Let A = begin{bmatrix} 0 & 1 1 & k end{bmatr | vedclass

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

Question

(B) Given $A = \begin{bmatrix} 0 & 1 \ 1 & k \end{bmatrix}$.First,calculate $A^2 = \begin{bmatrix} 0 & 1 \ 1 & k \end{bmatrix} \begin{bmatrix} 0 & 1

Vedclass · Accepted Answer

$74$

Vedclass · Answer

(B) Given $A = \begin{bmatrix} 0 & 1 \ 1 & k \end{bmatrix}$.First,calculate $A^2 = \begin{bmatrix} 0 & 1 \ 1 & k \end{bmatrix} \begin{bmatrix} 0 & 1 \ 1 & k \end{bmatrix} = \begin{bmatrix} 1 & k \ k & 1+k^2 \end{bmatrix}$.Next,calculate $A^3 = A^2 \cdot A = \begin{bmatrix} 1 & k \ k & 1+k^2 \end{bmatrix} \begin{bmatrix} 0 & 1 \ 1 & k \end{bmatrix} = \begin{bmatrix} k & 1+k^2 \ 1+k^2 & k+k(1+k^2) \end{bmatrix} = \begin{bmatrix} k & 1+k^2 \ 1+k^2 & k+k+k^3 \end{bmatrix} = \begin{bmatrix} k & 1+k^2 \ 1+k^2 & 2k+k^3 \end{bmatrix}$.We are given $d = 228$,so $2k + k^3 = 228$.By testing values,if $k = 6$,then $2(6) + 6^3 = 12 + 216 = 228$. Thus,$k = 6$.Now,$b = 1 + k^2 = 1 + 6^2 = 1 + 36 = 37$.Also,$c = 1 + k^2 = 1 + 36 = 37$.Therefore,$b + c = 37 + 37 = 74$.

Let $A = \begin{bmatrix} 0 & 1 \\ 1 & k \end{bmatrix}$,$k \in R$ and $A^3 = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. If $d = 228$,then $b + c =$

Similar Questions

If $A = \begin{bmatrix} 3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3 \end{bmatrix}$,then $A^3 = $ . . . . . . (in $A$)

Let $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$,$B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$,and $C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$. Find $A - B$.

If $A = \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix}$,then the value of $A^{40}$ is

If $A$ and $B$ are square matrices of order $2$,then $(A + B)^2 = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module