Note that
A–1 = −
−
−




1
11
2
5
3
2
Therefore
X = A–1B = – 1
11
2
5
3
2
71
−
−








i e 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