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To solve the expression \(2A - B^t\), we first need to compute two steps:

1. **Multiply matrix \(A\) by 2**: 
   Given:
   \[
   A = \begin{pmatrix} 
   3 & 2 & 1 \\ 
   3 & 2 & 2 
   \end{pmatrix}
   \]

   We compute \(2A\):
   \[
   2A = 2 \times \begin{pmatrix} 
   3 & 2 & 1 \\ 
   3 & 2 & 2 
   \end{pmatrix} = \begin{pmatrix} 
   6 & 4 & 2 \\ 
   6 & 4 & 4 
   \end{pmatrix}
   \]

2. **Transpose matrix \(B\)**: 
   Given:
   \[
   B = \begin{pmatrix} 
   -4 & 1 \\ 
   2 & 2 \\ 
   1 & -1 
   \end{pmatrix}
   \]

   The transpose of \(B\), denoted as \(B^t\), is achieved by swapping rows and columns:
   \[
   B^t = \begin{pmatrix} 
   -4 & 2 & 1 \\ 
   1 & 2 & -1 
   \end{pmatrix}
   \]

Now, we subtract \(B^t\) from \(2A\):

\[
2A - B^t = \begin{pmatrix} 
6 & 4 & 2 \\ 
6 & 4 & 4 
\end{pmatrix} - \begin{pmatrix} 
-4 & 2 & 1 \\ 
1 & 2 & -1 
\end{pmatrix} = \begin{pmatrix} 
6 - (-4) & 4 - 2 & 2 - 1 \\ 
6 - 1 & 4 - 2 & 4 - (-1) 
\end{pmatrix}
\]

\[
2A - B^t = \begin{pmatrix} 
10 & 2 & 1 \\ 
5 & 2 & 5 
\end{pmatrix}
\]

Thus, the result of \(2A - B^t\) is:

\[
\begin{pmatrix} 
10 & 2 & 1 \\ 
5 & 2 & 5 
\end{pmatrix}
\]

Would you like further details or have any questions?

### Related Questions:
1. What is the result of multiplying a matrix by a scalar?
2. How do you find the transpose of a matrix?
3. Can matrices of different dimensions be subtracted?
4. What are some properties of matrix transpose?
5. How can matrix operations be applied in real-world problems?

### Tip:
Always verify the dimensions of matrices before performing operations like addition or subtraction, as they must be compatible.

Diketahui matriks A dan B, maka 2A - B^t = ..
A = [[3, 2, 1], [3, 2, 2]], B = [[-4, 1], [2, 2], [1, -1]]

Solve the matrix operation 2A - B^t where A and B are given matrices. This solution covers scalar multiplication, matrix transposition, and matrix subtraction with step-by-step explanations. Suitable for Grades 9-12.

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