If \(A = \begin{bmatrix}4&-1\\2&1\end{bmatrix}\) and I is the identity matrix of order 2, then \(A^{-1}\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{6}(5I - A)\)
Let us find a relation between \(A^2\), A and I.
\( \displaystyle A^2 = \begin{bmatrix}16 - 2&-4 - 1\\8 + 2&-2 + 1\end{bmatrix} = \begin{bmatrix}14&-5\\10&-1\end{bmatrix} \)
\( \displaystyle A^2 - 5A = \begin{bmatrix}14 - 20&-5 + 5\\10 - 10&-1 - 5\end{bmatrix} = \begin{bmatrix}-6&0\\0&-6\end{bmatrix} \)
So
\(A^2 - 5A + 6I = O\)
Multiplying by \(A^{-1}\) (it exists because \(|A| = 4 + 2 = 6 \ne 0\)):
\(A - 5I + 6A^{-1} = O\)
\( \displaystyle A^{-1} = \frac{1}{6}(5I - A) \)
Check: \(5I - A = \begin{bmatrix}1&1\\-2&4\end{bmatrix}\), which is the adjoint of A, and \(|A| = 6\).
Shortcut: every \(2 \times 2\) matrix satisfies \(A^2 - (\text{sum of diagonal elements})A + |A|\,I = O\). Here the diagonal sum is 5 and \(|A| = 6\).