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Mathematics Question on Matrices

If P=[53 −1−2]P = \begin{bmatrix} 5 & 3 \\\ -1 & -2 \end{bmatrix} satisfies the equation P2−3P−7I=0P^2 - 3P - 7I = 0, where II is an identity matrix of order 2, then P−1P^{-1} is:

A

17[23 −1−5]\frac{1}{7} \begin{bmatrix} 2 & 3 \\\ -1 & -5 \end{bmatrix}

B

[23 −1−5]\begin{bmatrix} 2 & 3 \\\ -1 & -5 \end{bmatrix}

C

17[23 −1−1]\frac{1}{7} \begin{bmatrix} 2 & 3 \\\ -1 & -1 \end{bmatrix}

D

17[25 −1−1]\frac{1}{7} \begin{bmatrix} 2 & 5 \\\ -1 & -1 \end{bmatrix}

Answer

17[23 −1−5]\frac{1}{7} \begin{bmatrix} 2 & 3 \\\ -1 & -5 \end{bmatrix}

Explanation

Solution

Given that P2−3P−7I=0P^2 - 3P - 7I = 0, we can rearrange this as:

P2=3P+7I.P^2 = 3P + 7I.

Multiplying both sides by P−1P^{-1}, we get:

P=3I+7P−1.P = 3I + 7P^{-1}.

Rearranging for P−1P^{-1}:

P−1=17(P−3I).P^{-1} = \frac{1}{7}(P - 3I).

Substituting P=[53 −1−2]P = \begin{bmatrix} 5 & 3 \\\ -1 & -2 \end{bmatrix} and I=[10 01]I = \begin{bmatrix} 1 & 0 \\\ 0 & 1 \end{bmatrix}:

P−3I=[53 −1−2]−3×[10 01]=[23 −1−5].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=17[23 −1−5].P^{-1} = \frac{1}{7} \begin{bmatrix} 2 & 3 \\\ -1 & -5 \end{bmatrix}.