Let `A=[(5,5alpha,alpha),(0,alpha,5alpha),(0,0,5)]` If `|A^2| =25` , then `|alpha|` equals :

Options

A. `5^2`
B. 1
C. `1/5`
D. 5

Answer

To solve the problem, we need to find the value of \(|\alpha|\) given that the determinant of the matrix \(A^2\) is equal to 25. The matrix \(A\) is given as:

\[ A = \begin{pmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{pmatrix} \]

### Step 1: Calculate \(A^2\)

To find \(A^2\), we multiply matrix \(A\) by itself:

\[ A^2 = A \cdot A = \begin{pmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{pmatrix} \cdot \begin{pmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{pmatrix} \]

### Step 2: Perform the multiplication

Calculating each element of \(A^2\):

- **First row:**
- First column: \(5 \cdot 5 + 5\alpha \cdot 0 + \alpha \cdot 0 = 25\)
- Second column: \(5 \cdot 5\alpha + 5\alpha \cdot \alpha + \alpha \cdot 0 = 25\alpha + 5\alpha^2\)
- Third column: \(5 \cdot \alpha + 5\alpha \cdot 5\alpha + \alpha \cdot 5 = \alpha + 25\alpha^2 + 5\alpha = 30\alpha^2 + \alpha\)

- **Second row:**
- First column: \(0 \cdot 5 + \alpha \cdot 0 + 5\alpha \cdot 0 = 0\)
- Second column: \(0 \cdot 5\alpha + \alpha \cdot \alpha + 5\alpha \cdot 0 = \alpha^2\)
- Third column: \(0 \cdot \alpha + \alpha \cdot 5\alpha + 5\alpha \cdot 5 = 5\alpha^2 + 25\alpha = 5\alpha^2 + 25\alpha\)

- **Third row:**
- First column: \(0 \cdot 5 + 0 \cdot 0 + 5 \cdot 0 = 0\)
- Second column: \(0 \cdot 5\alpha + 0 \cdot \alpha + 5 \cdot 0 = 0\)
- Third column: \(0 \cdot \alpha + 0 \cdot 5\alpha + 5 \cdot 5 = 25\)

Thus, we have:

\[ A^2 = \begin{pmatrix} 25 & 25\alpha + 5\alpha^2 & 30\alpha^2 + \alpha \\ 0 & \alpha^2 & 5\alpha^2 + 25\alpha \\ 0 & 0 & 25 \end{pmatrix} \]

### Step 3: Calculate the determinant \(|A^2|\)

Using the property of determinants for upper triangular matrices, the determinant is the product of the diagonal entries:

\[ |A^2| = 25 \cdot \alpha^2 \cdot 25 = 625\alpha^2 \]

### Step 4: Set the determinant equal to 25

We know from the problem statement that:

\[ |A^2| = 625\alpha^2 = 25 \]

### Step 5: Solve for \(|\alpha|\)

Dividing both sides by 625:

\[ \alpha^2 = \frac{25}{625} = \frac{1}{25} \]

Taking the square root:

\[ |\alpha| = \sqrt{\frac{1}{25}} = \frac{1}{5} \]

### Final Answer

Thus, the value of \(|\alpha|\) is:

\[ |\alpha| = \frac{1}{5} \]