If \(A = \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}\) and \(2A^{-1} = KI - A\), where \(K\) is a real number and \(I\) is the identity matrix of order 2, then the value of \(K\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 9
For a 2 × 2 matrix, \(A^{-1} = \frac{1}{|A|}\,\mathrm{adj}\,A\), where the adjoint swaps the diagonal entries and changes the sign of the other two.
\(|A| = 2(7) - (-3)(-4) = 14 - 12 = 2\)
\( \displaystyle \mathrm{adj}\,A = \begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix} \)
\( \displaystyle 2A^{-1} = 2 \cdot \frac{1}{2}\,\mathrm{adj}\,A = \begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix} \)
Also,
\( \displaystyle KI - A = \begin{bmatrix} K - 2 & 3 \\ 4 & K - 7 \end{bmatrix} \)
Comparing the diagonal entries of the two equal matrices:
\(K - 2 = 7 \Rightarrow K = 9\)
\(K - 7 = 2 \Rightarrow K = 9\)
Shortcut: By the Cayley-Hamilton theorem, \(A^2 - (\mathrm{tr}\,A)A + |A|I = O\). Here \(\mathrm{tr}\,A = 9\) and \(|A| = 2\), so \(A^2 - 9A + 2I = O\). Multiplying by \(A^{-1}\) gives \(A - 9I + 2A^{-1} = O\), that is, \(2A^{-1} = 9I - A\). So \(K = 9\).