Question

If A = $\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ or $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, then which of the following holds for all n ³ 1, by

principle of mathematical induction

• A. Aⁿ = nA – (n– 1) I
• B. Aⁿ = 2^n–1 A– (n–1) I
• C. Aⁿ = nA + (n –1) I
• D. Aⁿ = 2^n–1 A+ (n–1) I

Answer 1

Answer: (A)

Aⁿ = nA – (n– 1) I

Answer 2

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

̃ Aⁿ = nA – (n –1)I

̃ A = nA – (n –1) A = A which is true

If A = $\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ ̃ A² = $\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$= A

Using Mathematical Induction,

A^m+1 = A^m. A= (mA –(m–1)I) A

= mA² –mA + A

= m $\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$–mA + $\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$

= m $\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$– m $\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$+ $\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ + $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$m – m $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

= $\begin{bmatrix} m - m + 1 + m & 0 \\ 2m - m + 1 + 0 & m - m + 1 + m \end{bmatrix}$– m $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

= $\begin{bmatrix} m + 1 & 0 \\ m + 1 & m + 1 \end{bmatrix}$– m $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

= (m +1)A –mA which is also true.

Thus choice (1) is true for both values of A.

If (1) is possible then (3) can’t be true. Again (2) and (4) are not possible (they have no symmetricity).

Choice (1) is correct.