NDA Math Matrices

(A)
Given |

x	−3i	1
y	1	i
0	2i	−i

|= 6+ 11i
|

x	−3i	1
y	1	i
0	2i	−i

| = x (−i−(2 i) (i)) − y((−3 i) (−i)−(2 i) ) + 0
= x (−i+2) − y(−3−2 i)
= (2x + 3 y ) + (2 y−x )i = 6+11i (∴given)
⇒ 2 x + 3 y = 6 …(i)
2 y − x = 11 …(ii)
Solving (i) and (ii),
we get x = −3 and y = 4.