Question

Let $A=\left[\begin{array}{cc}2& 1\\ 0& 3\end{array}\right]$ be a matrix. If $A^{10}=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ then

• A. number of divisors of $a$ is $11$
• B. $(a+c+d)$ is an even integer
• C. $(a+c+d)$ is an odd integer
• D. $a+d$ is a multiple of $13$

Answer 1

Answer: (A), (C), (D)

number of divisors of $a$ is $11$
$a+d$ is a multiple of $13$
$(a+c+d)$ is an odd integer

$A=\left[\begin{array}{cc}2& 1\\ 0& 3\end{array}\right]A^2=A\times A=\left[\begin{array}{cc}2& 1\\ 0& 3\end{array}\right]\times \left[\begin{array}{cc}2& 1\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}4+0& 2+3\\ 0+0& 0+9\end{array}\right]=\left[\begin{array}{cc}4& 5\\ 0& 9\end{array}\right]$

$A^4=A^2\times A^2=\left[\begin{array}{cc}4& 5\\ 0& 9\end{array}\right]\times \left[\begin{array}{cc}4& 5\\ 0& 9\end{array}\right]=\left[\begin{array}{cc}16+0& 20+45\\ 0+0& 0+81\end{array}\right]=\left[\begin{array}{cc}16& 65\\ 0& 81\end{array}\right]$

$A^8=A^4\times A^4=\left[\begin{array}{cc}16& 65\\ 0& 81\end{array}\right]\times \left[\begin{array}{cc}16& 65\\ 0& 81\end{array}\right]=\left[\begin{array}{cc}256+0& 1040+5265\\ 0+0& 0+6561\end{array}\right]=\left[\begin{array}{cc}256& 6305\\ 0& 6561\end{array}\right]$

$A^{10}=A^8\times A^2=\left[\begin{array}{cc}256& 6305\\ 0& 6561\end{array}\right]\times \left[\begin{array}{cc}4& 5\\ 0& 9\end{array}\right]=\left[\begin{array}{cc}1024+0& 1280+56745\\ 0+0& 0+59049\end{array}\right]=\left[\begin{array}{cc}1024& 58025\\ 0& 59049\end{array}\right].$

$\therefore a=1024;b=58025;c=0;d=59049$

Divisors of $a=1024=\{1,2,4,8,16,32,64,128,256,512,1024\}$. Number of divisors $=11$.

$a+c+d=1024+0+59049=60073$ which is an odd integer.

$a+d=1024+59049=60073$ which can be written as $13\times 4621$. So, $a+d$ is a multiple of $13$.

Hence, options $A,C,D$ are correct.