Let's work through the problem step by step.

Given Matrices:

The matrices provided are: $A = \begin{bmatrix} 3 & 0 \\ -1 & 2 \\ 1 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 4 & -1 \\ 0 & 2 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & 4 & 2 \\ 3 & 1 & 5 \end{bmatrix}$

Problem Statements:

1. Determine $4B C + 2C$.
2. Find the transpose of matrix $A$.

Step-by-Step Solution:

1. Compute $4B$

To find $4B$, we multiply each element in matrix $B$ by 4: $4B = 4 \times \begin{bmatrix} 4 & -1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 16 & -4 \\ 0 & 8 \end{bmatrix}$

2. Compute $4B C$

Next, we calculate the product of $4B$ and $C$. The resulting matrix from this multiplication will have dimensions $2 \times 3$ because $4B$ is $2 \times 2$ and $C$ is $2 \times 3$.

$4B C = \begin{bmatrix} 16 & -4 \\ 0 & 8 \end{bmatrix} \begin{bmatrix} 1 & 4 & 2 \\ 3 & 1 & 5 \end{bmatrix}$

Let's calculate each element:

• First row, first column: $16 \times 1 + (-4) \times 3 = 16 - 12 = 4$

• First row, second column: $16 \times 4 + (-4) \times 1 = 64 - 4 = 60$

• First row, third column: $16 \times 2 + (-4) \times 5 = 32 - 20 = 12$

• Second row, first column: $0 \times 1 + 8 \times 3 = 0 + 24 = 24$

• Second row, second column: $0 \times 4 + 8 \times 1 = 0 + 8 = 8$

• Second row, third column: $0 \times 2 + 8 \times 5 = 0 + 40 = 40$

So, $4B C = \begin{bmatrix} 4 & 60 & 12 \\ 24 & 8 & 40 \end{bmatrix}$

3. Compute $2C$

Now, we find $2C$ by multiplying each element in $C$ by 2: $2C = 2 \times \begin{bmatrix} 1 & 4 & 2 \\ 3 & 1 & 5 \end{bmatrix} = \begin{bmatrix} 2 & 8 & 4 \\ 6 & 2 & 10 \end{bmatrix}$

4. Compute $4B C + 2C$

Now, we add $4B C$ and $2C$ element-wise: $4B C + 2C = \begin{bmatrix} 4 & 60 & 12 \\ 24 & 8 & 40 \end{bmatrix} + \begin{bmatrix} 2 & 8 & 4 \\ 6 & 2 & 10 \end{bmatrix} = \begin{bmatrix} 6 & 68 & 16 \\ 30 & 10 & 50 \end{bmatrix}$

So, $4B C + 2C = \begin{bmatrix} 6 & 68 & 16 \\ 30 & 10 & 50 \end{bmatrix}$

5. Transpose of Matrix $A$

The transpose of matrix $A$, denoted as $A^T$, is obtained by swapping its rows and columns: $A = \begin{bmatrix} 3 & 0 \\ -1 & 2 \\ 1 & 1 \end{bmatrix} \Rightarrow A^T = \begin{bmatrix} 3 & -1 & 1 \\ 0 & 2 & 1 \end{bmatrix}$

Final Answers:

1. $4B C + 2C = \begin{bmatrix} 6 & 68 & 16 \\ 30 & 10 & 50 \end{bmatrix}$
2. The transpose of $A$, $A^T = \begin{bmatrix} 3 & -1 & 1 \\ 0 & 2 & 1 \end{bmatrix}$

Would you like any further details on these steps, or have any questions?

Additional Related Questions:

1. How do you perform matrix multiplication in general?
2. What are the properties of matrix transposition?
3. How does scalar multiplication work in matrices?
4. Can we generalize this process for any arbitrary matrices $B$ and $C$?
5. How is matrix addition different from matrix multiplication?

Tip:

When performing matrix multiplication, always check the dimensions first to ensure compatibility.