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If A=[abc1+bca]A = \begin{bmatrix} a & b \\ c & \frac{1+bc}{a} \end{bmatrix}, then aA−1aA^{-1} is equal to which of the following? [I is the identity matrix of order 2]

Solution

✅ Correct Option: 4

∣A∣=a⋅1+bca−bc=1|A| = a\cdot\frac{1+bc}{a} - bc = 1, so A−1=adjA=[1+bca−b−ca]A^{-1} = \text{adj}A = \begin{bmatrix} \frac{1+bc}{a} & -b \\ -c & a \end{bmatrix} and aA−1=[1+bc−ab−aca2]aA^{-1} = \begin{bmatrix} 1+bc & -ab \\ -ac & a^2 \end{bmatrix}. Expanding (a2+bc+1)I−aA(a^2+bc+1)I - aA gives exactly this matrix.

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