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

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

Question

(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

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

Vedclass · Answer

(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:

Similar Questions

Let $A = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$. Then the number of $3 \times 3$ matrices $B$ with entries from the set $\{1, 2, 3, 4, 5\}$ and satisfying $AB = BA$ is $....$

If $\left[\begin{array}{cc}x-1 & 2y \\ x+y & 3\end{array}\right]=\left[\begin{array}{cc}3x-7 & y^2-3 \\ 6 & y\end{array}\right]$,then $\{(x, y)\} = $ . . . . . .

If $A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$ is such that $A^{2} = I$,then

If $A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 4 & 0 \\ 1 & -2 \\ 0 & 3 \end{bmatrix}$,then $AB =$ . . . . . . .

If $A = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}$,then $A^3 - A^2$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module