CUET UG 2025 — Mathematics Algebra

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

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

Our matrix is:

$$A = \begin{bmatrix} 1 & 2 & -1 \\ 3 & 2 & 3 \\ 4 & -1 & 0 \end{bmatrix}$$A=​134​22−1​−130​​

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Checking Option (A): $M_{32} = 6$M32​=6

Delete row 3 and column 2:

$M_{32} = \begin{vmatrix} 1 & -1 \\ 3 & 3 \end{vmatrix} = (1)(3) - (-1)(3) = 3 + 3 = 6$M32​=​13​−13​​=(1)(3)−(−1)(3)=3+3=6 ✅

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Checking Option (B): $M_{23} = 9$M23​=9

Delete row 2 and column 3:

$M_{23} = \begin{vmatrix} 1 & 2 \\ 4 & -1 \end{vmatrix} = (1)(-1) - (2)(4) = -1 - 8 = -9$M23​=​14​2−1​​=(1)(−1)−(2)(4)=−1−8=−9 ❌

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Checking Option (C): $A_{32} = -6$A32​=−6

We found $M_{32} = 6$M32​=6, so:

$A_{32} = (-1)^{3+2} \cdot 6 = (-1)^5 \cdot 6 = -6$A32​=(−1)3+2⋅6=(−1)5⋅6=−6 ✅

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Checking Option (D): $A_{23} = -9$A23​=−9

We found $M_{23} = -9$M23​=−9, so:

$A_{23} = (-1)^{2+3} \cdot (-9) = (-1)^5 \cdot (-9) = 9$A23​=(−1)2+3⋅(−9)=(−1)5⋅(−9)=9 ❌

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Options (A) and (C) are correct.