UPSC NDA Math Model Paper 9

Let‌A=[

1	a	b
0	x	0
0	0	1

]
So, A−1=[

1	−a	−b
0	1	0
0	0	1

] We know that,
AA−1=I
Taking Determinant both sides,
Det(AA−1)=Det(I)
We know that Determinant of an Identity matrix is 1.
⇒|A||A−1|=1
⇒|

1	a	b
0	x	0
0	0	1

| |

1	−a	−b
0	1	0
0	0	1

|=‌1
Now, Expanding along C3
⇒ [0 – 0 + 1(x – 0)] [0 – 0 + 1(1 – 0)] = 1
⇒ x = 1
∴ Option 3 is correct.