Answer
$a.\qquad AH=\left[\begin{array}{rr}
-4 & 7\\
14 & -7
\end{array}\right]$
$b.\qquad HA=\left[\begin{array}{rr}
6 & 8\\
6 & -17
\end{array}\right]$
Work Step by Step
If $A$ is an $m\times n$ matrix and $B$ is an $n\times k$ matrix
(so the number of columns of $A$ is the same as the number of rows of $B$),
then the matrix product $AB$ is the $m\times k$ matrix
whose $ij$-entry is the inner product of the $i\mathrm{t}\mathrm{h}$ row of $A$ and the jth column of $B.$
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D is a 1$\times$2 matrix. H is a 2$\times$2 matrix
$a.$
$AH$ is defined, and is a 2$\times$2 matrix
$[AH]_{11}$= (row 1 in $A$) times (column 1 in $H$)$=2(3)+(-5)(2)=-4$
$[AH]_{12}$= (row 1 in $A$) times (column 2 in $H$)$=2(1)+(-5)(-1)=7$
$[AH]_{21}$= (row $2$ in $A$) times (column 1 in $H$)$=0(3)+7(2)=14$
$[AH]_{22}$= (row $2$ in $A$) times (column 2 in $H$)$=0(1)+7(-1)=-7$
$AH=\left[\begin{array}{ll}
-4 & 7\\
14 & -7
\end{array}\right]$
$b.$
$HA$ is defined, and is a 2$\times$2 matrix
$[HA]_{11}$= (row 1 in $H$) times (column 1 in $A$)$=3(2)+1(0)=6$
$[HA]_{12}$= (row 1 in $H$) times (column 2 in $A$)$=3(-5)+1(7)=8$
$[HA]_{21}$= (row $2$ in $H$) times (column 1 in $A$)$=2(2)+(-1)(0)=6$
$[HA]_{22}$= (row $2$ in $H$) times (column 2 in $A$)$=2(-5)+(-1)(7)=-17$
$HA=\left[\begin{array}{ll}
6 & 8\\
6 & -17
\end{array}\right]$
