Do I have to multiply it whole? Given that $$A=\begin{bmatrix}4 & 1\\ -9 & -2 \end{bmatrix}$$
and $$A^{100}=\begin{bmatrix}a & b\\ c & d \end{bmatrix}$$
What is $a$?
I tried to multiply it again and again but it seems lengthy. Is there a shorter method?
 A: I would advise you to try and google ''matrix diagonalization''. That's a method to write a matrix in a form of: $$A^n=P*D^n*P^{-1}$$ where $D$ is a diagonal matrix.
A: Uhh. Did you notice something like a series for $a$? It goes like 
$$4,7,10,13......$$
Compare it with an AP of first term=$4$, Common difference=$3$.
Thus $a=4+99*3$ which makes it $301$
A: From
$$
A^{k+1} = A^k A
$$
you have the recurrences
$$
a_1 = 4 \\
b_1 = 1 \\
a_{k+1} = 4 a_k - 9 b_k \\
b_{k+1} = a_k - 2 b_k
$$
for the upper elements of $A^{k+1}$.
Then
$$
2 a_{k+1} = 8 a_k - 18 b_k = - a_k + 9 b_{k+1} \\
9 b_{k+1} = 2 a_{k+1} + a_k
$$
and
$$
a_{k+2} 
= 4 a_{k+1} - 9 b_{k+1}
= 2 a_{k+1} - a_k
$$
which is an order $2$ homogeneous linear recurrence which we now can solve by the usual algorithm.
The characteristic polynomial is
$$
p(t) = t^2 - 2 t + 1 = (t - 1)^2
$$
with double root $r=1$ and solution
$$
a_n = k_1 r^n + k_2 n r^n = k_1 + k_2 n
$$
From the initial known sequence elements we have
$$
a_1 = k_1 + k_2 = 4 \\
a_2 = k_1 + 2 k_2 = 7
$$
so $k_2 = 3$ and $k_1 = 1$ which gives
$$
a_n = 1 + 3n
$$
In particular
$$
a_{100} = 1 + 3\cdot 100 = 301
$$
A: *

*Jewar, you give the green chevron to a proof which is not a proof; understanding mathematics is a long way...

*The good question would be: what is $A^{100}$ ?
Indeed $A=I_2+U$ where $U^2=0_2$. Then $A^k=I_2+k U$.
