https://file.aiquickdraw.com/mathbot/file/1724606113456hpiw6a3n.jpg

To solve the problem, we need to evaluate the value of \( X \) based on the provided expression:

\[
X = \text{det}(2A) - 2 \cdot \text{det}(A \cdot B^T)
\]

where:
\[
A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 0 & 4 \\ 3 & 2 & 6 \end{pmatrix}
\quad \text{and} \quad
B = \begin{pmatrix} 2 & 1 & 2 \\ -1 & 0 & 1 \\ 1 & 0 & 1 \end{pmatrix}
\]

### Step 1: Calculate \( \text{det}(A) \)
We first calculate the determinant of matrix \( A \):

\[
A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 0 & 4 \\ 3 & 2 & 6 \end{pmatrix}
\]

The determinant is given by:
\[
\text{det}(A) = 2 \cdot (0 \cdot 6 - 4 \cdot 2) - 1 \cdot (1 \cdot 6 - 4 \cdot 3) + 0 \cdot (1 \cdot 2 - 0 \cdot 3)
\]
\[
\text{det}(A) = 2 \cdot (0 - 8) - 1 \cdot (6 - 12)
\]
\[
\text{det}(A) = 2 \cdot (-8) - 1 \cdot (-6)
\]
\[
\text{det}(A) = -16 + 6 = -10
\]

### Step 2: Calculate \( \text{det}(2A) \)
Next, let's calculate the determinant of \( 2A \). If you multiply a matrix by a scalar, the determinant of the resulting matrix is the original determinant multiplied by that scalar raised to the power of the matrix size (in this case, 3x3 matrix):

\[
\text{det}(2A) = 2^3 \cdot \text{det}(A) = 8 \cdot (-10) = -80
\]

### Step 3: Calculate \( A \cdot B^T \)
Now, let's compute the product of \( A \) and \( B^T \) (the transpose of \( B \)):

First, \( B^T \) is:
\[
B^T = \begin{pmatrix} 2 & -1 & 1 \\ 1 & 0 & 0 \\ 2 & 1 & 1 \end{pmatrix}
\]

Then \( A \cdot B^T \) is:
\[
A \cdot B^T = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 0 & 4 \\ 3 & 2 & 6 \end{pmatrix} \cdot \begin{pmatrix} 2 & -1 & 1 \\ 1 & 0 & 0 \\ 2 & 1 & 1 \end{pmatrix}
\]

Calculating the multiplication:
\[
A \cdot B^T = \begin{pmatrix} 2(2) + 1(1) + 0(2) & 2(-1) + 1(0) + 0(1) & 2(1) + 1(0) + 0(1) \\ 1(2) + 0(1) + 4(2) & 1(-1) + 0(0) + 4(1) & 1(1) + 0(0) + 4(1) \\ 3(2) + 2(1) + 6(2) & 3(-1) + 2(0) + 6(1) & 3(1) + 2(0) + 6(1) \end{pmatrix}
\]
\[
A \cdot B^T = \begin{pmatrix} 4 + 1 + 0 & -2 + 0 + 0 & 2 + 0 + 0 \\ 2 + 0 + 8 & -1 + 0 + 4 & 1 + 0 + 4 \\ 6 + 2 + 12 & -3 + 0 + 6 & 3 + 0 + 6 \end{pmatrix}
\]
\[
A \cdot B^T = \begin{pmatrix} 5 & -2 & 2 \\ 10 & 3 & 5 \\ 20 & 3 & 9 \end{pmatrix}
\]

### Step 4: Calculate \( \text{det}(A \cdot B^T) \)
Now we calculate the determinant of the resulting matrix:

\[
\text{det}(A \cdot B^T) = 5 \cdot (3 \cdot 9 - 5 \cdot 3) - (-2) \cdot (10 \cdot 9 - 5 \cdot 20) + 2 \cdot (10 \cdot 3 - 3 \cdot 20)
\]
\[
\text{det}(A \cdot B^T) = 5 \cdot (27 - 15) + 2 \cdot (90 - 100) + 2 \cdot (30 - 60)
\]
\[
\text{det}(A \cdot B^T) = 5 \cdot 12 + 2 \cdot (-10) + 2 \cdot (-30)
\]
\[
\text{det}(A \cdot B^T) = 60 - 20 - 60 = -20
\]

### Step 5: Calculate \( X \)
Finally, substitute into the expression for \( X \):
\[
X = -80 - 2(-20) = -80 + 40 = -40
\]

So, the correct answer is not listed in the given options, meaning there might be an error in the options or the problem setup. But according to our calculations, the correct \( X \) value should be \(-40\). 

Let me know if you want to go over any part in more detail or if you have any questions!

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Here are some related questions:

1. How do you find the determinant of a 3x3 matrix in general?
2. What is the geometric interpretation of the determinant of a matrix?
3. How does matrix transposition affect the determinant?
4. What happens to the determinant if a matrix is scaled by a scalar factor?
5. How do you multiply two matrices together?

**Tip:** Always check your matrix dimensions before multiplying two matrices to ensure the operation is valid!

Learn how to solve the determinant and matrix multiplication problem using matrices A and B. This problem involves calculating the determinant of a 3x3 matrix, scalar multiplication, and matrix multiplication steps. Suitable for advanced high school students or college-level introductory linear algebra courses.

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