Finding a, b, c, d by Equating Matrices
Question:
Find the values of \( a, b, c, d \) if
\[ \begin{bmatrix} 2a + b & 1 – 2b \\ 5c – d & 4c + 3d \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ 11 & 24 \end{bmatrix} \]
Concept Used
Two matrices are equal if their corresponding elements are equal.
Step 1: Equate Corresponding Elements
\[ 2a + b = 4 \quad …(1) \]
\[ 1 – 2b = -3 \quad …(2) \]
\[ 5c – d = 11 \quad …(3) \]
\[ 4c + 3d = 24 \quad …(4) \]
Step 2: Solve for a and b
From (2):
\[ 1 – 2b = -3 \Rightarrow -2b = -4 \Rightarrow b = 2 \]
Substitute into (1):
\[ 2a + 2 = 4 \Rightarrow 2a = 2 \Rightarrow a = 1 \]
Step 3: Solve for c and d
From (3):
\[ d = 5c – 11 \]
Substitute into (4):
\[ 4c + 3(5c – 11) = 24 \]
\[ 4c + 15c – 33 = 24 \Rightarrow 19c = 57 \Rightarrow c = 3 \]
\[ d = 5(3) – 11 = 4 \]
Final Answer
\[ a = 1,\quad b = 2,\quad c = 3,\quad d = 4 \]