# Find the Formula for the Power of a Matrix

## Problem 383

Let
$A=\begin{bmatrix} 1 & 1 & 1 \\ 0 &0 &1 \\ 0 & 0 & 1 \end{bmatrix}$ be a $3\times 3$ matrix. Then find the formula for $A^n$ for any positive integer $n$.

## Proof.

We first compute several powers of $A$ and guess the general formula.
We have
\begin{align*}
A^2=\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
=\begin{bmatrix}
1 & 1 & 3 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix},
\end{align*}
\begin{align*}
A^3=A^2A=\begin{bmatrix}
1 & 1 & 3 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
=\begin{bmatrix}
1 & 1 & 5 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix},
\end{align*}
\begin{align*}
A^4=A^3A=\begin{bmatrix}
1 & 1 & 5 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}=\begin{bmatrix}
1 & 1 & 7 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}.
\end{align*}

From these computations, we guess the general formula of $A^n$ is
$A^n=\begin{bmatrix} 1 & 1 & 2n-1 \\ 0 &0 &1 \\ 0 & 0 & 1 \end{bmatrix}.$

We prove this formula by mathematical induction on $n$.
The base case $n=1$ follows from the definition of $A$.

Suppose that the formula is true for $n=k$.
We prove the formula for $n=k+1$.
We have
\begin{align*}
A^{k+1}&=A^{k}A\\
&=\begin{bmatrix}
1 & 1 & 2k-1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
&& \text{by the induction hypothesis}\\
&=\begin{bmatrix}
1 & 1 & 2k+1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}\\
&=\begin{bmatrix}
1 & 1 & 2(k+1)-1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}.
\end{align*}

Thus the formula holds for $n=k+1$.
Hence the formula is true for any positive integer $n$ by induction.

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