If `A={:[(0,2),(3,-4)]:}and kA={:[(0,3a),(2b,24)]:}`, then the values of k,a,b are respectively.

Options

A. `-6,-12,-18`
B. `-6,4,9`
C. `-6,-4,-9`
D. `-6,12,18`

Explanation

To solve the problem, we need to find the values of \( k \), \( a \), and \( b \) given the matrices \( A \) and \( kA \).

1. **Identify the matrices**:
\[ A = \begin{pmatrix} 0 & 2 \\ 3 & -4 \end{pmatrix} \]
\[ kA = \begin{pmatrix} 0 & 3a \\ 2b & 24 \end{pmatrix} \]

2. **Multiply matrix \( A \) by \( k \)**:
When we multiply matrix \( A \) by \( k \), we get:
\[ kA = k \begin{pmatrix} 0 & 2 \\ 3 & -4 \end{pmatrix} = \begin{pmatrix} 0 & 2k \\ 3k & -4k \end{pmatrix} \]

3. **Set the corresponding elements equal**:
Since \( kA \) is given as \( \begin{pmatrix} 0 & 3a \\ 2b & 24 \end{pmatrix} \), we can set the corresponding elements equal to each other:
- From the first row, first column:
\[ 0 = 0 \quad \text{(This is always true)} \]
- From the first row, second column:
\[ 2k = 3a \quad \text{(Equation 1)} \]
- From the second row, first column:
\[ 3k = 2b \quad \text{(Equation 2)} \]
- From the second row, second column:
\[ -4k = 24 \quad \text{(Equation 3)} \]

4. **Solve Equation 3 for \( k \)**:
\[ -4k = 24 \implies k = \frac{24}{-4} = -6 \]

5. **Substitute \( k \) into Equation 1 to find \( a \)**:
\[ 2(-6) = 3a \implies -12 = 3a \implies a = \frac{-12}{3} = -4 \]

6. **Substitute \( k \) into Equation 2 to find \( b \)**:
\[ 3(-6) = 2b \implies -18 = 2b \implies b = \frac{-18}{2} = -9 \]

7. **Final values**:
The values of \( k \), \( a \), and \( b \) are:
\[ k = -6, \quad a = -4, \quad b = -9 \]

### Summary of the solution:
The values of \( k \), \( a \), and \( b \) are respectively:
\[ \boxed{-6}, \quad \boxed{-4}, \quad \boxed{-9} \]