📘 Question
If
\[
A =
\begin{bmatrix}
0 & 2 \\
3 & -4
\end{bmatrix}
\]
and
\[
kA =
\begin{bmatrix}
0 & 3a \\
2b & 24
\end{bmatrix}
\]
Find \(k, a, b\).
✏️ Step-by-Step Solution
Step 1: Multiply matrix by scalar
\[
kA =
k
\begin{bmatrix}
0 & 2 \\
3 & -4
\end{bmatrix}
=
\begin{bmatrix}
0 & 2k \\
3k & -4k
\end{bmatrix}
\]
Step 2: Compare elements
\[ \begin{bmatrix} 0 & 2k \\ 3k & -4k \end{bmatrix} = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix} \]- \(2k = 3a\)
- \(3k = 2b\)
- \(-4k = 24\)
Step 3: Solve for \(k\)
\[
-4k = 24 \Rightarrow k = -6
\]
Step 4: Find \(a\) and \(b\)
From \(2k = 3a\):
\[
2(-6) = 3a \Rightarrow a = -4
\]
From \(3k = 2b\):
\[
3(-6) = 2b \Rightarrow b = -9
\]
✅ Final Answer
\[
\boxed{k = -6,\; a = -4,\; b = -9}
\]
💡 Key Concept
Scalar multiplication multiplies each element of the matrix. Then compare corresponding entries to solve unknowns.