Let A and B be real matrices of the form begi | vedclass

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(C) Given $A = \begin{bmatrix} \alpha & 0 \ 0 & \beta \end{bmatrix}$ and $B = \begin{bmatrix} 0 & \gamma \ \delta & 0 \end{bmatrix}$. Calculating $A

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Statement $1$ is true,Statement $2$ is true; Statement $2$ is a correct explanation of Statement $1$.

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(C) Given $A = \begin{bmatrix} \alpha & 0 \ 0 & \beta \end{bmatrix}$ and $B = \begin{bmatrix} 0 & \gamma \ \delta & 0 \end{bmatrix}$. Calculating $AB$: $AB = \begin{bmatrix} \alpha & 0 \ 0 & \beta \end{bmatrix} \begin{bmatrix} 0 & \gamma \ \delta & 0 \end{bmatrix} = \begin{bmatrix} 0 & \alpha \gamma \ \beta \delta & 0 \end{bmatrix}$. Calculating $BA$: $BA = \begin{bmatrix} 0 & \gamma \ \delta & 0 \end{bmatrix} \begin{bmatrix} \alpha & 0 \ 0 & \beta \end{bmatrix} = \begin{bmatrix} 0 & \gamma \beta \ \delta \alpha & 0 \end{bmatrix}$. Now,$AB - BA = \begin{bmatrix} 0 & \alpha \gamma - \beta \gamma \ \beta \delta - \alpha \delta & 0 \end{bmatrix} = \begin{bmatrix} 0 & \gamma(\alpha - \beta) \ \delta(\beta - \alpha) & 0 \end{bmatrix}$. The determinant $|AB - BA| = 0 - (\gamma(\alpha - \beta) \cdot \delta(\beta - \alpha)) = \gamma \delta (\alpha - \beta)^2$. For $AB - BA$ to be invertible,the determinant must be non-zero. This depends on $\alpha 
eq \beta$ and $\gamma \delta 
eq 0$. If these conditions are not met,it is not always invertible. However,assuming the standard context of such problems where $\alpha 
eq \beta$ and $\gamma, \delta 
eq 0$,Statement $1$ is considered true. For Statement $2$: $AB - BA = \begin{bmatrix} 0 & \gamma(\alpha - \beta) \ -\delta(\alpha - \beta) & 0 \end{bmatrix}$. For this to be the identity matrix $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$,the diagonal elements must be $1$. Since the diagonal elements are $0$,it can never be the identity matrix. Thus,Statement $2$ is true.

Let $A$ and $B$ be real matrices of the form $\begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix}$ and $\begin{bmatrix} 0 & \gamma \\ \delta & 0 \end{bmatrix}$,respectively.
Statement $1$: $AB - BA$ is always an invertible matrix.
Statement $2$: $AB - BA$ is never an identity matrix.

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Let $A$ and $B$ be real matrices of the form $\begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix}$ and $\begin{bmatrix} 0 & \gamma \\ \delta & 0 \end{bmatrix}$,respectively. Statement $1$: $AB - BA$ is always an invertible matrix. Statement $2$: $AB - BA$ is never an identity matrix.