Question

If $A=\left(\begin{array}{cc}1& 0\\ 1& 2\end{array}\right)$ Find $A^8=$

• A. $255A-254I$.
• B. $256A-255I$.
• C. $255A-255I$.
• D. None of these.

Answer 1

The correct option is A $255A-254I$.

$A=\left(\begin{array}{cc}1& 0\\ 1& 2\end{array}\right)$
$\Rightarrow A^2=\left(\begin{array}{cc}1& 0\\ 1& 2\end{array}\right)\left(\begin{array}{cc}1& 0\\ 1& 2\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 3& 4\end{array}\right)$
Now, $3A-2I=\left(\begin{array}{cc}1& 0\\ 1& 2\end{array}\right)-2\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$

$=\left(\begin{array}{cc}3-2& 0\\ 3-0& 6-2\end{array}\right)$

$=\left(\begin{array}{cc}1& 0\\ 3& 4\end{array}\right)$
Hence, $A^2=3A-2I$
Now,
$A^4=(A^2{)}^2=(3A-2I)2$
$=9A^2-12AI+4I^2$
$=9A^2-12A+4I$
$=9(3A-2I)-12A+4I$
$=15A-14I$
$\therefore A^8=(A^4{)}^2=(15A-14I{)}^2$
$=225A^2-420AI+196I^2$
$=225(3A-2I)-420A+196I$
$=255A-254I$