## College Algebra (6th Edition)

a. $AB=\left[\begin{array}{lll} 6 & 8 & 16\\ 11 & 16 & 24\\ 1 & -1 & 12 \end{array}\right]$ b. $BA=\left[\begin{array}{ll} 38 & 27\\ -16 & 4 \end{array}\right]$
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) $c_{11}=4(2)+2(-1)=8-2=6$ $c_{12}=4(3)+2(-2)=12-4=8$ and so on $AB=\left[\begin{array}{lll} 8-2 & 12-4 & 16+0\\ 12-1 & 18-2 & 24+0\\ 6-5 & 9-10 & 12+0 \end{array}\right]$ $=\left[\begin{array}{lll} 6 & 8 & 16\\ 11 & 16 & 24\\ 1 & -1 & 12 \end{array}\right]$ 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) $d_{11}=2(4)+3(6)+4(3)=8+18+12=38,$ $d_{12}=2(2)+3(1)+4(5)=4+3+20=27$ $d_{21}=-1(4)+(-2)(6)+0(3)=-4-12+0=-16$ $d_{22}=-1(2)+(-2)(1)+0(5)=-2-2+0=-4$ $BA=\left[\begin{array}{ll} 38 & 27\\ -16 & 4 \end{array}\right]$