`A = |(2a,3r,x),(4b,6s,2y),(-2c,-3t,-z)| = lambda |(a,r,x),(b,s,y),(c,t,z)|`, then what is the value of `lambda` ?

Options

A. A) 12
B. B) -12
C. C) 7
D. D) -7

Answer

To find the value of \( \lambda \) in the equation

\[ A = \begin{vmatrix} 2a & 3r & x \\ 4b & 6s & 2y \\ -2c & -3t & -z \end{vmatrix} = \lambda \begin{vmatrix} a & r & x \\ b & s & y \\ c & t & z \end{vmatrix}, \]

we will manipulate the determinant \( A \) step by step.

### Step 1: Factor out constants from the columns
We can factor out constants from each column of the determinant \( A \).

1. From the first column, we can factor out \( 2 \).
2. From the second column, we can factor out \( 3 \).
3. From the third column, we can factor out \( -1 \) from the third row.

This gives us:

\[ A = 2 \cdot 3 \cdot (-1) \begin{vmatrix} a & r & x \\ 2b & 2s & y \\ -c & -3t & -z \end{vmatrix}. \]

### Step 2: Factor out constants from the second row
Next, we can factor out \( 2 \) from the second row:

\[ A = 2 \cdot 3 \cdot (-1) \cdot 2 \begin{vmatrix} a & r & x \\ b & s & y \\ c & t & z \end{vmatrix}. \]

### Step 3: Calculate the product of the constants
Now, we compute the product of the constants:

\[ A = 2 \cdot 3 \cdot 2 \cdot (-1) \begin{vmatrix} a & r & x \\ b & s & y \\ c & t & z \end{vmatrix} = -12 \begin{vmatrix} a & r & x \\ b & s & y \\ c & t & z \end{vmatrix}. \]

### Step 4: Compare with the given equation
Now, we have:

\[ A = -12 \begin{vmatrix} a & r & x \\ b & s & y \\ c & t & z \end{vmatrix}. \]

Comparing this with the original equation \( A = \lambda \begin{vmatrix} a & r & x \\ b & s & y \\ c & t & z \end{vmatrix} \), we see that:

\[ \lambda = -12. \]

### Final Answer
Thus, the value of \( \lambda \) is

\[ \boxed{-12}. \]

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