Given the matrices M = [[-3, 4], [5, 6]] and N = [[1, 0], [-5, -7]], find MN and M^T + N^T.

To multiply two matrices, we multiply the rows of the first matrix by the columns of the second matrix.

$$MN = \begin{pmatrix} (-3)(1) + (4)(-5) & (-3)(0) + (4)(-7) \\ (5)(1) + (6)(-5) & (5)(0) + (6)(-7) \end{pmatrix}$$

$$MN = \begin{pmatrix} -3 - 20 & 0 - 28 \\ 5 - 30 & 0 - 42 \end{pmatrix}$$ $$MN = \begin{pmatrix} -23 & -28 \\ -25 & -42 \end{pmatrix}$$

The product $MN$ is:

$$\boxed{\begin{pmatrix} -23 & -28 \\ -25 & -42 \end{pmatrix}}$$

The transpose of a matrix is obtained by interchanging its rows and columns.

$$M^T = \begin{pmatrix} -3 & 5 \\ 4 & 6 \end{pmatrix}$$

To add matrices, we add their corresponding elements.

$$M^T + N^T = \begin{pmatrix} -3 & 5 \\ 4 & 6 \end{pmatrix} + \begin{pmatrix} 1 & -5 \\ 0 & -7 \end{pmatrix}$$ $$M^T + N^T = \begin{pmatrix} -3+1 & 5+(-5) \\ 4+0 & 6+(-7) \end{pmatrix}$$ $$M^T + N^T = \begin{pmatrix} -2 & 0 \\ 4 & -1 \end{pmatrix}$$

The sum $M^T + N^T$ is:

$$\boxed{\begin{pmatrix} -2 & 0 \\ 4 & -1 \end{pmatrix}}$$

For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is $ad - bc$.
For $N = \begin{pmatrix} 1 & 0 \\ -5 & -7 \end{pmatrix}$:

$$\det(N) = (1)(-7) - (0)(-5)$$ $$\det(N) = -7 - 0$$ $$\det(N) = -7$$

For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the inverse is $\frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.

$$N^{-1} = \frac{1}{-7} \begin{pmatrix} -7 & -0 \\ -(-5) & 1 \end{pmatrix}$$ $$N^{-1} = -\frac{1}{7} \begin{pmatrix} -7 & 0 \\ 5 & 1 \end{pmatrix}$$

$$N^{-1} = \begin{pmatrix} (-\frac{1}{7})(-7) & (-\frac{1}{7})(0) \\ (-\frac{1}{7})(5) & (-\frac{1}{7})(1) \end{pmatrix}$$ $$N^{-1} = \begin{pmatrix} 1 & 0 \\ -\frac{5}{7} & -\frac{1}{7} \end{pmatrix}$$

The inverse of N is:

$$\boxed{\begin{pmatrix} 1 & 0 \\ -\frac{5}{7} & -\frac{1}{7} \end{pmatrix}}$$