Q. Let $a$ and $b$ be two real numbers such that $a > 1\ ,\ b > 1.$ If $A = \ \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$, then ${\lim_{n \rightarrow \infty}A}^{- 1}$ is

A) Unit matrix
B) Null matrix
C) $2I$
D) None of these
Explanation

(b)
$A = \begin{pmatrix}
a & 0 \\
0 & b
\end{pmatrix}$
$\Rightarrow A^{2} = \begin{pmatrix}
a & 0 \\
0 & b
\end{pmatrix}\begin{pmatrix}
a & 0 \\
0 & b
\end{pmatrix} = \begin{pmatrix}
a^{2} & 0 \\
0 & b^{2}
\end{pmatrix}$
$\Rightarrow A^{3} = \begin{pmatrix}
a^{2} & 0 \\
0 & b^{2}
\end{pmatrix}\begin{pmatrix}
a & 0 \\
0 & b
\end{pmatrix} = \begin{pmatrix}
a^{3} & 0 \\
0 & b^{3}
\end{pmatrix}$
$\Rightarrow A^{n} = \begin{pmatrix}
a^{n} & 0 \\
0 & b^{n}
\end{pmatrix}$
$\Rightarrow \left( A^{n} \right)^{- 1} = \frac{1}{a^{n}b^{n}}\begin{pmatrix}
a^{n} & 0 \\
0 & b^{n}
\end{pmatrix} = \begin{pmatrix}
a^{- n} & 0 \\
0 & b^{- n}
\end{pmatrix}$
$\Rightarrow \lim_{n \rightarrow \infty}\left( A^{n} \right)^{- 1} = \begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}$ as $a > 1\$and$\ b > 1$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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