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Question

If A = $open square brackets table row 1 0 row 1 1 end table close square brackets$ and I = $open square brackets table row 1 0 row 0 1 end table close square brackets$ , then which one of the following holds for all n $greater or equal than$ 1, by the principle of mathematical induction -

Aⁿ = nA – (n – 1) I
Aⁿ = 2^n–1 A – (n – 1) I
Aⁿ = nA + (n – I)I
Aⁿ = 2^n–1A + (n – 1) I

hint Hint:

$A equals open square brackets table row 1 0 row 1 1 end table close square brackets space A to the power of n equals n A minus left parenthesis n minus 1 right parenthesis I A squared equals open square brackets table row cell 1 x 1 plus 1 x 0 end cell cell 1 x 0 plus 1 x 0 end cell row cell 1 x 1 plus 1 x 1 end cell cell 1 x 0 plus 1 x 1 end cell end table close square brackets equals open square brackets table row 1 0 row 2 1 end table close square brackets A to the power of m plus 1 end exponent equals A to the power of m. A equals left parenthesis m A minus left parenthesis m minus 1 right parenthesis A right parenthesis A equals m A squared minus m A plus A equals m open square brackets table row 1 0 row 2 1 end table close square brackets minus m A plus open square brackets table row 1 0 row 1 1 end table close square brackets space equals open square brackets table row cell m plus 1 end cell 0 row cell m plus 1 end cell cell m plus 1 end cell end table close square brackets minus m space open square brackets table row 1 0 row 1 1 end table close square brackets space equals space left parenthesis m space plus space 1 right parenthesis A space – space m A space equals space A$

The correct answer is: Aⁿ = nA – (n – 1) I

Given A = $open square brackets table row 1 0 row 1 1 end table close square brackets$ and I = $open square brackets table row 1 0 row 0 1 end table close square brackets$ , then which one of the following holds for all n $greater or equal than$ 1, by the principle of mathematical induction.

Therefore, An = nA – (n – 1) I is the correct option.

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