Matrices — Class 12 Maths Solution

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We have, $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&1\\0&0\end{array}} \right]$and ${(aI + bA)^n} = {a^n}I + n{a^{n - 1}}bA$ ...(i)

For n = 1, ${(aI + bA)^1} + {a^1}I + 1{a^{1 - 1}}bA$
$\Rightarrow$ $al + bA = al + bA$

So, it is true for n = 1
Let us assume that (i) is true for n = k,
i.e.,

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

Then ${(aI + bA)^{k + 1}} = {(aI + bA)^k} + (aI + bA)$

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

$= {a^{k + 1}}I \times I + k{a^k}bAI + {a^k}bAI + k{a^{k - 1}}{b^2}A \cdot A$

$= {a^{k + 1}} + k{a^k}bA + {a^k}bA + k{a^{k - 1}}{b^2} \times O$

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

$\Rightarrow$

(i) is true for n = k + 1

Hence, by mathematical induction it is true for all n $\in$ N.