Question

If $A=\left[\begin{array}{cc}0& 1\\ 2& 0\end{array}\right]$ and $(A^8+A^6+A^4+A^2+I)V=\left[\begin{array}{c}31\\ 62\end{array}\right]$.
(Where I is the $(2\times 2)$ identity matrix), then the matrix of elements of matrix $V$ is

• A. $V=\left[\begin{array}{c}31\\ 62\end{array}\right]{\left[\begin{array}{c}310\\ 031\end{array}\right]}^{-1}$
• B. $I$
• C. $V=\left[\begin{array}{c}31\\ 62\end{array}\right]\left[\begin{array}{c}310\\ 031\end{array}\right]$
• D. None of these

Answer 1

Answer: (A)

$V=\left[\begin{array}{c}31\\ 62\end{array}\right]{\left[\begin{array}{c}310\\ 031\end{array}\right]}^{-1}$

$A=\left[\begin{array}{c}01\\ 20\end{array}\right]$

$A^2=\left[\begin{array}{c}20\\ 02\end{array}\right]$

$A^4=\left[\begin{array}{c}40\\ 04\end{array}\right]$

$A^6=\left[\begin{array}{c}80\\ 08\end{array}\right]$

$A^8=\left[\begin{array}{c}160\\ 016\end{array}\right]$

According to question,

$(A^8+A^6+A^4+A^2+I)V=\left[\begin{array}{c}31\\ 62\end{array}\right]$.

$(\left[\begin{array}{c}160\\ 016\end{array}\right]+\left[\begin{array}{c}80\\ 08\end{array}\right]+\left[\begin{array}{c}40\\ 04\end{array}\right]+\left[\begin{array}{c}20\\ 02\end{array}\right]+\left[\begin{array}{c}10\\ 01\end{array}\right])\times V=\left[\begin{array}{c}31\\ 62\end{array}\right]$

$V=\left[\begin{array}{c}31\\ 62\end{array}\right]{\left[\begin{array}{c}310\\ 031\end{array}\right]}^{-1}$