Question

If $X=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$, then value of $X^n$ is, (where n is natural number)

• A. $\left[\begin{array}{cc}3n& -4n\\ n& -n\end{array}\right]$
• B. $\left[\begin{array}{cc}2+n& 5-n\\ n& -n\end{array}\right]$
• C. 0
• D. $\left[\begin{array}{cc}2n+1& -4n\\ n& -(2n-1)\end{array}\right]$

Answer 1

The correct option is D $\left[\begin{array}{cc}2n+1& -4n\\ n& -(2n-1)\end{array}\right]$

Given $X=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$

$X^2=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$

$=\left[\begin{array}{cc}9-4& -12+4\\ 3-1& -4+1\end{array}\right]$

$=\left[\begin{array}{cc}5& -8\\ 2& -3\end{array}\right]$

Again $X^3=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]\left[\begin{array}{cc}5& -8\\ 2& -3\end{array}\right]$

$=\left[\begin{array}{cc}15-8& -24+12\\ 5-2& -8+3\end{array}\right]$

$=\left[\begin{array}{cc}7& -12\\ 3& -5\end{array}\right]$

$\therefore \left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right],\left[\begin{array}{cc}5& -8\\ 2& -3\end{array}\right],\left[\begin{array}{cc}7& -12\\ 3& -5\end{array}\right]$ are in A.P

Consider the terms $3,5,7$ whose first term is $3$ and common difference$=5-3=2$

$\therefore n^{th}$ term $T_n=a+(n-1)d=3+(n-1)2=2n+1$

$-4,-8,-12$ whose first term is $-4$ and common difference$=-8-(-4)=-4$

$\therefore n^{th}$ term $T_n=a+(n-1)d=-4+(n-1)(-4)=-4n$

$1,2,3$ whose first term is $1$ and common difference$=2-1=1$

$\therefore n^{th}$ term $T_n=a+(n-1)d=1+(n-1)1=n$

$-1,-3,-5$ whose first term is $-1$ and common difference$=-3-(-1)=-3+1=-2$

$\therefore n^{th}$ term $T_n=a+(n-1)d=-1+(n-1)\times -2=-(2n-1)$

$\therefore X^n=\left[\begin{array}{cc}2n+1& -4n\\ n& -(2n-1)\end{array}\right]$