Question

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:

• A. \(\frac{1}{7} \begin{bmatrix} 2 & 3 \\ -1 & -5 \end{bmatrix}\)
• B. \(\begin{bmatrix} 2 & 3 \\ -1 & -5 \end{bmatrix}\)
• C. \(\frac{1}{7} \begin{bmatrix} 2 & 3 \\ -1 & -1 \end{bmatrix}\)
• D. \(\frac{1}{7} \begin{bmatrix} 2 & 5 \\ -1 & -1 \end{bmatrix}\)

Answer 1

The Correct Option is A

Given that \( P^2 - 3P - 7I = 0 \), we can rearrange this as:

\[ P^2 = 3P + 7I. \]

Multiplying both sides by \( P^{-1} \), we get:

\[ P = 3I + 7P^{-1}. \]

Rearranging for \( P^{-1} \):

\[ P^{-1} = \frac{1}{7}(P - 3I). \]

Substituting \( P = \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \):

\[ P - 3I = \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix} - 3 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ -1 & -5 \end{bmatrix}. \]

Therefore:

\[ P^{-1} = \frac{1}{7} \begin{bmatrix} 2 & 3 \\ -1 & -5 \end{bmatrix}. \]