Question

If $A=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]$ and $f\left(x\right)=\frac{1+x}{1-x}$ find the value of $xf\left(A\right)$.

• 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. None of these

Answer 1

Answer: (B)

$\left[\begin{array}{cc}-1& -1\\ -1& -1\end{array}\right]$

We have, $f\left(x\right)=\frac{1+x}{1-x}$
Therefore, $f\left(A\right)=\frac{I+A}{I-A}$
$=(I+A)(I-A{)}^{-1}$
$\left[\begin{array}{cc}2& 2\\ 2& 2\end{array}\right]{\left[\begin{array}{cc}0& -2\\ -2& 0\end{array}\right]}^{-1}$
Thus $f\left(A\right)=\left[\begin{array}{cc}2& 2\\ 2& 2\end{array}\right]\left[\begin{array}{cc}0& -1/2\\ 1/2& 0\end{array}\right]$
$=\left[\begin{array}{cc}-1& -1\\ -1& -1\end{array}\right]$