A) \[\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} -1 & -1 \\ -1 & -1 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 2 & 2 \\ 2 & 2 \\ \end{matrix} \right]\]
D) None of these
Correct Answer: B
Solution :
We have,\[f(x)=\frac{1+x}{1-x}\] \[\therefore \] \[f(A)=\frac{I+A}{I-A}\] \[=(I+A){{(I-A)}^{-1}}\] \[=\left[ \begin{matrix} 2 & 2 \\ 2 & 2 \\ \end{matrix} \right]{{\left[ \begin{matrix} 0 & -2 \\ -2 & 0 \\ \end{matrix} \right]}^{-1}}\] \[\Rightarrow \]\[f(A)=\left[ \begin{matrix} 2 & 2 \\ 2 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & -1/2 \\ -1/2 & 0 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} -1 & -1 \\ -1 & -1 \\ \end{matrix} \right]\]
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