## Algebra and Trigonometry 10th Edition

$\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{bmatrix}$ AB is a 3×3 matrix
A is a 3×3 matrix, and B is a 3×3 matrix. The number of columns of A is equal to the number of rows of B. So, it is possible to find AB, where AB is a 3×3 matrix. $\begin{bmatrix} 0(6) + 0(8) + 5(0) & 0(-11) + 0(16) + 5(0) & 0(4) + 0(4) + 5(0)\\ 0(6) + 0(8) + (-3)(0) & 0(-11) + 0(16) + (-3)(0) & 0(4) + 0(4) + (-3)(0)\\ 0(6) + 0(8) + 4(0) & 0(-11) + 0(16) + 4(0) & 0(4) + 0(4) + 4(0)\\ \end{bmatrix}$ = $\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{bmatrix}$