To solve the problem, we need to find the matrix \( D \) given the matrices \( A \), \( B \), and \( C \) and the equation \( AB - CD = 0 \).
### Step 1: Define the matrices
Given:
\[
A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} 5 & 2 \\ 7 & 4 \end{pmatrix}, \quad C = \begin{pmatrix} 2 & 5 \\ 3 & 8 \end{pmatrix}
\]
Assume:
\[
D = \begin{pmatrix} a & b \\ c & d \end{pmatrix}
\]
### Step 2: Set up the equation
From the problem, we have:
\[
AB - CD = 0 \implies AB = CD
\]
### Step 3: Calculate \( AB \)
To find \( AB \):
\[
AB = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 5 & 2 \\ 7 & 4 \end{pmatrix}
\]
Calculating the elements:
- First row, first column: \( 2 \times 5 + (-1) \times 7 = 10 - 7 = 3 \)
- First row, second column: \( 2 \times 2 + (-1) \times 4 = 4 - 4 = 0 \)
- Second row, first column: \( 3 \times 5 + 4 \times 7 = 15 + 28 = 43 \)
- Second row, second column: \( 3 \times 2 + 4 \times 4 = 6 + 16 = 22 \)
Thus,
\[
AB = \begin{pmatrix} 3 & 0 \\ 43 & 22 \end{pmatrix}
\]
### Step 4: Calculate \( CD \)
Now, we need to calculate \( CD \):
\[
CD = \begin{pmatrix} 2 & 5 \\ 3 & 8 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix}
\]
Calculating the elements:
- First row, first column: \( 2a + 5c \)
- First row, second column: \( 2b + 5d \)
- Second row, first column: \( 3a + 8c \)
- Second row, second column: \( 3b + 8d \)
Thus,
\[
CD = \begin{pmatrix} 2a + 5c & 2b + 5d \\ 3a + 8c & 3b + 8d \end{pmatrix}
\]
### Step 5: Set up the equations
Since \( AB = CD \), we have:
\[
\begin{pmatrix} 3 & 0 \\ 43 & 22 \end{pmatrix} = \begin{pmatrix} 2a + 5c & 2b + 5d \\ 3a + 8c & 3b + 8d \end{pmatrix}
\]
From this, we can set up the following equations:
1. \( 2a + 5c = 3 \) (1)
2. \( 2b + 5d = 0 \) (2)
3. \( 3a + 8c = 43 \) (3)
4. \( 3b + 8d = 22 \) (4)
### Step 6: Solve equations (1) and (3)
From equation (1):
\[
2a + 5c = 3 \implies 2a = 3 - 5c \implies a = \frac{3 - 5c}{2}
\]
Substituting \( a \) into equation (3):
\[
3\left(\frac{3 - 5c}{2}\right) + 8c = 43
\]
Multiplying through by 2 to eliminate the fraction:
\[
3(3 - 5c) + 16c = 86
\]
\[
9 - 15c + 16c = 86 \implies c = 86 - 9 = 77
\]
Now substituting \( c = 77 \) back into equation (1):
\[
2a + 5(77) = 3 \implies 2a + 385 = 3 \implies 2a = 3 - 385 = -382 \implies a = -191
\]
### Step 7: Solve equations (2) and (4)
From equation (2):
\[
2b + 5d = 0 \implies 2b = -5d \implies b = -\frac{5d}{2}
\]
Substituting \( b \) into equation (4):
\[
3\left(-\frac{5d}{2}\right) + 8d = 22
\]
Multiplying through by 2:
\[
-15d + 16d = 44 \implies d = 44
\]
Now substituting \( d = 44 \) back into equation (2):
\[
2b + 5(44) = 0 \implies 2b + 220 = 0 \implies 2b = -220 \implies b = -110
\]
### Final Step: Write the matrix \( D \)
Thus, we have:
\[
D = \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} -191 & -110 \\ 77 & 44 \end{pmatrix}
\]
### Summary of the Solution
The matrix \( D \) is:
\[
D = \begin{pmatrix} -191 & -110 \\ 77 & 44 \end{pmatrix}
\]
