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
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
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