## College Algebra (6th Edition)

a. $AB=\left[\begin{array}{lll} 2 & -8 & 20\\ 8 & 3 & 5\\ 10 & 2 & 10 \end{array}\right]$ b. $BA=\left[\begin{array}{ll} 12 & 14\\ 9 & 3 \end{array}\right]$
A product of two matrices exists if the first has as many columns as the second matrix has rows. The product of an $m\times\underline{n}$ matrix $A$ and an $\underline{n}\times p$ matrix $B$ is an $m\times p$ matrix $AB$. a. C=AB exists because A is a $3\times\underline{2}$ matrix and B is a $\underline{2}\times 3$ matrix. C is a 3$\times$3 matrix, $C=[c_{ij}]$, where $c_{ij}$ is the (ith row of A) times (jth column of B) $c_{11}=2(3)+4(-1)=6-4=2$ $c_{12}=2(2)+4(-3)=4-12=-8$ and so on $AB=\left[\begin{array}{lll} 6-4 & 4-12 & 0+20\\ 9-1 & 6-3 & 0+5\\ 12-2 & 8-6 & 0+10 \end{array}\right]$ $=\left[\begin{array}{lll} 2 & -8 & 20\\ 8 & 3 & 5\\ 10 & 2 & 10 \end{array}\right]$ b. D=BA exists because B is a $2\times\underline{3}$ matrix and A is a $\underline{3}\times 2$ matrix. D is a $2\times 2$ matrix, $D=[d_{ij}]$, where $d_{ij}$ is the (ith row of B) times (jth column of A) $d_{11}=3(2)+2(3)+0(4)=12$ $d_{12}=3(4)+2(1)+0(2)=14$ $d_{21}=-1(2)+(-3)(3)+5(4)=9$ $d_{22}=-1(4)+(-3)(1)+5(2)=3$ $BA=\left[\begin{array}{ll} 12 & 14\\ 9 & 3 \end{array}\right]$