Question

Let

$A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$

Find a formula for $A^n$, whenever n is a positive integer.

Short answer

$A^n=\left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right)$

Step-by-step solution

Step 1

Given:

$A=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$

$A^n$ is the product of A with itself (n - 1) times (n factors in total)

A is a matrix Their product of A with itself is then also a 2 x 2 matrix.

The (i,j) th element of the product AB is the sum of the products of each element in the ith row of A with the corresponding element in the j th column of B .

Step 2

Letus first determine $A^2$

$A^2=AA=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]=\left[\begin{array}{l}1\left(1\right)+1\left(0\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\left(1\right)+1\left(1\right)\\ 0\left(1\right)+1\left(0\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\left(1\right)+1\left(1\right)\end{array}\right]=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]$

Next determine $A^3$

$A^3=A^{2}A=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]=\left[\begin{array}{l}1\left(1\right)+2\left(0\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\left(1\right)+2\left(1\right)\\ 0\left(1\right)+1\left(0\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\left(1\right)+1\left(1\right)\end{array}\right]=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]$

Next determine $A^4$

$A^4=A^{3}A=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]=\left[\begin{array}{l}1\left(1\right)+3\left(0\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\left(1\right)+3\left(1\right)\\ 0\left(1\right)+1\left(0\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\left(1\right)+1\left(1\right)\end{array}\right]=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]$

We then note that the matrix has the elements$A^n$on the diagonal, 0 in the bottom left corner and n in the upper right corner.

data-custom-editor="chemistry" $A^n=\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]$