Question

If `A=[[1,1],[0,1]] ,` Prove that `A=[[1,n],[0,1]]` for all positive integers 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 powers of matrix, we have

\[A^1 = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix} = A\]
\[\]

So, the result is true for n = 1.
Step 2: Let the result be true for n = m. Then,

\[A^m = \begin{bmatrix}1 & m \\ 0 & 1\end{bmatrix}\] ... (1)

Now, we shall show that the result is true for

\[n = m + 1\]

Here,

\[A^{m + 1} = \begin{bmatrix}1 & m + 1 \\ 0 & 1\end{bmatrix}\] 

By definition of integral power of matrix, we have

\[A^{m + 1} = A^m A\]
 = \begin{bmatrix}1 & m \\ 0 & 1\end{bmatrix}\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}

 [From eq .(1)]
= \begin{bmatrix}1 + 0 & 1 + m \\ 0 + 0 & 0 + 1\end{bmatrix}
= \begin{bmatrix}1 & 1 + m \\ 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.