Question

If $A=\left[\begin{array}{cc}1& 0\\ \frac{1}{2}& 1\end{array}\right]$ and $B=\left[b_{ij}\right]$ is a matrix such that $A^n-A^2+I=B$ (where $n\in N$). If $2(b_{21}+b_{22})=7$ then, $n$ is

• A. $9$
• B. $5$
• C. $7$
• D. $11$

Answer 1

The correct option is C $7$

$A=\left[\begin{array}{cc}1& 0\\ \frac{1}{2}& 1\end{array}\right]\Rightarrow A^2=\left[\begin{array}{cc}1& 0\\ \frac{1}{2}& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ \frac{1}{2}& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$

$\Rightarrow A^3=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ \frac{1}{2}& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 3\left(\frac{1}{2}\right)& 1\end{array}\right]$

$\Rightarrow A^n=\left[\begin{array}{cc}1& 0\\ n\left(\frac{1}{2}\right)& 1\end{array}\right]$

$\Rightarrow A^n-A^2+I=\left[\begin{array}{cc}1& 0\\ \frac{n}{2}& 1\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=B$

$\Rightarrow b_{21}=\frac{n}{2}-1;b_{22}=1$

$\because 2(b_{21}+b_{22})=7$

$\Rightarrow n=7$