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 $\underset{n\to \infty }{lim}(A^{-1}{)}^n$ is

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

Answer 1

The correct option is B null matrix

$A=\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)$
$\Rightarrow A^2=\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)=\left(\begin{array}{cc}{a}^2& 0\\ 0& b^2\end{array}\right)$
$\Rightarrow A^3=\left(\begin{array}{cc}{a}^2& 0\\ 0& b^2\end{array}\right)\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)=\left(\begin{array}{cc}{a}^3& 0\\ 0& b^3\end{array}\right)$
$⋮$
$A^n=\left(\begin{array}{cc}{a}^n& 0\\ 0& b^n\end{array}\right)$
$\Rightarrow (A^n{)}^{-1}=\frac{1}{a^n{b}^n}\left(\begin{array}{cc}{b}^n& 0\\ 0& a^n\end{array}\right)=\left(\begin{array}{cc}{a}^{-n}& 0\\ 0& b^{-n}\end{array}\right)$

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