Q1.    Let $A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$, show that $(aI + bA)^n = a^n I + na^{n-1} bA$, where I is the identity matrix of order 2 and $n \in N$.

Answers (1)
S seema garhwal

Given :

$A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$

To prove :   $(aI + bA)^n = a^n I + na^{n-1} bA$

For  n=1, $aI + bA = a I + a^{0} bA =a I + bA$

The result is true for  n=1.

Let result be true for n=k,

$(aI + bA)^k = a^k I + ka^{k-1} bA$

Now, we prove that the result is true for n=k+1,

$(aI + bA)^k^+^1 = (aI + bA)^k^(aI + bA)$

$= (a^k I + ka^{k-1} bA)$$(aI + bA)$

$=a^{k+1}I+Ka^{k}bAI+a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$=a^{k+1}I+(k+1)a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$A^{2} = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}=0$

Put the value of $A^{2}$ in above equation,

$(aI + bA)^k^+^1$$=a^{k+1}I+(k+1)a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$(aI + bA)^k^+^1$$=a^{k+1}I+(k+1)a^{k}bAI+0$

$=a^{k+1}I+(k+1)a^{k}bAI$

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

Thus, we have $(aI + bA)^n = a^n I + na^{n-1} bA$  where $A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$,$n \in N$.

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