−
−
−




−

So
X =
–1
1
5
1
8
1
A B =
8
6
9
1
17
10
1
7
4
−
−
−

  

  
−
−
−

  

  
−
  
i e