Q. If $A\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$ and $f(x) = \frac{1 + x}{1 - x}$, then $f(A)$ is

A) $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$
B) $\begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$
C) $\begin{bmatrix} - 1 & - 1 \\ - 1 & - 1 \end{bmatrix}$
D) None of these
Explanation

(c)
$f(x) = \frac{1 + x}{1 - x}$
$\Rightarrow (1 - x)f(x) = 1 + x$
$\Rightarrow (I - A)f(A) = (I + A)$
$\Rightarrow f(A) = (I - A)^{- 1}(I + A)$
$= \left( \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} - \begin{bmatrix}
1 & 2 \\
2 & 1
\end{bmatrix} \right)^{- 1}\left( \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} + \begin{bmatrix}
1 & 2 \\
2 & 1
\end{bmatrix} \right)$
$\Rightarrow f(A) = \left( \begin{bmatrix}
0 & - 2 \\
- 2 & 0
\end{bmatrix}^{- 1}\begin{bmatrix}
2 & 2 \\
2 & 2
\end{bmatrix} \right)$
$= \frac{\begin{bmatrix}
0 & 2 \\
2 & 0
\end{bmatrix}\begin{bmatrix}
2 & 2 \\
2 & 2
\end{bmatrix}}{- 4}$
$= \frac{\begin{bmatrix}
4 & 4 \\
4 & 4
\end{bmatrix}}{- 4}$
$= \begin{bmatrix}
- 1 & - 1 \\
- 1 & - 1
\end{bmatrix}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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