`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. 12
B. -12
C. 7
D. -7

Explanation

To solve the problem, we need to find the value of \( \lambda \) in the equation:

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

### Step 1: Calculate the determinant \( A \)

We will calculate the determinant \( A \) using the properties of determinants. The determinant of a matrix can be simplified by factoring out constants from the rows or columns.

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

We can factor out the constants from each column:

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

Thus, we have:

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

This simplifies to:

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

### Step 2: Set up the equation

Now we can set this equal to \( \lambda \) times the determinant of the second matrix:

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

### Step 3: Solve for \( \lambda \)

Assuming that the determinant \( \begin{vmatrix} a & r & x \\ b & s & y \\ c & t & z \end{vmatrix} \) is not zero, we can divide both sides by this determinant:

\[ -6 = \lambda \]

### Final Answer

Thus, the value of \( \lambda \) is:

\[ \lambda = -6 \]

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