The matrix A = begin{bmatrix} 1/sqrt{2} & 1/s | vedclass

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

Question

(C) To determine the type of matrix $A$,we calculate $A^2$: $A^2 = A \cdot A = \begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{2} \ -1/\sqrt{2} & -1/\sqrt{2} \e

Vedclass · Accepted Answer

Nilpotent

Vedclass · Answer

(C) To determine the type of matrix $A$,we calculate $A^2$: $A^2 = A \cdot A = \begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{2} \ -1/\sqrt{2} & -1/\sqrt{2} \end{bmatrix} \begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{2} \ -1/\sqrt{2} & -1/\sqrt{2} \end{bmatrix}$ $= \begin{bmatrix} (1/\sqrt{2})(1/\sqrt{2}) + (1/\sqrt{2})(-1/\sqrt{2}) & (1/\sqrt{2})(1/\sqrt{2}) + (1/\sqrt{2})(-1/\sqrt{2}) \ (-1/\sqrt{2})(1/\sqrt{2}) + (-1/\sqrt{2})(-1/\sqrt{2}) & (-1/\sqrt{2})(1/\sqrt{2}) + (-1/\sqrt{2})(-1/\sqrt{2}) \end{bmatrix}$ $= \begin{bmatrix} 1/2 - 1/2 & 1/2 - 1/2 \ -1/2 + 1/2 & -1/2 + 1/2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} = O$ Since $A^2 = O$,the matrix $A$ is a nilpotent matrix of order $2$.

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

Similar Questions

If $A = \begin{bmatrix} 3 & x-1 \\ 2x+3 & x+2 \end{bmatrix}$ is a symmetric matrix, then the value of $x$ is

If $A = \begin{bmatrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$,verify that $(A+B)^{\prime} = A^{\prime} + B^{\prime}$.

Let $A$ and $B$ be any two $3 \times 3$ matrices. If $A$ is symmetric and $B$ is skew-symmetric,then the matrix $AB - BA$ is

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & -1 & 4 \end{bmatrix}$, $A = B + C$, $B = B^T$ and $C = -C^T$, then $C = $

If $A$ and $B$ are square matrices of the same order,then which of the following properties holds true for the transpose of their product?

Vedclass Test Series

Exam Paper Generator

Online Exam Module