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

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

Question

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

Vedclass · Accepted Answer

$A^2 + I = A(A^2 - I)$

Vedclass · Answer

(A) Given $A = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$.First,calculate $A^2$:$A^2 = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix} = -I$.Now,calculate higher powers:$A^3 = A^2 \cdot A = -I \cdot A = -A = \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix}$.$A^4 = (A^2)^2 = (-I)^2 = I$.Check the options:$A$. $A^2 + I = -I + I = 0$. $A(A^2 - I) = A(-I - I) = A(-2I) = -2A$. Since $0 
eq -2A$,this is incorrect.$B$. $A^4 - I = I - I = 0$. $A^2 + I = -I + I = 0$. Thus $0 = 0$ (Correct).$C$. $A^3 + I = -A + I$. $A(A^3 - I) = A(-A - I) = -A^2 - A = I - A$. Since $-A + I = I - A$,this is correct.$D$. $A^3 - I = -A - I$. $A(A - I) = A^2 - A = -I - A$. Since $-A - I = -I - A$,this is correct.Therefore,the statement in option $A$ is not correct.

If $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$,then which one of the following statements is not correct?

Similar Questions

$A$ 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:

If matrix $A = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$,then:

If $A = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}$ satisfies the equation $x^2 + 4x - p = 0$,then $p$ is equal to

If $AB = A$ and $BA = B$,then

For $A = \begin{bmatrix} 0 & 0 & -2 \\ 0 & -2 & 0 \\ -2 & 0 & 0 \end{bmatrix}$,which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module