Question

Let $a$ and $b$ be two real numbers such that $a>1,b>1.$ If $A=\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right),$ then $\underset{n\to \infty }{lim}{A}^{-n}$ is

• A. unit matrix
• B. null matrix
• C. $2I,$ where $I$ is identity matrix of order $2$
• D. none of these

Answer 1

The correct option is B null matrix

$A=\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)$

$\Rightarrow A^2=\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)=\left(\begin{array}{cc}{a}^2& 0\\ 0& b^2\end{array}\right)$

$\Rightarrow A^3=\left(\begin{array}{cc}{a}^2& 0\\ 0& b^2\end{array}\right)\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)=\left(\begin{array}{cc}{a}^3& 0\\ 0& b^3\end{array}\right)$

$⋮⋮$
$A^n=\left(\begin{array}{cc}{a}^n& 0\\ 0& b^n\end{array}\right)$

$(A^n{)}^{-1}=\frac{1}{a^n{b}^n}\left(\begin{array}{cc}{b}^n& 0\\ 0& a^n\end{array}\right)=\left(\begin{array}{cc}{a}^{-n}& 0\\ 0& b^{-n}\end{array}\right)$

$\Rightarrow \underset{n\to \infty }{lim}(A^n{)}^{-1}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$ as $a>1$ and $b>1$