The problem is asking you to work with matrices $A = \begin{pmatrix} 3 & -4 \\ 1 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 2 \\ 1 & 0 \end{pmatrix}$, to compute the following:

1. $B^T$ (the transpose of $B$)
2. $A^{-1}$ (the inverse of $A$)
3. $A^T + 3B$ (the transpose of $A$ added to 3 times $B$)

Solutions:

1. $B^T$

The transpose of a matrix is obtained by swapping rows and columns. Therefore: $B = \begin{pmatrix} 3 & 2 \\ 1 & 0 \end{pmatrix}, \quad B^T = \begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix}.$

2. $A^{-1}$

The inverse of $A$ is computed as: $A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}, \text{ where } \text{det}(A) = ad - bc.$ For $A = \begin{pmatrix} 3 & -4 \\ 1 & 2 \end{pmatrix}$: $\text{det}(A) = (3)(2) - (1)(-4) = 6 + 4 = 10.$ So: $A^{-1} = \frac{1}{10} \begin{pmatrix} 2 & 4 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 0.2 & 0.4 \\ -0.1 & 0.3 \end{pmatrix}.$

3. $A^T + 3B$

The transpose of $A$, $A^T$, is: $A^T = \begin{pmatrix} 3 & 1 \\ -4 & 2 \end{pmatrix}.$ Now, calculate $3B$: $3B = 3 \cdot \begin{pmatrix} 3 & 2 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 9 & 6 \\ 3 & 0 \end{pmatrix}.$ Finally, compute $A^T + 3B$: $A^T + 3B = \begin{pmatrix} 3 & 1 \\ -4 & 2 \end{pmatrix} + \begin{pmatrix} 9 & 6 \\ 3 & 0 \end{pmatrix} = \begin{pmatrix} 12 & 7 \\ -1 & 2 \end{pmatrix}.$

Final Results:

1. $B^T = \begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix}$
2. $A^{-1} = \begin{pmatrix} 0.2 & 0.4 \\ -0.1 & 0.3 \end{pmatrix}$
3. $A^T + 3B = \begin{pmatrix} 12 & 7 \\ -1 & 2 \end{pmatrix}$

Would you like a detailed explanation of any of these steps?

Expanding Your Understanding:

1. How do you calculate the determinant of a matrix?
2. What is the geometric meaning of a transpose of a matrix?
3. How does the inverse of a matrix relate to solving linear equations?
4. Why does the determinant of a matrix need to be nonzero for the inverse to exist?
5. How does scalar multiplication affect each element of a matrix?

Tip: When finding the inverse of a matrix, always double-check the determinant value—it ensures the calculation is valid!