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Matrices & Determinants — JEE Maths practice question

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

Question

A = \begin{pmatrix}-1 & \dfrac{3}{2}\\[4pt] -\dfrac{1}{2} & \dfrac{1}{2}\end{pmatrix}

, then

I + A + A^2 + \cdots\infty

is equal to:

\begin{pmatrix}1 & -3\\ 1 & 4\end{pmatrix}

\dfrac{2}{7}\begin{pmatrix}1 & -3\\ -1 & 4\end{pmatrix}

\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 + A^2 + \cdots = (I - A)^{-1}

provided all eigenvalues of

A

have modulus less than

1

. Step 2: Verify finiteness of series The characteristic equation of

A

gives

|\lambda|^2 = \det(A) = \frac{1}{4}

, so

|\lambda| = \frac{1}{2} < 1

. The series finite Step 3: Compute

I - A

and its inverse

I - A = \begin{pmatrix}2 & -\dfrac{3}{2}\\[4pt] \dfrac{1}{2} & \dfrac{1}{2}\end{pmatrix}

|I - A| = 2\cdot\frac{1}{2} - \left(-\frac{3}{2}\right)\cdot\frac{1}{2} = 1 + \frac{3}{4} = \frac{7}{4}

\text{adj}(I-A) = \begin{pmatrix}\dfrac{1}{2} & \dfrac{3}{2}\\[4pt] -\dfrac{1}{2} & 2\end{pmatrix}

(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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