Question

Let $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$, show that ${(aI+bA)}^n=a^{n}I+n{a}^{n-1}bA$, where $I$ is the identity matrix of order $2$ and $n\in N$

Answer 1

It is given that $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$
To show: $P\left(n\right):{(aI+bA)}^n=A^{n}I+n{a}^{n-1}bA,n\in N$
We shall prove the result by using the principle of mathematical induction.
For $n=1$, we have:
$P\left(I\right):(aI+bA)=aI+b{a}^{0}A=aI+bA$
Therefore, the result is true for $n=k$.
That is,
$P\left(k\right):{(aI+bA)}^k=a^{k}I+k{a}^{k-1}bA$
Now, we prove that the result is true for $n=k+1$
Consider
${(aI+bA)}^{k+1}={(aI+bA)}^k(aI+bA)$
$=(a^{k}I+k{a}^{k-1}bA)(aI+bA)$
$=a^{k+1}I+k{a}^{k}bAI+a^{k}bIA+k{a}^{k-1}{b}^2{A}^2$
$=a^{k+1}I+(k+1)a^{k}bA+k{a}^{k-1}{b}^2{A}^2...........\left(1\right)$
Now,
$A^2=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0$
From (1), we have:
${(aI+bA)}^{k+1}=a^{k+1}I+(k+1)a^{k}bA+O$
$=a^{k+1}I+(k+1)a^{k}bA$
Therefore, the result is true for $n=k+1$.
Thus, by the principle of mathematical induction, we have:
${(aI+nA)}^n=a^{n}I+n{a}^{n-1}bA$
where $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],n\in N$