Question

If\[A = \begin{bmatrix}a & b \\ 0 & 1\end{bmatrix}\], prove that\[A^n = \begin{bmatrix}a^n & b( a^n - 1)/a - 1 \\ 0 & 1\end{bmatrix}\] for every positive integer n .

Answer 1

We shall prove the result by the principle of mathematical induction on n.

Step 1: If n = 1, by definition of integral power of a matrix, we have

\[A^1 = \begin{bmatrix}a^1 & b\left( a^1 - 1 \right)/a - 1 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}a & b \\ 0 & 1\end{bmatrix} = A\]

So, the result is true for n = 1.

Step 2: Let the result be true for n = m. Then,

 

\[A^m = \begin{bmatrix}a^m & b\left( a^m - 1 \right)/a - 1 \\ 0 & 1\end{bmatrix}\]      ...(1)

Now, we shall show that the result is true for

\[n = m + 1\]

Here,

\[A^{m + 1} = \begin{bmatrix}a^{m + 1} & b\left( a^{m + 1} - 1 \right)/a - 1 \\ 0 & 1\end{bmatrix}\]

By definition of integral power of matrix, we have

\[A^{m + 1} = A^m A\]

\[ \Rightarrow A^{m + 1} = \begin{bmatrix}a^m & b\left( a^m - 1 \right)/a - 1 \\ 0 & 1\end{bmatrix}\begin{bmatrix}a & b \\ 0 & 1\end{bmatrix} \left[\text{From eq .} \left( 1 \right) \right]\]

 

 

\[ \Rightarrow A^{m + 1} = \begin{bmatrix}a^m a + 0 & \left\{ a^m b + b\left( a^m - 1 \right) \right\}/a - 1 \\ 0 + 0 & 0 + 1\end{bmatrix}\]

\[ \Rightarrow A^{m + 1} = \begin{bmatrix}a^{m + 1} & \left( a^{m + 1} b - a^m b + a^m b - b \right)/a - 1 \\ 0 & 1\end{bmatrix}\]

\[ \Rightarrow A^{m + 1} = \begin{bmatrix}a^{m + 1} & b\left( a^{m + 1} - 1 \right)/a - 1 \\ 0 & 1\end{bmatrix}\]

\[\]

This shows that when the result is true for n = m, it is also true for n = m +1.

Hence, by the principle of mathematical induction, the result is valid for any positive integer n.