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Let I=[1001]I = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} and A=[1−224]A = \begin{bmatrix}1 & -2\\2 & 4\end{bmatrix} be two matrices such that 8A−1=kI−A,k∈R8A^{-1} = kI - A, k \in \mathbb{R}, then the value of kk is:

Solution

✅ Correct Option: 3

∣A∣=4+4=8|A| = 4+4 = 8, so 8A−1=adj A=[42−21]8A^{-1} = adj\,A = \begin{bmatrix}4 & 2\\-2 & 1\end{bmatrix}.

Also kI−A=[k−12−2k−4]kI - A = \begin{bmatrix}k-1 & 2\\-2 & k-4\end{bmatrix}.

Comparing entries, k−1=4k-1=4 and k−4=1k-4=1, so k=5k=5.

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