If $\text{A}$ is a $3\times 3$ matrix such that ${\text{A}}^2=\text{A}$ and $\text{I}$ is the identity matrix, then the value of $(\text{I}+\text{A}{)}^n$ is equal to:

Solution

1. Given that ${\text{A}}^2=\text{A}$, it follows by induction that ${\text{A}}^k=\text{A}$ for all positive integers $k\ge 1$.
2. Using the binomial expansion, we have $$\begin{array}{rl}(\text{I}+\text{A}{)}^n& =\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\text{A}}^k\\ & =\text{I}+\sum _{k=1}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\text{A}}^k\end{array}$$.
3. Substituting ${\text{A}}^k=\text{A}$ into the summation, we get $(\text{I}+\text{A}{)}^n=\text{I}+({\sum }_{k=1}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right))\text{A}$.
4. Since ${\sum }_{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)=2^n$, it follows that ${\sum }_{k=1}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)=2^n-1$.
5. Substituting this sum back into the expression yields $(\text{I}+\text{A}{)}^n=\text{I}+(2^n-1)\text{A}$.
6. Hence the answer is (A).