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The inverse of the matrix A=[2312]A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} is:

Solution

✅ Correct Option: 2

Determinant ∣A∣=2(2)−3(1)=1|A| = 2(2) - 3(1) = 1. Adjoint =[2−3−12]= \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}. Inverse A−1=(1/∣A∣)adj(A)=[2−3−12]A^{-1} = (1/|A|)\text{adj}(A) = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}.

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