The correct option is B 1,−1
A=⎡⎢⎣0121233a1⎤⎥⎦ and A−1=⎡⎢⎣1/21/21/2−43c5/2−3/21/2⎤⎥⎦
Here, |A|=−1(−8)+2(a−6)=2a−4
adjA=CT=⎡⎢⎣2−3a8a−62a−1−63−12−1⎤⎥⎦T
⇒adjA=⎡⎢⎣2−3a2a−1−18−62a−63−1⎤⎥⎦
Hence, A−1=12a−4⎡⎢⎣2−3a2a−1−18−62a−63−1⎤⎥⎦
Comparing with given A−1,
−12a−4=12
⇒a=1
Also, 22a−4=c
⇒c=−1
Thus the values of a and c are 1 and −1 respectively.