af3811…: “Upon a still quadrilateral”

$\newenvironment {prompt}{}{} \newcommand {\DeclareMathOperator }[2]{\@OldDeclareMathOperator {##1}{##2}\immediate \write \myfile {\unexpanded {\DeclareMathOperator }{\unexpanded {##1}}{\unexpanded {##2}}}} \newcommand {\amatrix }[1]{\left [\begin {array}{@{}*{##1}{c}|c@{}} } \newcommand {\RR }[0]{\bf R} \newcommand {\reserved@a }[2]{} \newcommand {\reserved@a }[2]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{} \newcommand {\AppendGraphicsExtensions }[0]{\@ifundefined {Gin@extensions}{\let \Gin@extensions \@empty }{}\@ifstar {\grfext@Append \grfext@Check }{\grfext@Append \grfext@@Add }} \newcommand {\PrependGraphicsExtensions }[0]{\@ifundefined {Gin@extensions}{\let \Gin@extensions \@empty }{}\@ifstar {\grfext@Prepend \grfext@Check }{\grfext@Prepend \grfext@@Add }} \newcommand {\RemoveGraphicsExtensions }[1]{\@ifundefined {Gin@extensions}{\def \Gin@extensions {}}{\edef \grfext@tmp {\zap@space ##1 \@empty }\@for \grfext@ext :=\grfext@tmp \do {\def \grfext@next {\let \grfext@tmp \Gin@extensions \@expandtwoargs \@removeelement \grfext@ext \Gin@extensions \Gin@extensions \ifx \grfext@tmp \Gin@extensions \let \grfext@next \relax \fi \grfext@next }\grfext@next }}\grfext@Print \RemoveGraphicsExtensions } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} \newcommand {\epstopdfcall }[1]{\ifETE@InsideSetfile \expandafter \@firstoftwo \else \expandafter \@secondoftwo \fi {‘##1}{\Gin@base \Gin@ext }} \newcommand {\epstopdfDeclareGraphicsRule }[4]{\ifx \\##4\\\@PackageError {epstopdf-base}{Conversion command is missing}\@ehc \else \begingroup \@ifundefined {Gin@rule@##1}{}{\@PackageInfo {epstopdf-base}{Redefining graphics rule for ‘##1’}}\endgroup \@namedef {Gin@rule@##1}####1{{##2}{##3}{\epstopdfcall {##4}}}\fi } \newcommand {\mdf@patchamsthm }[0]{\let \mdf@deferred@thm@head \deferred@thm@head \pretocmd {\deferred@thm@head }{\@inlabelfalse }{\mdf@PackageInfo {mdframed detected package amsthm changed the theorem header of amsthm\MessageBreak }}{\mdf@PackageError {mdframed detected package amsthm changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\GetTitleString }[0]{\ifGTS@expand \expandafter \GetTitleStringExpand \else \expandafter \GetTitleStringNonExpand \fi } \newcommand {\GetTitleStringExpand }[1]{\def \GetTitleStringResult {##1}\begingroup \GTS@DisablePredefinedCmds \GTS@DisableHook \edef \x {\endgroup \noexpand \def \noexpand \GetTitleStringResult {\GetTitleStringResult }}\x } \newcommand {\GetTitleStringNonExpand }[1]{\def \GetTitleStringResult {##1}\global \let \GTS@GlobalString \GetTitleStringResult \begingroup \GTS@RemoveLeft \GTS@RemoveRight \endgroup \let \GetTitleStringResult \GTS@GlobalString } \newcommand {\GTS@DisableHook }[0]{} \newcommand {\label@hook }[0]{} \newcommand {\@currentlabelname }[0]{} \newcommand {\reserved@a }[0]{} \newcommand {\@safe@activestrue }[0]{} \newcommand {\@safe@activesfalse }[0]{} \newcommand {\nameref }[0]{\@ifstar \T@nameref \T@nameref } \newcommand {\Sectionformat }[2]{##1} \newcommand {\vnameref }[1]{\unskip ~\nameref {##1}\@vpageref [\unskip ]{##1}} \newcommand {\ref }[0]{\@ifstar \@refstar \T@ref } \newcommand {\pageref }[0]{\@ifstar \@pagerefstar \T@pageref } \newcommand {\nameref }[0]{\@ifstar \@namerefstar \T@nameref } \newcommand {\reserved@a }[2]{} \newcommand {\reserved@a }[2]{} \newcommand {\reserved@a }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{}$

Determine which of the following products are defined.

$AB= \begin {bmatrix} 1&3&4 \end {bmatrix} \begin {bmatrix} 0\\-3\\2 \end {bmatrix}$

defined undefined

$AB= \begin {bmatrix} 0\\-3\\2 \end {bmatrix} \begin {bmatrix} 1&3&4 \end {bmatrix}$

defined undefined

$AB= \begin {bmatrix} 1&3\\ 0&4 \end {bmatrix} \begin {bmatrix} 0 &7\\-3&1\\2&0 \end {bmatrix}$

defined undefined

Let $A = \begin {bmatrix} 11& 0 &12\\ 7&6&-1\\ 0 &-3 &14\end {bmatrix}$ and $B = \begin {bmatrix} 1 &2\\ 0 &-1 \\ 3& 1 \end {bmatrix}$ . Compute the following matrices.

$AB = \left [ \begin {array}{cc} \answer {47} & \answer {34} \\ \answer {4} &\answer {7}\\ \answer {42}&\answer {17} \end {array}\right ]$

$A^T = \left [ \begin {array}{ccc} \answer {11} & \answer {7} & \answer {0} \\ \answer {0} &\answer {6}& \answer {-3}\\ \answer {12}&\answer {-1} & \answer {14}\end {array}\right ]$

$B^T = \left [ \begin {array}{ccc} \answer {1} & \answer {0}& \answer {3} \\ \answer {2} &\answer {-1} & \answer {1} \end {array}\right ]$

$4B = \left [\begin {array}{cc} \answer {4} & \answer {8} \\ \answer {0} &\answer {-4}\\ \answer {12}&\answer {4} \end {array}\right ]$