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

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

Question

(B) Given $A = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix}$.To find $A^2$,we multiply $A$ by itself:$A^2 = A 	imes A = \begi

Vedclass · Accepted Answer

Unit matrix

Vedclass · Answer

(B) Given $A = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix}$.To find $A^2$,we multiply $A$ by itself:$A^2 = A 	imes A = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix}$Performing matrix multiplication:$A^2 = \begin{bmatrix} (-1)(-1) + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 0 \ 0 + 0 + 0 & 0 + (-1)(-1) + 0 & 0 + 0 + 0 \ 0 + 0 + 0 & 0 + 0 + 0 & 0 + 0 + (-1)(-1) \end{bmatrix}$$A^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$Thus,$A^2$ is the unit matrix (identity matrix) $I$.

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

Similar Questions

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

If $A = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$,then $A^{4}$ is equal to

The number of all possible matrices of order $3 \times 3$ with each entry $0$ or $1$ 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 $A = \begin{bmatrix} 1 & -2 \\ 4 & 5 \end{bmatrix}$ and $f(t) = t^2 - 3t + 7$,then $f(A) + \begin{bmatrix} 3 & 6 \\ -12 & -9 \end{bmatrix}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module