Let `A=[(5,5alpha,alpha),(0,alpha,5alpha),(0,0,5)]`. If `|A|^2=25`, then `|alpha|` is equal to :

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 the matrix \(A\) and the condition \(|A|^2 = 25\).

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 the determinant of matrix \(A\)

To find the determinant of \(A\), we can use the formula for the determinant of a \(3 \times 3\) matrix:
\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

For our matrix \(A\):
- \(a = 5\), \(b = 5\alpha\), \(c = \alpha\)
- \(d = 0\), \(e = \alpha\), \(f = 5\alpha\)
- \(g = 0\), \(h = 0\), \(i = 5\)

Now, substituting into the determinant formula:
\[ |A| = 5(\alpha \cdot 5 - 5\alpha \cdot 0) - 5\alpha(0 \cdot 5 - 5\alpha \cdot 0) + \alpha(0 \cdot 0 - 0 \cdot \alpha) \]
This simplifies to:
\[ |A| = 5(5\alpha) - 0 + 0 = 25\alpha \]

### Step 2: Use the condition \(|A|^2 = 25\)

We are given that \(|A|^2 = 25\). Therefore:
\[ (25\alpha)^2 = 25 \]
This simplifies to:
\[ 625\alpha^2 = 25 \]

### Step 3: Solve for \(\alpha^2\)

Dividing both sides by 625:
\[ \alpha^2 = \frac{25}{625} = \frac{1}{25} \]

### Step 4: Find \(|\alpha|\)

Taking the square root of both sides gives:
\[ |\alpha| = \sqrt{\frac{1}{25}} = \frac{1}{5} \]

Thus, the final answer is:
\[ |\alpha| = \frac{1}{5} \]