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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 its diagonal elements:$|A| = k 	imes \alpha 	imes k = \alpha k^2$.We are given that $|A^2| = k^2$.Using the property of determinants $|A^2| = |A|^2$, we have:$|A|^2 = k^2$.Substituting the value of $|A|$:$(\alpha k^2)^2 = k^2$.$\alpha^2 k^4 = k^2$.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 on both sides:$|\alpha| = \sqrt{\frac{1}{k^2}} = \frac{1}{k}$.

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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