If \(P=\begin{bmatrix}5 & 3\\ -1 & -2\end{bmatrix}\) satisfies the equation \(P^2-3P-7I=0\), where I is an identity matrix of order 2, then \(P^{-1}\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{7}\begin{bmatrix}2 & 3\\ -1 & -5\end{bmatrix}\)
\(|P| = 5(-2) - 3(-1) = -7 \ne 0\), so \(P^{-1}\) exists. Here 0 in the equation is the zero matrix O. Multiplying the given equation by \(P^{-1}\), with \(P^{-1}P^2 = P\) and \(P^{-1}P = I\):
\(P^2 - 3P - 7I = O\)
\(P - 3I - 7P^{-1} = O\)
\( \displaystyle P^{-1} = \frac{1}{7}(P - 3I) \)
\( \displaystyle = \frac{1}{7}\begin{bmatrix}5 - 3 & 3\\ -1 & -2 - 3\end{bmatrix} \)
\( \displaystyle = \frac{1}{7}\begin{bmatrix}2 & 3\\ -1 & -5\end{bmatrix} \)
Check with the adjoint: \(\mathrm{adj}\,P = \begin{bmatrix}-2 & -3\\ 1 & 5\end{bmatrix}\), and dividing by |P| = −7 gives the same matrix.
NCERT Exemplar:
Matrices
