Question

If $A=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]$ and $f\left(x\right)=\frac{1+x}{1-x}$, then $f\left(A\right)$ is

• A. $\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}-1& -1\\ -1& -1\end{array}\right]$
• C. $\left[\begin{array}{cc}2& 2\\ 2& 2\end{array}\right]$
• D. $noneofthese$

Answer 1

The correct option is B $\left[\begin{array}{cc}-1& -1\\ -1& -1\end{array}\right]$

$f\left(A\right)=\frac{I+A}{I-A}$

$(I-A)f\left(A\right)=(I+A)$

multiplying both sides by $(I-A{)}^{-1}$

$\Rightarrow f\left(A\right)=(I+A)(I-A{)}^{-1}$

$I+A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}2& 2\\ 2& 2\end{array}\right]$

$I-A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}0& -2\\ -2& 0\end{array}\right]$

For inverse of $(I-A)$

$|I-A|=0-(-2)(-2)=-4$

$\mathrm{adj}(I-A)=\left[\begin{array}{cc}0& 2\\ 2& 0\end{array}\right]$

$(I-A{)}^{-1}=-\frac{1}{4}\left[\begin{array}{cc}0& 2\\ 2& 0\end{array}\right]$

$f\left(A\right)=\left[\begin{array}{cc}2& 2\\ 2& 2\end{array}\right]\times \left[\begin{array}{cc}0& -1/2\\ -1/2& 0\end{array}\right]$

$f\left(A\right)=\left[\begin{array}{cc}-1& -1\\ -1& -1\end{array}\right]$

Answer: option (B)