For a matrix \( A=\begin{bmatrix}a & -1 & b\\-2 & 1 & -1\\1 & 0 & 2\end{bmatrix} \), \( \mathrm{C_{32}}=-1 \) and \( \mathrm{C_{33}}=3 \), where \( \mathrm{C}_{ij} \) is the cofactor of element \( a_{ij} \) in \( A \). Then
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( a=5 \) & \( b=3 \)
The cofactor \( C_{ij}=(-1)^{i+j}M_{ij} \), where \( M_{ij} \) is the determinant left after deleting row i and column j.
\( \displaystyle C_{33}=(+1)\begin{vmatrix}a & -1\\-2 & 1\end{vmatrix} \)
\( \displaystyle =a-2=3\ \Rightarrow\ a=5 \)
\( \displaystyle C_{32}=(-1)\begin{vmatrix}a & b\\-2 & -1\end{vmatrix} \)
\( \displaystyle =-(-a+2b)=a-2b \)
\( \displaystyle a-2b=-1\ \Rightarrow\ 5-2b=-1\ \Rightarrow\ b=3 \)
So a = 5 and b = 3.
Why the other options are incorrect
Option (1) \( a=5 \) & \( b=2 \) — Incorrect. This comes from leaving out the sign \( (-1)^{3+2}=-1 \) of \( C_{32} \): \( -a+2b=-1 \) with a = 5 gives b = 2. With b = 2 the true cofactor is \( C_{32}=5-4=1 \), not −1.
Option (3) \( a=2 \) & \( b=5 \) — Incorrect. With a = 2, \( C_{33}=2-2=0 \), not 3.
Option (4) \( a=3 \) & \( b=5 \) — Incorrect. With a = 3, \( C_{33}=3-2=1 \), not 3.