If A = begin{bmatrix} alpha - 1 0 0 end{bma | vedclass

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(C) Given $A = \begin{bmatrix} \alpha - 1 \ 0 \ 0 \end{bmatrix}$ and $B = \begin{bmatrix} \alpha + 1 \ 0 \ 0 \end{bmatrix}$.The transpose of matri

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$1$

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(C) Given $A = \begin{bmatrix} \alpha - 1 \ 0 \ 0 \end{bmatrix}$ and $B = \begin{bmatrix} \alpha + 1 \ 0 \ 0 \end{bmatrix}$.The transpose of matrix $B$ is $B^T = \begin{bmatrix} \alpha + 1 & 0 & 0 \end{bmatrix}$.Now,calculate the product $AB^T$:$AB^T = \begin{bmatrix} \alpha - 1 \ 0 \ 0 \end{bmatrix} \begin{bmatrix} \alpha + 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} (\alpha - 1)(\alpha + 1) & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} \alpha^2 - 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$.For $AB^T$ to be a non-zero matrix,at least one element must be non-zero.Thus,$\alpha^2 - 1 
eq 0$,which implies $\alpha^2 
eq 1$.Taking the square root on both sides,we get $|\alpha| 
eq 1$.

If $A = \begin{bmatrix} \alpha - 1 \\ 0 \\ 0 \end{bmatrix}$ and $B = \begin{bmatrix} \alpha + 1 \\ 0 \\ 0 \end{bmatrix}$ are two matrices,then $AB^T$ is a non-zero matrix for $|\alpha|$ not equal to:

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