The given problem involves matrix operations. Let's analyze and solve each part step by step.

We have the matrices: $A = \begin{bmatrix} 1 & -4 \\ 2 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$

Step 1: Compute Transposes

The transpose of a matrix $M$ is obtained by swapping its rows and columns.

• Transpose of $A$ (denoted $A^T$): $A^T = \begin{bmatrix} 1 & 2 \\ -4 & 3 \end{bmatrix}$

• Transpose of $B$ (denoted $B^T$): $B^T = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

Step 2: Solve (a) $B^T - A$

$B^T - A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 1 & -4 \\ 2 & 3 \end{bmatrix}$

Perform element-wise subtraction:

$B^T - A = \begin{bmatrix} 1-1 & 2-(-4) \\ 3-2 & 4-3 \end{bmatrix} = \begin{bmatrix} 0 & 6 \\ 1 & 1 \end{bmatrix}$

Step 3: Solve (b) $A^T + B$

$A^T + B = \begin{bmatrix} 1 & 2 \\ -4 & 3 \end{bmatrix} + \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$

Perform element-wise addition:

$A^T + B = \begin{bmatrix} 1+1 & 2+3 \\ -4+2 & 3+4 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ -2 & 7 \end{bmatrix}$

Final Answers:

• $B^T - A = \begin{bmatrix} 0 & 6 \\ 1 & 1 \end{bmatrix}$
• $A^T + B = \begin{bmatrix} 2 & 5 \\ -2 & 7 \end{bmatrix}$

Would you like further clarification on any steps?

Here are five related questions to deepen your understanding:

1. How do you find the inverse of a $2 \times 2$ matrix?
2. What is the determinant of matrix $A$ and $B$?
3. How do you multiply two matrices together?
4. What are the properties of transpose operations in matrices?
5. How can matrices be used to solve systems of linear equations?

Tip: When dealing with matrices, always check for properties like symmetry, determinant, and rank to understand their behavior in mathematical operations.