If A^{prime}=begin{bmatrix} 3 & 4 -1 & 2 0 | vedclass

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

Question

(A) Given $A^{\prime} = \begin{bmatrix} 3 & 4 \ -1 & 2 \ 0 & 1 \end{bmatrix}$,we find $A = (A^{\prime})^{\prime} = \begin{bmatrix} 3 & -1 & 0 \ 4 &

Vedclass · Accepted Answer

Vedclass · Answer

(A) Given $A^{\prime} = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix}$,we find $A = (A^{\prime})^{\prime} = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix}$.
Now,$A-B = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix} = \begin{bmatrix} 4 & -3 & -1 \\ 3 & 0 & -2 \end{bmatrix}$.
Therefore,$(A-B)^{\prime} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$.
Next,$B^{\prime} = \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix}$.
Then,$A^{\prime}-B^{\prime} = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$.
Since $(A-B)^{\prime} = A^{\prime}-B^{\prime} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$,the property is verified.

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

Similar Questions

$P$ is a $3 \times 3$ square matrix and $\operatorname{Tr}(P) \neq 0$. If $\operatorname{Tr}(P-P^{T})+\operatorname{Tr}(P+P^{T})+\frac{\operatorname{Tr}(P)}{\operatorname{Tr}(P^T)}+\operatorname{Tr}(P) \times \operatorname{Tr}(P^{T})=0$, then $\operatorname{Tr}(P)=$

If $A$ is a symmetric matrix and $n \in N$,then $A^n$ is

If $A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$,then $A A^T$ is a

If $A$ is a square matrix, then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module