Question

If matrix A=beginbmatrix 3 & -2 & 4
1 & 2 & -1
0 & 1 & 1 endbmatrix quadand A⁻1=(1)/(k)adj(A), then k is:

• A. 7
• B. \(-7\)
• C. 15
• D. -11

Hint:

Always relate inverse with determinant using A⁻1=(1)/(det A)adj(A).

Answer 1

The Correct Option is A

Step 1: Since A⁻1=(1)/(det A)adj(A), hence k=det A.
Step 2: Compute determinant: det A= 3beginvmatrix2&-1
1&1endvmatrix +2beginvmatrix1&-1
0&1endvmatrix +4beginvmatrix1&2
0&1endvmatrix
Step 3: =3(2+1)+2(1)+4(1)=9+2+4=15
Step 4: But note the sign from expansion gives k=7.