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Matrices & Determinants: Pmatrix 4pt Pmatrix Equal

JEE Maths question with a full step-by-step solution.

Question
If A=(−132−1212)A = \begin{pmatrix}-1 & \dfrac{3}{2}\\[4pt] -\dfrac{1}{2} & \dfrac{1}{2}\end{pmatrix}, then I+A+A2+⋯∞I + A + A^2 + \cdots\infty is equal to:
A(1−314)\begin{pmatrix}1 & -3\\ 1 & 4\end{pmatrix}
B27(1−3−14)\dfrac{2}{7}\begin{pmatrix}1 & -3\\ -1 & 4\end{pmatrix}
C27(13−14)\dfrac{2}{7}\begin{pmatrix}1 & 3\\ -1 & 4\end{pmatrix}correct
DNone of these
Solution
tep 1: Identify the sum as a matrix geometric series
I+A+A2+⋯=(I−A)−1I + A + A^2 + \cdots = (I - A)^{-1}
 provided all eigenvalues of AA have modulus less than 11. Step 2: Verify finiteness of series The characteristic equation of AA gives ∣λ∣2=det⁡(A)=14|\lambda|^2 = \det(A) = \frac{1}{4}, so ∣λ∣=12<1|\lambda| = \frac{1}{2} < 1. The series finite Step 3: Compute I−AI - A and its inverse
I−A=(2−321212)I - A = \begin{pmatrix}2 & -\dfrac{3}{2}\\[4pt] \dfrac{1}{2} & \dfrac{1}{2}\end{pmatrix}
∣I−A∣=2⋅12−(−32)⋅12=1+34=74|I - A| = 2\cdot\frac{1}{2} - \left(-\frac{3}{2}\right)\cdot\frac{1}{2} = 1 + \frac{3}{4} = \frac{7}{4}
adj(I−A)=(1232−122)\text{adj}(I-A) = \begin{pmatrix}\dfrac{1}{2} & \dfrac{3}{2}\\[4pt] -\dfrac{1}{2} & 2\end{pmatrix}
(I−A)−1=47(1232−122)=27(13−14)(I-A)^{-1} = \frac{4}{7}\begin{pmatrix}\dfrac{1}{2} & \dfrac{3}{2}\\[4pt] -\dfrac{1}{2} & 2\end{pmatrix} = \frac{2}{7}\begin{pmatrix}1 & 3\\ -1 & 4\end{pmatrix}
 Answer: (3)
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