If A = begin{bmatrix} -1 & 0 0 & 2 end{bmatr | vedclass

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

Question

(A) Given that,$A = \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix} \dots (i)$ First,we calculate $A^2$: $A^2 = A \cdot A = \begin{bmatrix} -1 & 0 \ 0

Vedclass · Accepted Answer

$2A$

Vedclass · Answer

(A) Given that,$A = \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix} \dots (i)$ First,we calculate $A^2$: $A^2 = A \cdot A = \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix} \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} (-1)(-1) + (0)(0) & (-1)(0) + (0)(2) \ (0)(-1) + (2)(0) & (0)(0) + (2)(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 4 \end{bmatrix}$ Next,we calculate $A^3$: $A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 0 \ 0 & 4 \end{bmatrix} \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} (1)(-1) + (0)(0) & (1)(0) + (0)(2) \ (0)(-1) + (4)(0) & (0)(0) + (4)(2) \end{bmatrix} = \begin{bmatrix} -1 & 0 \ 0 & 8 \end{bmatrix}$ Now,calculate $A^3 - A^2$: $A^3 - A^2 = \begin{bmatrix} -1 & 0 \ 0 & 8 \end{bmatrix} - \begin{bmatrix} 1 & 0 \ 0 & 4 \end{bmatrix} = \begin{bmatrix} -1-1 & 0-0 \ 0-0 & 8-4 \end{bmatrix} = \begin{bmatrix} -2 & 0 \ 0 & 4 \end{bmatrix}$ Since $A = \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix}$,we can see that $2A = 2 \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 0 \ 0 & 4 \end{bmatrix}$. Therefore,$A^3 - A^2 = 2A$.

If $A = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}$,then $A^3 - A^2$ is equal to

Similar Questions

If $A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$,then for all $n \in N$,find $A^n$.

If $A$ is a matrix of order $m \times n$ and $B$ is a matrix such that $AB^{\prime}$ and $B^{\prime}A$ are both defined,then the order of the matrix $B$ is

Compute the indicated product: $\begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$

Let $M$ and $N$ be two matrices over $\mathbb{R}$ of order $2$. Then,$MN = NM$ if .......

If $A = \begin{vmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{vmatrix}$,then $1 + A^2 =$ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module