If $$A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -3 & 2 & 1 \end{bmatrix}$$ and $$B = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 7 & -2 & 1 \end{bmatrix}$$ then $$AB$$ equals

Write the two matrices explicitly:
$$A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -3 & 2 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 7 & -2 & 1 \end{bmatrix}$$

To find $$AB$$ we multiply rows of $$A$$ with columns of $$B$$.

Row 1 of $$A$$ with each column of $$B$$
Row 1 of $$A$$ is $$\begin{bmatrix} 1 & 0 & 0 \end{bmatrix}$$.
Column 1 of $$B$$: $$1(1) + 0(-2) + 0(7) = 1$$
Column 2 of $$B$$: $$1(0) + 0(1) + 0(-2) = 0$$
Column 3 of $$B$$: $$1(0) + 0(0) + 0(1) = 0$$
Hence the first row of $$AB$$ is $$\begin{bmatrix} 1 & 0 & 0 \end{bmatrix}$$.

Row 2 of $$A$$ with each column of $$B$$
Row 2 of $$A$$ is $$\begin{bmatrix} 2 & 1 & 0 \end{bmatrix}$$.
Column 1 of $$B$$: $$2(1) + 1(-2) + 0(7) = 2 - 2 = 0$$
Column 2 of $$B$$: $$2(0) + 1(1) + 0(-2) = 1$$
Column 3 of $$B$$: $$2(0) + 1(0) + 0(1) = 0$$
Hence the second row of $$AB$$ is $$\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}$$.

Row 3 of $$A$$ with each column of $$B$$
Row 3 of $$A$$ is $$\begin{bmatrix} -3 & 2 & 1 \end{bmatrix}$$.
Column 1 of $$B$$: $$-3(1) + 2(-2) + 1(7) = -3 - 4 + 7 = 0$$
Column 2 of $$B$$: $$-3(0) + 2(1) + 1(-2) = 2 - 2 = 0$$
Column 3 of $$B$$: $$-3(0) + 2(0) + 1(1) = 1$$
Hence the third row of $$AB$$ is $$\begin{bmatrix} 0 & 0 & 1 \end{bmatrix}$$.

Collecting the three rows, we obtain
$$AB = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I_3$$

Therefore $$AB$$ equals the identity matrix.

Option A which is: $$I$$