## Algebra and Trigonometry 10th Edition

Part A: $\begin{bmatrix} 7 & 7 & 14\\ 8 & 8 & 16\\ -1 & -1 & -2\\ \end{bmatrix}$ Part B: $\begin{bmatrix} 13\\ \end{bmatrix}$ Part C: Not possible.
A has 3 rows and 1 column (3 x 1). B has 1 row and 3 columns (1 x 3). Part A: $\begin{bmatrix} 7(1) & 7(1) & 7(2)\\ 8(1) & 8(1) & 8(2)\\ (-1)(1) & (-1)(1) & (-1)(2)\\ \end{bmatrix}$ = $\begin{bmatrix} 7 & 7 & 14\\ 8 & 8 & 16\\ -1 & -1 & -2\\ \end{bmatrix}$ Part B: $\begin{bmatrix} 1(7) + 1(8) + 2(-1)\\ \end{bmatrix}$ = $\begin{bmatrix} 7 + 8 - 2\\ \end{bmatrix}$ = $\begin{bmatrix} 13\\ \end{bmatrix}$ Part C: Since the number of columns in A does not equal the number of rows in A, AA is not possible.