Let $M_{ij}$ and $A_{ij}$ denote respectively minors and co-factors of the element in the $i^{th}$ row and $j^{th}$ column of the matrix $A = \begin{bmatrix} 1 & 2 & -1 \\ 3 & 2 & 3 \\ 4 & -1 & 0 \end{bmatrix}$. Then:

(A) $M_{32} = 6$

(B) $M_{23} = 9$

(C) $A_{32} = -6$

(D) $A_{23} = -9$

Choose the correct answer from the options given below:

A minor $M_{ij}$ is the determinant obtained after deleting row $i$ and column $j$ from the matrix.

A cofactor $A_{ij}$ is defined as $A_{ij} = (-1)^{i+j} \cdot M_{ij}$

Our matrix is:

$A = \begin{bmatrix} 1 & 2 & -1 \\ 3 & 2 & 3 \\ 4 & -1 & 0 \end{bmatrix}$

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Checking Option (A): $M_{32} = 6$

Delete row 3 and column 2:

$M_{32} = \begin{vmatrix} 1 & -1 \\ 3 & 3 \end{vmatrix} = (1)(3) - (-1)(3) = 3 + 3 = 6$ ✅

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Checking Option (B): $M_{23} = 9$

Delete row 2 and column 3:

$M_{23} = \begin{vmatrix} 1 & 2 \\ 4 & -1 \end{vmatrix} = (1)(-1) - (2)(4) = -1 - 8 = -9$ ❌

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Checking Option (C): $A_{32} = -6$

We found $M_{32} = 6$, so:

$A_{32} = (-1)^{3+2} \cdot 6 = (-1)^5 \cdot 6 = -6$ ✅

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Checking Option (D): $A_{23} = -9$

We found $M_{23} = -9$, so:

$A_{23} = (-1)^{2+3} \cdot (-9) = (-1)^5 \cdot (-9) = 9$ ❌

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Options (A) and (C) are correct.