CUET UG 2025 — Mathematics Algebra

For any invertible matrix B:

$$\det(B^{-1}) = \frac{1}{\det(B)}$$det(B−1)=det(B)1​

For a matrix A of order n:

$$\det(\text{adj } A) = (\det A)^{n-1}$$det(adj A)=(detA)n−1

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Since A is a 2×2 matrix (order 2):

$$\det(\text{adj } A) = (\det A)^{2-1}$$det(adj A)=(detA)2−1

$$\det(\text{adj } A) = (\det A)^{1}$$det(adj A)=(detA)1

$$\det(\text{adj } A) = \det A$$det(adj A)=detA

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Applying the inverse determinant property:

$$\det((\text{adj } A)^{-1}) = \frac{1}{\det(\text{adj } A)}$$det((adj A)−1)=det(adj A)1​

Substituting the value from above:

$$\det((\text{adj } A)^{-1}) = \frac{1}{\det A}$$det((adj A)−1)=detA1​

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Therefore, $\det((\text{adj } A)^{-1}) = \dfrac{1}{\det A}$det((adj A)−1)=detA1​