To solve the problem, we need to find the determinant of the expression \( A + A^2B^2 + A^3 + A^4B^4 + \ldots \) up to 20 terms, where \( A \) and \( B \) are given matrices.
### Step 1: Calculate \( A^2 \)
We start by calculating \( A^2 \) which is \( A \times A \).
Given:
\[
A = \begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}
\]
Calculating \( A^2 \):
\[
A^2 = A \times A = \begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix} \times \begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}
\]
Calculating each element:
- First row, first column: \( 2 \cdot 2 + (-2) \cdot (-1) + (-4) \cdot 1 = 4 + 2 - 4 = 2 \)
- First row, second column: \( 2 \cdot (-2) + (-2) \cdot 3 + (-4) \cdot (-2) = -4 - 6 + 8 = -2 \)
- First row, third column: \( 2 \cdot (-4) + (-2) \cdot 4 + (-4) \cdot (-3) = -8 - 8 + 12 = -4 \)
- Second row, first column: \( -1 \cdot 2 + 3 \cdot (-1) + 4 \cdot 1 = -2 - 3 + 4 = -1 \)
- Second row, second column: \( -1 \cdot (-2) + 3 \cdot 3 + 4 \cdot (-2) = 2 + 9 - 8 = 3 \)
- Second row, third column: \( -1 \cdot (-4) + 3 \cdot 4 + 4 \cdot (-3) = 4 + 12 - 12 = 4 \)
- Third row, first column: \( 1 \cdot 2 + (-2) \cdot (-1) + (-3) \cdot 1 = 2 + 2 - 3 = 1 \)
- Third row, second column: \( 1 \cdot (-2) + (-2) \cdot 3 + (-3) \cdot (-2) = -2 - 6 + 6 = -2 \)
- Third row, third column: \( 1 \cdot (-4) + (-2) \cdot 4 + (-3) \cdot (-3) = -4 - 8 + 9 = -3 \)
So,
\[
A^2 = \begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}
\]
### Step 2: Calculate \( B^2 \)
Next, we calculate \( B^2 \).
Given:
\[
B = \begin{pmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{pmatrix}
\]
Calculating \( B^2 \):
\[
B^2 = B \times B = \begin{pmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{pmatrix} \times \begin{pmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{pmatrix}
\]
Calculating each element:
- First row, first column: \( -4 \cdot -4 + (-3) \cdot 1 + (-3) \cdot 4 = 16 - 3 - 12 = 1 \)
- First row, second column: \( -4 \cdot -3 + (-3) \cdot 0 + (-3) \cdot 4 = 12 + 0 - 12 = 0 \)
- First row, third column: \( -4 \cdot -3 + (-3) \cdot 1 + (-3) \cdot 3 = 12 - 3 - 9 = 0 \)
- Second row, first column: \( 1 \cdot -4 + 0 \cdot 1 + 1 \cdot 4 = -4 + 0 + 4 = 0 \)
- Second row, second column: \( 1 \cdot -3 + 0 \cdot 0 + 1 \cdot 4 = -3 + 0 + 4 = 1 \)
- Second row, third column: \( 1 \cdot -3 + 0 \cdot 1 + 1 \cdot 3 = -3 + 0 + 3 = 0 \)
- Third row, first column: \( 4 \cdot -4 + 4 \cdot 1 + 3 \cdot 4 = -16 + 4 + 12 = 0 \)
- Third row, second column: \( 4 \cdot -3 + 4 \cdot 0 + 3 \cdot 4 = -12 + 0 + 12 = 0 \)
- Third row, third column: \( 4 \cdot -3 + 4 \cdot 1 + 3 \cdot 3 = -12 + 4 + 9 = 1 \)
So,
\[
B^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}
\]
This is the identity matrix \( I \).
### Step 3: Analyze the Series
Now we can analyze the series:
\[
A + A^2B^2 + A^3 + A^4B^4 + \ldots
\]
Since \( A^2 = A \) and \( B^2 = I \), we can simplify the series:
\[
A + A + A + A + \ldots \text{ (20 terms)}
\]
This simplifies to:
\[
20A
\]
### Step 4: Calculate the Determinant
Now we need to find the determinant:
\[
\text{det}(20A) = 20^3 \cdot \text{det}(A)
\]
where the factor \( 20^3 \) comes from the property of determinants that states \( \text{det}(kA) = k^n \cdot \text{det}(A) \) for an \( n \times n \) matrix \( A \).
### Step 5: Calculate \( \text{det}(A) \)
Now we calculate \( \text{det}(A) \):
\[
A = \begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}
\]
Using the determinant formula for a \( 3 \times 3 \) matrix:
\[
\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)
\]
where \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \).
Calculating:
\[
\text{det}(A) = 2(3 \cdot (-3) - 4 \cdot (-2)) - (-2)(-1 \cdot (-3) - 4 \cdot 1) + (-4)(-1 \cdot (-2) - 3 \cdot 1)
\]
Calculating each term:
- First term: \( 2(-9 + 8) = 2(-1) = -2 \)
- Second term: \( -2(3 - 4) = -2(-1) = 2 \)
- Third term: \( -4(2 - 3) = -4(-1) = 4 \)
So,
\[
\text{det}(A) = -2 + 2 + 4 = 4
\]
### Step 6: Final Calculation
Now substituting back:
\[
\text{det}(20A) = 20^3 \cdot 4 = 8000
\]
### Conclusion
Thus, the value of the determinant \( A + A^2B^2 + A^3 + A^4B^4 + \ldots \) up to 20 terms is:
\[
\boxed{8000}
\]
