If A = begin{bmatrix} 1 & 0 1 & 1 end{bmatri | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $A = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix}$.Calculate $A^2 = A 	imes A = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} \begin{bmatrix

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given $A = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix}$.Calculate $A^2 = A 	imes A = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 2 & 1 \end{bmatrix}$.Calculate $A^3 = A^2 	imes A = \begin{bmatrix} 1 & 0 \ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 3 & 1 \end{bmatrix}$.By observation,$A^n = \begin{bmatrix} 1 & 0 \ n & 1 \end{bmatrix}$ for all $n \ge 1$.Now,evaluate $nA - (n - 1)I$:$nA = n \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} n & 0 \ n & n \end{bmatrix}$.$(n - 1)I = (n - 1) \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} n - 1 & 0 \ 0 & n - 1 \end{bmatrix}$.$nA - (n - 1)I = \begin{bmatrix} n - (n - 1) & 0 - 0 \ n - 0 & n - (n - 1) \end{bmatrix} = \begin{bmatrix} 1 & 0 \ n & 1 \end{bmatrix}$.Since $A^n = \begin{bmatrix} 1 & 0 \ n & 1 \end{bmatrix}$,it follows that $A^n = nA - (n - 1)I$.

If $A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,then which one of the following holds for all $n \ge 1$ (by the principle of mathematical induction)?

Similar Questions

Let $S = \{ m \in \mathbb{Z} : A^{m^2} + A^m = 3I - A^{-6} \}$,where $A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}$. Then $n(S)$ is equal to

The number of $3 \times 2$ matrices $A$,which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^{T}A$ is $5$,is . . . . . . .

$A$ determinant is chosen at random from the set of all determinants of order $2$ with elements $0$ or $1$ only. The probability that the value of the chosen determinant is positive is:

Let $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$ and $B$ be two matrices such that $A^{100} = 100B + I$. Then the sum of all the elements of $B^{100}$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module