If matrix P `= [(0,2),(3,-4)] and kP = [ (0,3a),(2b,24)]` where k, a and b are constants, then the values of k, a, b respectively are

Options

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

Answer

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

### Given:
- Matrix \( P = \begin{pmatrix} 0 & 2 \\ 3 & -4 \end{pmatrix} \)
- Matrix \( kP = \begin{pmatrix} 0 & 3a \\ 2b & 24 \end{pmatrix} \)

### Step-by-Step Solution:

1. **Calculate \( kP \)**:
To find \( kP \), we multiply each element of matrix \( P \) by \( k \):
\[ kP = k \begin{pmatrix} 0 & 2 \\ 3 & -4 \end{pmatrix} = \begin{pmatrix} 0 \cdot k & 2 \cdot k \\ 3 \cdot k & -4 \cdot k \end{pmatrix} = \begin{pmatrix} 0 & 2k \\ 3k & -4k \end{pmatrix} \]

2. **Set up the equations**:
Now, we can equate the elements of \( kP \) with those of the given matrix \( kP \):
\[ \begin{pmatrix} 0 & 2k \\ 3k & -4k \end{pmatrix} = \begin{pmatrix} 0 & 3a \\ 2b & 24 \end{pmatrix} \]
This gives us the following equations:
- From the first element: \( 0 = 0 \) (always true)
- From the second element: \( 2k = 3a \) (Equation 1)
- From the third element: \( 3k = 2b \) (Equation 2)
- From the fourth element: \( -4k = 24 \) (Equation 3)

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

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

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

### Final Values:
Thus, we have:
- \( k = -6 \)
- \( a = -4 \)
- \( b = -9 \)

### Conclusion:
The values of \( k \), \( a \), and \( b \) are \( -6 \), \( -4 \), and \( -9 \) respectively.