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Question: If the matrix $A = \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix}$ and $A^{-1} = xA + yI$, when I is ...

If the matrix A=[12−51]A = \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix} and A−1=xA+yIA^{-1} = xA + yI, when I is a unit matrix of order 2, then the value of 2x + 3y is

A

811\frac{8}{11}

B

411\frac{4}{11}

C

−811\frac{-8}{11}

D

−411\frac{-4}{11}

Answer

411\frac{4}{11}

Explanation

Solution

First, find A−1A^{-1} directly.

Determinant: det⁡A=1⋅1−2(−5)=11\det A =1\cdot1-2(-5)=11.

Adjugate: adj⁡(A)=(1−251)\operatorname{adj}(A)=\begin{pmatrix} 1&-2\\5&1\end{pmatrix}.

Thus, A−1=111(1−251)A^{-1}=\frac{1}{11}\begin{pmatrix} 1&-2\\5&1\end{pmatrix}.

Write xA+yI=(x+y2x−5xx+y)xA+yI = \begin{pmatrix} x+y &2x\\-5x &x+y \end{pmatrix}. Equate entries:

x+y=111x+y=\frac{1}{11} and 2x=−2112x=\frac{-2}{11} ⟹ x=−111x=-\frac{1}{11}.

Then, y=111−x=111+111=211y=\frac{1}{11} - x=\frac{1}{11}+\frac{1}{11}=\frac{2}{11}.

Hence, 2x+3y=2(−111)+3(211)=−2+611=4112x+3y=2\Bigl(-\frac{1}{11}\Bigr)+3\Bigl(\frac{2}{11}\Bigr)=\frac{-2+6}{11}=\frac{4}{11}.