Calculate \(\left[\begin{array}{rr}2 & -1 \\ 3 & 2\end{array}\right]\left[\begin{array}{l}2 \\ 1\end{array}\right]\) a) \(\left[\begin{array}{l}8 \\ 3\end{array}\right]\) b) \(\left[\begin{array}{l}3 \\ 8\end{array}\right]\) c) \(\left[\begin{array}{l}-3 \\ -8\end{array}\right]\) d) \([3,8]\)

Identify the given input matrices. Matrix A: \(\left[\begin{array}{rr}2 & -1 \\\ 3 & 2\end{array}\right]\). Matrix B: \(\left[\begin{array}{l}2 \\\ 1\end{array}\right]\).

Multiply the given matrices as follows: $$(Matrix A) * (Matrix B) = (2x2) * (2x1) = (2x1)$$ To perform the multiplication, we will take the first row of Matrix A and multiply it, element-wise, with the first column of Matrix B. Then, we will add them together, placing the sum in the first position of the result: $$\left[\begin{array}{rr}2 & -1\end{array}\right] * \left[\begin{array}{l}2 \\\ 1\end{array}\right] = (2 * 2) + (-1 * 1) = 4 - 1 = 3$$ Next, we'll do the same thing with the second row of Matrix A: $$\left[\begin{array}{rr}3 & 2\end{array}\right] * \left[\begin{array}{l}2 \\\ 1\end{array}\right] = (3 * 2) + (2 * 1) = 6 + 2 = 8$$ So, the resulting matrix (let's call it Matrix C) is: $$Matrix C = \left[\begin{array}{l}3 \\\ 8\end{array}\right]$$ Thus, the correct answer is: b) \(\left[\begin{array}{l}3 \\\ 8\end{array}\right]\)