Let's break down the problem step by step:

You provided the matrix \( C \):

\[
C = \begin{bmatrix}
0 & 1 & 4 \\
-4 & -1 & 2
\end{bmatrix}
\]

### Step 1: Find the transpose of matrix \( C \) (denoted as \( C^T \)).

The transpose of a matrix is obtained by swapping rows and columns. So the transpose of \( C \), which is a \( 2 \times 3 \) matrix, will become a \( 3 \times 2 \) matrix:

\[
C^T = \begin{bmatrix}
0 & -4 \\
1 & -1 \\
4 & 2
\end{bmatrix}
\]

### Step 2: Multiply the transpose \( C^T \) by \( -2 \).

Now we need to multiply each element of \( C^T \) by \( -2 \):

\[
-2C^T = -2 \times \begin{bmatrix}
0 & -4 \\
1 & -1 \\
4 & 2
\end{bmatrix} = \begin{bmatrix}
0 & 8 \\
-2 & 2 \\
-8 & -4
\end{bmatrix}
\]

### Final Answer:
\[
-2C^T = \begin{bmatrix}
0 & 8 \\
-2 & 2 \\
-8 & -4
\end{bmatrix}
\]

Would you like any further details on the steps?

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Here are five related questions you might find interesting:
1. What is the determinant of a square matrix and how do you calculate it?
2. How do you multiply two matrices together?
3. What is the inverse of a matrix, and how is it computed?
4. Can you explain the significance of matrix transposition in geometry?
5. How do eigenvalues and eigenvectors relate to matrix transformations?

**Tip:** When multiplying matrices by a scalar, every element in the matrix is scaled by the same factor, but the overall structure of the matrix remains the same.

Solve the problem involving matrix C = [[0, 1, 4], [-4, -1, 2]] by finding the transpose C^T and performing scalar multiplication with -2. This step-by-step guide covers core concepts like matrix transposition and scalar multiplication, suitable for students in Grades 10-12.

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