Question

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

• A. \( -6, -12, -18 \)
• B. \( -6, -4, -9 \)
• C. \( -6, 4, 9 \)
• D. \( -6, 12, 18 \)

Hint:

When dealing with scalar multiplication of matrices, compare corresponding elements systematically to find unknown values.

Answer 1

The Correct Option is B

Step 1: Understanding Scalar Multiplication Since \( kA \) represents the matrix \( A \) multiplied by the scalar \( k \), we equate each entry: \[ k \begin{bmatrix} 0 & 2 \\ 3 & -4 \end{bmatrix} = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}. \] This means: \[ \begin{bmatrix} 0 & 2k \\ 3k & -4k \end{bmatrix} = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}. \] Step 2: Solving for \( k \) Comparing the bottom-right elements: \[ -4k = 24. \] Solving for \( k \): \[ k = -6. \] Step 3: Solving for \( a \) From the top-right elements: \[ 2k = 3a. \] Substituting \( k = -6 \): \[ 2(-6) = 3a \quad \Rightarrow \quad -12 = 3a \quad \Rightarrow \quad a = -4. \] Step 4: Solving for \( b \) From the bottom-left elements: \[ 3k = 2b. \] Substituting \( k = -6 \): \[ 3(-6) = 2b \quad \Rightarrow \quad -18 = 2b \quad \Rightarrow \quad b = -9. \] Thus, the correct values are \( k = -6 \), \( a = -4 \), and \( b = -9 \), which matches option (B).