Find the value of the expression-

Question

Consider a matrix $A=\begin{bmatrix}3 & 1\\-6 & -2\end{bmatrix}$, then $(I+A)^{99}$ equals ?

So how can I expand this ?

The solution paper gives the answer as $I+(2^{99}-1)A$

Answer 1

Notice that $A^2=A$ so $A^k=A,\; k\ge1$ and then by the binomial formula we get

$$(I+A)^{99}=\sum_{k=0}^{99}\binom{99}{k}A^kI^{99-k}=I+A\sum_{k=1}^{99}\binom{99}{k}=I+(2^{99}-1)A$$