If a square matrix A is such that left(A^T-fr | vedclass

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

Question

(C) Given,$\left(A^T-\frac{1}{2} I\right)\left(A-\frac{1}{2} I\right) = \left(A^T+\frac{1}{2} I\right)\left(A+\frac{1}{2} I\right) = I$. Expanding bot

Vedclass · Accepted Answer

skew-symmetric matrix

Vedclass · Answer

(C) Given,$\left(A^T-\frac{1}{2} I\right)\left(A-\frac{1}{2} I\right) = \left(A^T+\frac{1}{2} I\right)\left(A+\frac{1}{2} I\right) = I$. Expanding both sides: $A^T A - \frac{1}{2} A^T - \frac{1}{2} A + \frac{1}{4} I = A^T A + \frac{1}{2} A^T + \frac{1}{2} A + \frac{1}{4} I$. Subtracting $A^T A + \frac{1}{4} I$ from both sides,we get: $-\frac{1}{2} A^T - \frac{1}{2} A = \frac{1}{2} A^T + \frac{1}{2} A$. Rearranging the terms: $0 = A^T + A$. Therefore,$A^T = -A$. This is the condition for a skew-symmetric matrix. Hence,$A$ is a skew-symmetric matrix.

If a square matrix $A$ is such that $\left(A^T-\frac{1}{2} I\right)\left(A-\frac{1}{2} I\right) = \left(A^T+\frac{1}{2} I\right)\left(A+\frac{1}{2} I\right) = I$,where $I$ is a unit matrix,then $A$ is

Similar Questions

If $A$ is a square matrix and $A + A^T$ is a symmetric matrix,then $A - A^T$ is:

If $A = \begin{bmatrix} -2 \\ 4 \\ 5 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 3 & -6 \end{bmatrix}$,verify that $(AB)^{\prime} = B^{\prime} A^{\prime}$.

If $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 3 & 2 \\ 3 & 4 & 5\end{array}\right]$,then $(A+A^T)(A-A^T)=$

If $A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$,then $A' = $ . . . . . . .

The matrix $A = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module