If I is the identity matrix of order 2 and A | vedclass

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(A) Given $A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$.We calculate the powers of $A$:$A^2 = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \begin

Vedclass · Accepted Answer

$A^n = nA - (n-1)I$

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(A) Given $A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$.We calculate the powers of $A$:$A^2 = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix}$$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \ 0 & 1 \end{bmatrix}$By observation, we hypothesize that $A^n = \begin{bmatrix} 1 & n \ 0 & 1 \end{bmatrix}$.Now, evaluate the expression $nA - (n-1)I$:$nA - (n-1)I = n \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} - (n-1) \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$$= \begin{bmatrix} n & n \ 0 & n \end{bmatrix} - \begin{bmatrix} n-1 & 0 \ 0 & n-1 \end{bmatrix}$$= \begin{bmatrix} n - (n-1) & n - 0 \ 0 - 0 & n - (n-1) \end{bmatrix} = \begin{bmatrix} 1 & n \ 0 & 1 \end{bmatrix} = A^n$.Thus, $A^n = nA - (n-1)I$ is correct.

If $I$ is the identity matrix of order $2$ and $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$, then for $n \geq 1$, mathematical induction gives:

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