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(D) Given the matrix $A = \begin{bmatrix} k & k\alpha & \alpha \ 0 & \alpha & k\alpha \ 0 & 0 & k \end{bmatrix}$.Since $A$ is an upper triangular ma

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$\frac{1}{k}$

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(D) Given the matrix $A = \begin{bmatrix} k & k\alpha & \alpha \ 0 & \alpha & k\alpha \ 0 & 0 & k \end{bmatrix}$.Since $A$ is an upper triangular matrix,its determinant is the product of the diagonal elements:$|A| = k 	imes \alpha 	imes k = \alpha k^2$.We are given that the determinant of $A^2$ is $k^2$. Using the property $|A^2| = |A|^2$,we have:$|A|^2 = k^2$.Substituting $|A| = \alpha k^2$ into the equation:$(\alpha k^2)^2 = k^2$.$\alpha^2 k^4 = k^2$.Since $k > 1$,we can divide both sides by $k^4$:$\alpha^2 = \frac{k^2}{k^4} = \frac{1}{k^2}$.Taking the square root of both sides:$|\alpha| = \sqrt{\frac{1}{k^2}} = \frac{1}{|k|}$.Since $k > 1$,$|k| = k$,therefore $|\alpha| = \frac{1}{k}$.

If $k > 1$ and the determinant of the matrix $A^2$,where $A = \begin{bmatrix} k & k\alpha & \alpha \\ 0 & \alpha & k\alpha \\ 0 & 0 & k \end{bmatrix}$,is $k^2$,then $|\alpha|$ is equal to

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