1. Key Concepts and Formulas

• Determinant of an Upper Triangular Matrix: The determinant of an upper triangular matrix (where all elements below the main diagonal are zero) is the product of its diagonal elements. This property significantly simplifies determinant calculation for such matrices.
• Property of Determinant of a Power: For any square matrix $A$ and any positive integer $k$, the determinant of $A^k$ is equal to the $k$-th power of the determinant of $A$. Mathematically, $\left| {{A^k}} \right| = {\left| A \right|^k}$.

2. Step-by-Step Solution

Step 1: Calculate the Determinant of Matrix $A$

• What we are doing: Our first goal is to find an expression for the determinant of the given matrix $A$ in terms of the variable $\alpha$.
• Why this step? The problem provides a condition involving $\left| {{A^2}} \right|$. To effectively use this condition, we first need to determine $\left| A \right|$.
• The Math: The given matrix is: $A = \begin{vmatrix} 5 & {5\alpha} & \alpha \\ 0 & \alpha & {5\alpha} \\ 0 & 0 & 5 \end{vmatrix}$
• Reasoning: Observe that matrix $A$ is an upper triangular matrix. This is because all elements below the main diagonal (i.e., $a_{21}=0$, $a_{31}=0$, $a_{32}=0$) are zero. This observation is crucial as it allows for a much simpler calculation of the determinant compared to cofactor expansion. According to the property of determinants for triangular matrices, its determinant is simply the product of its diagonal elements. The diagonal elements of $A$ are $5$, $\alpha$, and $5$. Therefore, the determinant of $A$ is: $|A| = (5) \times (\alpha) \times (5)$ $|A| = 25\alpha$

Step 2: Utilize the Given Condition $\left| {{A^2}} \right| = 25$

• What we are doing: We will apply a fundamental property of determinants to relate $\left| {{A^2}} \right|$ to $\left| A \right|$.
• Why this step? This is the primary equation provided in the problem statement that links the matrix $A$ to a numerical value, allowing us to eventually solve for $\alpha$.
• The Math: We are given the condition $\left| {{A^2}} \right| = 25$. We use the determinant property that for any square matrix $A$ and positive integer $k$, $\left| {{A^k}} \right| = {\left| A \right|^k}$. Applying this property for $k=2$: $\left| {{A^2}} \right| = {\left| A \right|^2}$ Now, substitute the given value $\left| {{A^2}} \right| = 25$ into this equation: ${\left| A \right|^2} = 25$

Step 3: Solve for $|A|$

• What we are doing: We will solve the equation obtained in Step 2 to find the possible numerical values for $|A|$.
• Why this step? We now have a numerical equation for $|A|$. By equating this with the algebraic expression for $|A|$ (from Step 1), we can form an equation to solve for $\alpha$.
• The Math: We have the equation ${\left| A \right|^2} = 25$. To find the value of $|A|$, we take the square root of both sides: $|A| = \pm\sqrt{25}$ $|A| = \pm 5$

Step 4: Determine the Value of $|\alpha|$

• What we are doing: We will equate the algebraic expression for $|A|$ (from Step 1) with the numerical values for $|A|$ (from Step 3) to solve for $\alpha$.
• Why this step? This combines our derived expression for $|A|$ in terms of $\alpha$ with the numerical value obtained from the given condition, directly leading to the solution for $\alpha$.
• The Math: From Step 1, we found that $|A| = 25\alpha$. From Step 3, we found that $|A| = \pm 5$. Equating these two expressions for $|A|$: $25\alpha = \pm 5$ Now, divide by $25$ to solve for $\alpha$: $\alpha = \frac{\pm 5}{25}$ $\alpha = \pm \frac{1}{5}$ The question specifically asks for the value of $\left| \alpha \right|$. Taking the absolute value of $\alpha$: $\left| \alpha \right| = \left| \pm \frac{1}{5} \right|$ $\left| \alpha \right| = \frac{1}{5}$

3. Common Mistakes & Tips

• Forgetting the $\pm$ sign: When solving an equation of the form $x^2 = k$, remember that $x = \pm\sqrt{k}$. Although the final answer $|\alpha|$ makes the sign irrelevant, intermediate steps must be correct for full understanding.
• Miscalculating the determinant of a triangular matrix: Always double-check that you are multiplying only the elements on the main diagonal. This shortcut is a time-saver but requires correct identification of the matrix type.
• Reading the question carefully: Pay attention to what is asked. Here, it was $|\alpha|$, not $\alpha$. This distinction can be crucial in multiple-choice questions.

4. Summary

This problem is a straightforward application of fundamental determinant properties, commonly tested in JEE. The solution involves two main steps: first, efficiently calculating the determinant of the given upper triangular matrix, $A$, by multiplying its diagonal elements to get $|A| = 25\alpha$. Second, using the determinant property $\left| {{A^k}} \right| = {\left| A \right|^k}$ to relate the given condition $\left| {{A^2}} \right| = 25$ to $|A|$, which yields $|A| = \pm 5$. Equating these two expressions for $|A|$ allows us to solve for $\alpha = \pm 1/5$. Finally, taking the absolute value as requested by the question gives $\left| \alpha \right| = 1/5$. Mastering these basic properties is essential for competitive exams like JEE.

5. Final Answer

The final answer is $\boxed{1/5}$ which corresponds to option (A).