Solution We have 2A + 3X = 5B
or
2A + 3X – 2A = 5B – 2A
or
2A – 2A + 3X = 5B – 2A
(Matrix addition is commutative)
or
O + 3X = 5B – 2A
(– 2A is the additive inverse of 2A)
or
3X = 5B – 2A
(O is the additive identity)
or
X = 1
3  (5B – 2A)
or
2
2
8 0
1
X
5
4
2
2 4
2
3
5 1
3 6


−










=
−
−












−





 = 
10
10
16
0
1
20
10
8
4
3
25
5
6
12


−
−










+
−












−
−
−






Rationalised 2023-24
 48
MATHEMATICS
    = 
10
16
10
0
1
20
8
10
4
3
25
6
5 12
−
−
+




−
+




−
−
−


 = 
6
10
1
12
14
3
31
7
−
−








−
−


 = 
10
2
143
4
3
31
7
3
3
−


−









−
−






Example 9 Find  X and Y, if 
5
2
X
Y
0
9


+
= 



 and 
3
6
X
Y
0
1


−
= 
−

 Solution We have (
)
(
)
5
2
3
6
X
Y
X
Y
0
9
0
1




+
+
−
=
+



−



 or
(X + X) + (Y – Y) =
8
8
0
8






 ⇒ 
8
8
2X
0
8


= 



or
X =
8
8
4
4
1
0
8
0
4
2




=








Also
(X + Y) – (X – Y) =
5
2
3
6
0
9
0
1




−



−




or
(X – X) + (Y + Y) =
5
3
2
6
0
9
1
−
−



+


 ⇒ 
2
4
2Y
0
10
−


= 



or
Y =
2
4
1
2
1
0
10
0
5
2
−
−




=








Example 10 Find the values of x and y from the following equation:
5
3
4
2 7
3
1
2
x
y
−




+




−



  = 
7
6
15
14






Solution We have
   
5
3
4
2 7
3
1
2
x
y
−




+




−



  = 
7
6
15
14






 ⇒ 2
10
3
4
7
6
14
2
6
1
2
15
14
x
y
−






+
=






−






Rationalised 2023-24
MATRICES         49
or
2
3
10
4
14
1
2
6
2
x
y
+
−




+
−
+

  =
7
6
15 14





 ⇒ 2
3
6
7
6
15
2
4
15
14
x
y
+




=




−




or
2x + 3 = 7
and
2y – 4 = 14
(Why )
or
2x = 7 – 3
and
2y = 18
or
x = 4
2
and
y = 18
2
i