Question

If $A=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$, then $A^k$ is of the form $\left[\begin{array}{cc}1+ak& bk\\ ck& 1+dk\end{array}\right]$, where k is any positive integer, then $abcd$ is equal to

Answer 1

We have,
$A^2=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]=\left[\begin{array}{cc}5& -8\\ 2& -3\end{array}\right]=\left[\begin{array}{cc}1+2\times 2& -4\times 2\\ 2& 1-2\times 2\end{array}\right]$
and
$A^3=\left[\begin{array}{cc}5& -8\\ 2& -3\end{array}\right]\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]=\left[\begin{array}{cc}7& -12\\ 3& -5\end{array}\right]=\left[\begin{array}{cc}1+2\times 3& -4\times 3\\ 3& 1-2\times 3\end{array}\right]$
Thus, it is true for indices 2 and 3. Now, assume
$A^k=\left[\begin{array}{cc}1+2k& -4k\\ k& 1-2k\end{array}\right]$
Then,
$A^{k+1}=\left[\begin{array}{cc}1+2k& -4k\\ k& 1-2k\end{array}\right]\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$
$=\left[\begin{array}{cc}3+2k& -4(k+1)\\ k+1& -1-2k\end{array}\right]$
$=\left[\begin{array}{cc}1+2(k+1)& -4(k+1)\\ k+1& 1-2(k+1)\end{array}\right]$
Thus,
if the law is true for $A^k$, it is also true for $A^{k+1}$. But
it is true for k $=$ 2, 3. etc. Hence, by induction, the required
result follows.
Hence $abcd=16$