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Maths-

General

Easy

Question

If A = $open square brackets table row 1 0 row 1 1 end table close square brackets$ or $open square brackets table row 1 0 row 0 1 end table close square brackets$ , then which of the following holds of all n ≥ 1,

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

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

A = $open square brackets table row 1 0 row 1 1 end table close square brackets$ $rightwards double arrow$ Aⁿ = nA – (n – 1) I
$rightwards double arrow$ A = nA – (n – 1) A = A
A² = $open square brackets table row 1 0 row 2 1 end table close square brackets$
A^m+1 = A^m. A = (mA – (m– 1) A) A
= mA² – mA + A
= m $open square brackets table row 1 0 row 2 1 end table close square brackets$ – m A + $open square brackets table row 1 0 row 1 1 end table close square brackets$
= $open square brackets table row cell m plus 1 end cell 0 row cell m plus 1 end cell cell m plus 1 end cell end table close square brackets$ – m $open square brackets table row 1 0 row 0 1 end table close square brackets$
= (m + 1)A – mA = A
So (A) is true

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