Question

Let A=$\begin{bmatrix} 0 & 1 \\\ 0 & 0 \end{bmatrix}$,show that(aI+bA)n=anI+nan-1bA,where I is the identity matrix of order 2 and n∈N

Answer 1

It is given that A= $\begin{bmatrix} 0 & 1 \\\ 0 & 0 \end{bmatrix}$

To show: (aI+bA)n=anI+nan-1bA
We shall prove the result by using the principle of mathematical induction.

For n = 1, we have:
P(1):(aI+bA)=aI+ba0A=aI+bA

Therefore, the result is true for n =1.
Let the result be true for n = k.
That is,
P(k):(aI+bA)k=akI+kak-1bA
Now, we prove that the result is true for n = k + 1.
Consider:
(aI+bA)k+1=(aI+bA)k(aI+bA)
=(akI+kak-1bA)(aI+bA)
=ak+1+kakbAI+akbIA+kak-1b2A2
=ak+1I+(k+1)akbA+kak-1b2A2 ...(1)

Now A2=\begin{bmatrix} 0 & 1 \\\ 0 & 0 \end{bmatrix}$$\begin{bmatrix} 0 & 1 \\\ 0 & 0 \end{bmatrix}= $\begin{bmatrix} 0 & 0 \\\ 0 & 0 \end{bmatrix}$ = O

From (1), we have:
(aI+bA)k+1=ak+1I+(k+1)akbA+O
=ak+1I+(k+1)akbA

Therefore, the result is true for n = k + 1.

Thus, by the principle of mathematical induction, we have:

(aI+bA)n=anI+nan-1bA where A=$\begin{bmatrix} 0 & 1 \\\ 0 & 0 \end{bmatrix}$,n∈N