The figure to the right shows the letter $\mathrm{L}$ in a rectangular coordinate system. The figure can be represented by the matrix B, shown to the right. Each column in the matrix describes a point on the letter. The order of the columns shows the direction in which a pencil must move to draw the letter. The $\mathrm{L}$ is completed by connecting the last point in the matrix, $(0,6)$, to the starting point, $(0,0)$. Use these ideas to complete parts $(a)$ and $(b)$ below. \[ B=\left[\begin{array}{llllll} 0 & 4 & 4 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 6 & 6 \end{array}\right] \] a. If $A=\left[\begin{array}{rr}-1 & 0 \\ 0 & 1\end{array}\right]$, find $A B$. \[ A B=\square(\text { Simplify your answer.) } \]
Solution
Step 1 ：Given matrices A and B as follows: \[A=\left[\begin{array}{rr}-1 & 0 \ 0 & 1\end{array}\right]\] and \[B=\left[\begin{array}{llllll}0 & 4 & 4 & 1 & 1 & 0 \ 0 & 0 & 1 & 1 & 6 & 6\end{array}\right]\]
Step 2 ：We are asked to find the product of matrices A and B. Matrix multiplication is done element by element in a specific order. The element in the i-th row and j-th column of the resulting matrix is the sum of the product of corresponding elements from the i-th row of the first matrix and the j-th column of the second matrix.
Step 3 ：Performing the matrix multiplication, we get: \[AB = \left[\begin{array}{llllll}0 & -4 & -4 & -1 & -1 & 0 \ 0 & 0 & 1 & 1 & 6 & 6\end{array}\right]\]
Step 4 ：\(\boxed{AB = \left[\begin{array}{llllll}0 & -4 & -4 & -1 & -1 & 0 \ 0 & 0 & 1 & 1 & 6 & 6\end{array}\right]}\) is the final answer.
