Let `P=[(2alpha),(5),(-3alpha^(2))] and Q=[(2l,-m,5n)]` are two matrices, where `l, m, n, alpha in R`, then the value of determinant PQ is equal to

Options

A. 0
B. `-1`
C. 2
D. not possible

Answer

To solve the problem, we need to find the determinant of the product of two matrices \( P \) and \( Q \). Let's break down the solution step by step.

### Step 1: Define the matrices
We have:
\[ P = \begin{bmatrix} 2\alpha \\ 5 \\ -3\alpha^2 \end{bmatrix} \]
\[ Q = \begin{bmatrix} 2l & -m & 5n \end{bmatrix} \]

### Step 2: Multiply the matrices
The product \( PQ \) will be a \( 3 \times 3 \) matrix. To compute \( PQ \), we multiply each element of \( P \) by each element of \( Q \).

\[ PQ = \begin{bmatrix} 2\alpha \cdot 2l & 2\alpha \cdot (-m) & 2\alpha \cdot 5n \\ 5 \cdot 2l & 5 \cdot (-m) & 5 \cdot 5n \\ -3\alpha^2 \cdot 2l & -3\alpha^2 \cdot (-m) & -3\alpha^2 \cdot 5n \end{bmatrix} \]

Calculating each entry:
- First row:
- \( 4\alpha l \)
- \( -2\alpha m \)
- \( 10\alpha n \)

- Second row:
- \( 10l \)
- \( -5m \)
- \( 25n \)

- Third row:
- \( -6\alpha^2 l \)
- \( 3\alpha^2 m \)
- \( -15\alpha^2 n \)

So, we have:
\[ PQ = \begin{bmatrix} 4\alpha l & -2\alpha m & 10\alpha n \\ 10l & -5m & 25n \\ -6\alpha^2 l & 3\alpha^2 m & -15\alpha^2 n \end{bmatrix} \]

### Step 3: Calculate the determinant
To find the determinant of the matrix \( PQ \), we can use the property that if two columns (or rows) of a determinant are identical, the determinant is zero.

Let's observe the columns:
- The first column is \( \begin{bmatrix} 4\alpha l \\ 10l \\ -6\alpha^2 l \end{bmatrix} \)
- The second column is \( \begin{bmatrix} -2\alpha m \\ -5m \\ 3\alpha^2 m \end{bmatrix} \)
- The third column is \( \begin{bmatrix} 10\alpha n \\ 25n \\ -15\alpha^2 n \end{bmatrix} \)

Notice that:
- The first and third columns can be simplified and compared.

### Step 4: Factor out common terms
We can factor out \( 2l \) from the first column, \( m \) from the second column, and \( 5n \) from the third column:
\[ \text{Det}(PQ) = 2l \cdot m \cdot 5n \cdot \text{Det}\begin{bmatrix} 2\alpha & -1 & 5 \\ 5 & -1 & 5 \\ -3\alpha^2 & 3 & -3 \end{bmatrix} \]

### Step 5: Check for identical columns
After simplifying, we can see that if any two columns are identical, the determinant will be zero.

In our case, if we analyze the matrix formed, we can see that the first and third columns are multiples of each other, confirming that they are linearly dependent.

### Conclusion
Thus, the value of the determinant of \( PQ \) is:
\[ \text{Det}(PQ) = 0 \]