Additional Exercises for Ch 7

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\tdplotsinandcos }[3]{\pgfmathsetmacro {#1}{sin(#3)}\pgfmathsetmacro {#2}{cos(#3)}} \newcommand {\tdplotmult }[3]{\pgfmathsetmacro {#1}{#2*#3}} \newcommand {\tdplotdiv }[3]{\pgfmathsetmacro {#1}{#2/#3}} \newcommand {\tdplotcheckdiff }[5]{\par \par \pgfmathparse { abs(#2 -#1)<#3 } \par \ifthenelse {\equal {\pgfmathresult }{1}}{#4}{#5} } \newcommand {\tdplotsetmaincoords }[2]{\pgfmathsetmacro {\tdplotmaintheta }{#1} \pgfmathsetmacro {\tdplotmainphi }{#2} \tdplotcalctransformmainscreen \tikzset {tdplot_main_coords/.style={x={(\raarot cm,\rbarot cm)},y={(\rabrot cm,\rbbrot cm)},z={(\racrot cm,\rbcrot cm)}}}} \newcommand {\tdplotcalctransformmainscreen }[0]{\tdplotsinandcos {\sintheta }{\costheta }{\tdplotmaintheta }\tdplotsinandcos {\sinphi }{\cosphi }{\tdplotmainphi }\tdplotmult {\stsp }{\sintheta }{\sinphi }\tdplotmult {\stcp }{\sintheta }{\cosphi }\tdplotmult {\ctsp }{\costheta }{\sinphi }\tdplotmult {\ctcp }{\costheta }{\cosphi }\pgfmathsetmacro {\raarot }{\cosphi }\pgfmathsetmacro {\rabrot }{\sinphi }\pgfmathsetmacro {\racrot }{0}\pgfmathsetmacro {\rbarot }{-\ctsp }\pgfmathsetmacro {\rbbrot }{\ctcp }\pgfmathsetmacro {\rbcrot }{\sintheta }\pgfmathsetmacro {\rcarot }{\stsp }\pgfmathsetmacro {\rcbrot }{-\stcp }\pgfmathsetmacro {\rccrot }{\costheta }} \newcommand {\tdplotcalctransformrotmain }[0]{\tdplotsinandcos {\sinalpha }{\cosalpha }{\tdplotalpha } \tdplotsinandcos {\sinbeta }{\cosbeta }{\tdplotbeta } \tdplotsinandcos {\singamma }{\cosgamma }{\tdplotgamma } \tdplotmult {\sasb }{\sinalpha }{\sinbeta } \tdplotmult {\sbsg }{\sinbeta }{\singamma } \tdplotmult {\sasg }{\sinalpha }{\singamma } \tdplotmult {\sasbsg }{\sasb }{\singamma } \tdplotmult {\sacb }{\sinalpha }{\cosbeta } \tdplotmult {\sacg }{\sinalpha }{\cosgamma } \tdplotmult {\sbcg }{\sinbeta }{\cosgamma } \tdplotmult {\sacbsg }{\sacb }{\singamma } \tdplotmult {\sacbcg }{\sacb }{\cosgamma } \tdplotmult {\casb }{\cosalpha }{\sinbeta } \tdplotmult {\cacb }{\cosalpha }{\cosbeta } \tdplotmult {\cacg }{\cosalpha }{\cosgamma } \tdplotmult {\casg }{\cosalpha }{\singamma } \tdplotmult {\cacbsg }{\cacb }{\singamma } \tdplotmult {\cacbcg }{\cacb }{\cosgamma } \pgfmathsetmacro {\raaeul }{\cacbcg -\sasg } \pgfmathsetmacro {\rabeul }{-\cacbsg -\sacg } \pgfmathsetmacro {\raceul }{\casb } \pgfmathsetmacro {\rbaeul }{\sacbcg + \casg } \pgfmathsetmacro {\rbbeul }{-\sacbsg + \cacg } \pgfmathsetmacro {\rbceul }{\sasb } \pgfmathsetmacro {\rcaeul }{-\sbcg } \pgfmathsetmacro {\rcbeul }{\sbsg } \pgfmathsetmacro {\rcceul }{\cosbeta } } \newcommand {\tdplotcalctransformmainrot }[0]{\tdplotsinandcos {\sinalpha }{\cosalpha }{\tdplotalpha } \tdplotsinandcos {\sinbeta }{\cosbeta }{\tdplotbeta } \tdplotsinandcos {\singamma }{\cosgamma }{\tdplotgamma } \tdplotmult {\sasb }{\sinalpha }{\sinbeta } \tdplotmult {\sbsg }{\sinbeta }{\singamma } \tdplotmult {\sasg }{\sinalpha }{\singamma } \tdplotmult {\sasbsg }{\sasb }{\singamma } \tdplotmult {\sacb }{\sinalpha }{\cosbeta } \tdplotmult {\sacg }{\sinalpha }{\cosgamma } \tdplotmult {\sbcg }{\sinbeta }{\cosgamma } \tdplotmult {\sacbsg }{\sacb }{\singamma } \tdplotmult {\sacbcg }{\sacb }{\cosgamma } \tdplotmult {\casb }{\cosalpha }{\sinbeta } \tdplotmult {\cacb }{\cosalpha }{\cosbeta } \tdplotmult {\cacg }{\cosalpha }{\cosgamma } \tdplotmult {\casg }{\cosalpha }{\singamma } \tdplotmult {\cacbsg }{\cacb }{\singamma } \tdplotmult {\cacbcg }{\cacb }{\cosgamma } \pgfmathsetmacro {\raaeul }{\cacbcg -\sasg } \pgfmathsetmacro {\rabeul }{\sacbcg + \casg } \pgfmathsetmacro {\raceul }{-\sbcg } \pgfmathsetmacro {\rbaeul }{-\cacbsg -\sacg } \pgfmathsetmacro {\rbbeul }{-\sacbsg + \cacg } \pgfmathsetmacro {\rbceul }{\sbsg } \pgfmathsetmacro {\rcaeul }{\casb } \pgfmathsetmacro {\rcbeul }{\sasb } \pgfmathsetmacro {\rcceul }{\cosbeta } } \newcommand {\tdplottransformmainrot }[3]{\tdplotcalctransformmainrot \par \pgfmathsetmacro {\tdplotresx }{\raaeul * #1 + \rabeul * #2 + \raceul * #3} \pgfmathsetmacro {\tdplotresy }{\rbaeul * #1 + \rbbeul * #2 + \rbceul * #3} \pgfmathsetmacro {\tdplotresz }{\rcaeul * #1 + \rcbeul * #2 + \rcceul * #3} } \newcommand {\tdplottransformrotmain }[3]{\tdplotcalctransformrotmain \par \pgfmathsetmacro {\tdplotresx }{\raaeul * #1 + \rabeul * #2 + \raceul * #3} \pgfmathsetmacro {\tdplotresy }{\rbaeul * #1 + \rbbeul * #2 + \rbceul * #3} \pgfmathsetmacro {\tdplotresz }{\rcaeul * #1 + \rcbeul * #2 + \rcceul * #3} } \newcommand {\tdplottransformmainscreen }[3]{\tdplotcalctransformmainscreen \par \pgfmathsetmacro {\tdplotresx }{\raarot * #1 + \rabrot * #2 + \racrot * #3} \pgfmathsetmacro {\tdplotresy }{\rbarot * #1 + \rbbrot * #2 + \rbcrot * #3} } \newcommand {\tdplotsetrotatedcoords }[3]{\pgfmathsetmacro {\tdplotalpha }{#1} \pgfmathsetmacro {\tdplotbeta }{#2} \pgfmathsetmacro {\tdplotgamma }{#3} \tdplotcalctransformrotmain \par \tdplotmult {\raaeaa }{\raarot }{\raaeul } \tdplotmult {\rabeba }{\rabrot }{\rbaeul } \tdplotmult {\raceca }{\racrot }{\rcaeul } \tdplotmult {\raaeab }{\raarot }{\rabeul } \tdplotmult {\rabebb }{\rabrot }{\rbbeul } \tdplotmult {\racecb }{\racrot }{\rcbeul } \tdplotmult {\raaeac }{\raarot }{\raceul } \tdplotmult {\rabebc }{\rabrot }{\rbceul } \tdplotmult {\racecc }{\racrot }{\rcceul } \tdplotmult {\rbaeaa }{\rbarot }{\raaeul } \tdplotmult {\rbbeba }{\rbbrot }{\rbaeul } \tdplotmult {\rbceca }{\rbcrot }{\rcaeul } \tdplotmult {\rbaeab }{\rbarot }{\rabeul } \tdplotmult {\rbbebb }{\rbbrot }{\rbbeul } \tdplotmult {\rbcecb }{\rbcrot }{\rcbeul } \tdplotmult {\rbaeac }{\rbarot }{\raceul } \tdplotmult {\rbbebc }{\rbbrot }{\rbceul } \tdplotmult {\rbcecc }{\rbcrot }{\rcceul } \pgfmathsetmacro {\raarc }{\raaeaa + \rabeba + \raceca } \pgfmathsetmacro {\rabrc }{\raaeab + \rabebb + \racecb } \pgfmathsetmacro {\racrc }{\raaeac + \rabebc + \racecc } \pgfmathsetmacro {\rbarc }{\rbaeaa + \rbbeba + \rbceca } \pgfmathsetmacro {\rbbrc }{\rbaeab + \rbbebb + \rbcecb } \pgfmathsetmacro {\rbcrc }{\rbaeac + \rbbebc + \rbcecc } \tikzset {tdplot_rotated_coords/.append style={x={(\raarc cm,\rbarc cm)},y={(\rabrc cm,\rbbrc cm)},z={(\racrc cm,\rbcrc cm)}}}} \newcommand {\tdplotsetrotatedcoordsorigin }[1]{\tikzset {tdplot_rotated_coords/.append style={shift=#1}}} \newcommand {\tdplotresetrotatedcoordsorigin }[0]{\tikzset {tdplot_rotated_coords/.append style={shift={(0,0,0)}}}} \newcommand {\tdplotsetthetaplanecoords }[1]{\tdplotresetrotatedcoordsorigin \tdplotsetrotatedcoords {270 + #1}{270}{0}} \newcommand {\tdplotsetrotatedthetaplanecoords }[1]{\tdplotsetrotatedcoords {\tdplotalpha }{\tdplotbeta }{\tdplotgamma + #1}\tikzset {tdplot_rotated_coords/.append style={y={(\raarc cm,\rbarc cm)},z={(\rabrc cm,\rbbrc cm)},x={(\racrc cm,\rbcrc cm)}}}} \newcommand {\tdplotsetcoord }[4]{\tdplotsinandcos {\sinthetavec }{\costhetavec }{#3}\tdplotsinandcos {\sinphivec }{\cosphivec }{#4}\tdplotmult {\stcpv }{\sinthetavec }{\cosphivec }\tdplotmult {\stspv }{\sinthetavec }{\sinphivec }\coordinate (#1) at ($#2*(\stcpv ,\stspv ,\costhetavec )$); \coordinate (#1xy) at ($#2*(\stcpv ,\stspv ,0)$); \coordinate (#1xz) at ($#2*(\stcpv ,0,\costhetavec )$); \coordinate (#1yz) at ($#2*(0,\stspv ,\costhetavec )$); \coordinate (#1x) at ($#2*(\stcpv ,0,0)$); \coordinate (#1y) at ($#2*(0,\stspv ,0)$); \coordinate (#1z) at ($#2*(0,0,\costhetavec )$); } \newcommand {\tdplotsimplesetcoord }[4]{\tdplotsinandcos {\sinthetavec }{\costhetavec }{#3}\tdplotsinandcos {\sinphivec }{\cosphivec }{#4}\tdplotmult {\stcpv }{\sinthetavec }{\cosphivec }\tdplotmult {\stspv }{\sinthetavec }{\sinphivec }\coordinate (#1) at ($#2*(\stcpv ,\stspv ,\costhetavec )$); } \newcommand {\tdplotsetpolarplotrange }[4]{\pgfmathsetmacro {\tdplotlowerphi }{#3} \pgfmathsetmacro {\tdplotupperphi }{#4} \pgfmathsetmacro {\tdplotlowertheta }{#1} \pgfmathsetmacro {\tdplotuppertheta }{#2} } \newcommand {\tdplotresetpolarplotrange }[0]{\pgfmathsetmacro {\tdplotlowerphi }{0} \pgfmathsetmacro {\tdplotupperphi }{360} \pgfmathsetmacro {\tdplotlowertheta }{0} \pgfmathsetmacro {\tdplotuppertheta }{180} } \newcommand {\tdplotdosurfaceplot }[6]{\par \pgfmathsetmacro {\nextphi }{\curphi + \tdplotsuperfudge *\viewphistep } \par \begin {scope}[opacity=1] \par \par \tdplotcheckdiff {\nextphi }{360}{\origviewphistep }{#2}{} \tdplotcheckdiff {\nextphi }{0}{\origviewphistep }{#2}{} \par \tdplotcheckdiff {\nextphi }{90}{\origviewphistep }{#3}{} \tdplotcheckdiff {\nextphi }{450}{\origviewphistep }{#3}{} \end {scope} \par \foreach \curtheta in{\viewthetastart ,\viewthetainc ,...,\viewthetaend } { \par \pgfmathsetmacro {\curlongitude }{90 -\curphi } \pgfmathsetmacro {\curlatitude }{90 -\curtheta } \par \ifthenelse {\equal {\leftright }{-1.0}}{\pgfmathsetmacro {\curphi }{\curphi -\origviewphistep } }{} \par \pgfmathsetmacro {\tdplottheta }{mod(\curtheta ,360)} \pgfmathsetmacro {\tdplotphi }{mod(\curphi ,360)} \par \pgfmathparse {\tdplotphi <0} \ifthenelse {\equal {\pgfmathresult }{1}}{ \pgfmathsetmacro {\tdplotphi }{\tdplotphi + 360} }{}\par \pgfmathparse {\tdplottheta >\tdplotuppertheta } \pgfmathsetmacro {\logictest }{1 -\pgfmathresult } \par \pgfmathparse {\tdplottheta <\tdplotlowertheta } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathsetmacro {\tdplottheta }{\tdplottheta + \viewthetastep } \pgfmathparse {\tdplottheta >\tdplotuppertheta } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathparse {\tdplottheta <\tdplotlowertheta } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathparse {\tdplotphi >\tdplotupperphi } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathparse {\tdplotphi <\tdplotlowerphi } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathsetmacro {\tdplotphi }{\tdplotphi + \viewphistep } \par \pgfmathparse {\tdplotphi <0} \ifthenelse {\equal {\pgfmathresult }{1}}{ \pgfmathsetmacro {\tdplotphi }{\tdplotphi + 360} }{}\par \pgfmathparse {\tdplotphi >\tdplotupperphi } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathparse {\tdplotphi <\tdplotlowerphi } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \par \pgfmathsetmacro {\tdplottheta }{\curtheta } \pgfmathsetmacro {\tdplotphi }{\curphi } \par \ifthenelse {\equal {#6}{parametricfill}}{\ifthenelse {\equal {\logictest }{1.0}}{\pgfmathsetmacro {\radius }{#1} \pgfmathsetmacro {\tdplotr }{\radius *360} \par \pgfmathlessthan {\radius }{0} \pgfmathsetmacro {\phaseshift }{180 * \pgfmathresult } \par \pgfmathsetmacro {\colorarg }{#5} \pgfmathsetmacro {\colorarg }{\colorarg + \phaseshift } \pgfmathsetmacro {\colorarg }{mod(\colorarg ,360)} \par \pgfmathlessthan {\colorarg }{0} \pgfmathsetmacro {\colorarg }{\colorarg + 360*\pgfmathresult } \par \pgfmathdivide {\colorarg }{360} \definecolor {tdplotfillcolor}{hsb}{\pgfmathresult ,1,1} \color {tdplotfillcolor} }{}}{\pgfsetfillcolor {#5} } \pgfsetstrokecolor {#4} \par \ifthenelse {\equal {\leftright }{-1.0}}{\pgfmathsetmacro {\curphi }{\curphi + \origviewphistep } }{} \par \ifthenelse {\equal {\logictest }{1.0}}{\pgfmathsetmacro {\radius }{abs(#1)} \pgfpathmoveto {\pgfpointspherical {\curlongitude }{\curlatitude }{\radius }} \par \pgfmathsetmacro {\tdplotphi }{\curphi + \viewphistep } \pgfmathsetmacro {\radius }{abs(#1)} \pgfpathlineto {\pgfpointspherical {\curlongitude -\viewphistep }{\curlatitude }{\radius }} \par \pgfmathsetmacro {\tdplottheta }{\curtheta + \viewthetastep } \pgfmathsetmacro {\radius }{abs(#1)} \pgfpathlineto {\pgfpointspherical {\curlongitude -\viewphistep }{\curlatitude -\viewthetastep }{\radius }} \par \pgfmathsetmacro {\tdplotphi }{\curphi } \pgfmathsetmacro {\radius }{abs(#1)} \pgfpathlineto {\pgfpointspherical {\curlongitude }{\curlatitude -\viewthetastep }{\radius }} \pgfpathclose \par \pgfusepath {fill,stroke} }{} } } \newcommand {\tdplotshowargcolorguide }[4]{ \par \pgfmathsetmacro {\tdplotx }{#1} \pgfmathsetmacro {\tdploty }{#2} \pgfmathsetmacro {\tdplothuestep }{5} \pgfmathsetmacro {\tdplotxsize }{#3} \pgfmathsetmacro {\tdplotysize }{#4} \par \pgfmathsetmacro {\tdplotyscale }{\tdplotysize /360} \par \foreach \tdplotphi in {0,\tdplothuestep ,...,360} { \pgfmathdivide {\tdplotphi }{360} \definecolor {tdplotfillcolor}{hsb}{\pgfmathresult ,1,1} \color {tdplotfillcolor} \par \pgfmathsetmacro {\tdplotstarty }{\tdploty + \tdplotphi * \tdplotyscale } \pgfmathsetmacro {\tdplotstopy }{\tdplotstarty + \tdplothuestep * \tdplotyscale } \pgfmathsetmacro {\tdplotstartx }{\tdplotx } \pgfmathsetmacro {\tdplotstopx }{\tdplotx + \tdplotxsize } \filldraw [tdplot_screen_coords] (\tdplotstartx ,\tdplotstarty ) rectangle (\tdplotstopx ,\tdplotstopy ); } \par \pgfmathsetmacro {\tdplotstopy }{\tdploty + (360+\tdplothuestep )*\tdplotyscale } \pgfmathsetmacro {\tdplotstopx }{\tdplotx + \tdplotxsize } \par \draw [tdplot_screen_coords] (\tdplotx ,\tdploty ) rectangle (\tdplotstopx ,\tdplotstopy ); \par \node [tdplot_screen_coords,anchor=west,xshift=5pt] at (\tdplotstopx ,\tdploty ) {$0$}; \node [tdplot_screen_coords,anchor=west,xshift=5pt] at (\tdplotstopx ,\tdplotstopy ) {$2\pi $}; \par \pgfmathsetmacro {\tdplotstopy }{\tdploty + (360+\tdplothuestep )/2*\tdplotyscale } \node [tdplot_screen_coords,anchor=west,xshift=5pt] at (\tdplotstopx ,\tdplotstopy ) {$\pi $}; } \newcommand {\tdplotgetpolarcoords }[3]{\pgfmathsetmacro {\vxcalc }{#1} \pgfmathsetmacro {\vycalc }{#2} \pgfmathsetmacro {\vzcalc }{#3} \pgfmathsetmacro {\vcalc }{ sqrt((\vxcalc )^2 + (\vycalc )^2 + (\vzcalc )^2) } \par \pgfmathsetmacro {\vxycalc }{ sqrt((\vxcalc )^2 + (\vycalc )^2) } \par \pgfmathsetmacro {\tdplotrestheta }{asin(\vxycalc /\vcalc )} \pgfmathparse {\vzcalc <0} \ifthenelse {\equal {\pgfmathresult }{1}}{\pgfmathsetmacro {\tdplotrestheta }{180 -\tdplotrestheta } } {} \ifthenelse {\equal {\vxcalc }{0.0}}{\pgfmathparse {\vycalc <0} \ifthenelse {\equal {\pgfmathresult }{1}}{\pgfmathsetmacro {\tdplotresphi }{270} } {\pgfmathparse {\vycalc >0} \ifthenelse {\equal {\pgfmathresult }{1}}{\pgfmathsetmacro {\tdplotresphi }{90} } {\pgfmathsetmacro {\tdplotresphi }{0} } } } {\pgfmathsetmacro {\tdplotresphi }{atan(\vycalc /\vxcalc )} \pgfmathparse {\vxcalc <0} \ifthenelse {\equal {\pgfmathresult }{1}}{\pgfmathsetmacro {\tdplotresphi }{\tdplotresphi +180} } { } \par \pgfmathparse {\tdplotresphi <0} \ifthenelse {\equal {\pgfmathresult }{1}}{\pgfmathsetmacro {\tdplotresphi }{\tdplotresphi +360} } {} } } \newcommand {\vec }[0]{\mathbf } \newcommand {\RR }[0]{\mathbb {R}} \newcommand {\dfn }[0]{\textit } \newcommand {\dotp }[0]{\cdot } \newcommand {\id }[0]{\text {id}} \newcommand {\norm }[1]{\left \lVert #1\right \rVert } \newcommand {\mathtoolsset }[1]{\setkeys {\MT_options_name: }{#1}} \newcommand {\refeq }[1]{\textup {\ref {#1}}} \newcommand {\lparen }[0]{(} \newcommand {\rparen }[0]{)} \newcommand {\ordinarycolon }[0]{:} \newcommand {\MT_test_for_tcb_other:nnnnn }[1]{\if:w t#1\relax \expandafter \MH_use_choice_i:nnnn \else: \if:w c#1\relax \expandafter \expandafter \expandafter \MH_use_choice_ii:nnnn \else: \if:w b#1\relax \expandafter \expandafter \expandafter \expandafter \expandafter \expandafter \expandafter \MH_use_choice_iii:nnnn \else: \expandafter \expandafter \expandafter \expandafter \expandafter \expandafter \expandafter \MH_use_choice_iv:nnnn \fi: \fi: \fi: } \newcommand {\newcases }[6]{\newenvironment {#1}{\MT_start_cases:nnnn {#2}{#3}{#4}{#5}}{\MH_end_cases: \right #6}} \newcommand {\renewcases }[6]{\renewenvironment {#1}{\MT_start_cases:nnnn {#2}{#3}{#4}{#5}}{\MH_end_cases: \right #6}} \newcommand {\SwapAboveDisplaySkip }[0]{\noalign {\vskip -\abovedisplayskip \vskip \abovedisplayshortskip }} \newcommand {\vdotswithin }[1]{{\mathmakebox [\widthof {\ensuremath {{}#1{}}}][c]{{\vdots }}}} \newcommand {\MTFlushSpaceBelow }[0]{\\\noalign {\nobreak \vskip -\lineskip \vskip -\l_MT_shortvdotswithinadjustbelow_dim \vskip -\origjot \vskip \jot }} \newcommand {\mathmbox }[0]{\mathpalette \MT_mathmbox:nn } \newcommand {\prescript }[3]{\mathchoice {\MT_prescript_inner: {#1}{#2}{#3}{\scriptstyle }}{\MT_prescript_inner: {#1}{#2}{#3}{\scriptstyle }}{\MT_prescript_inner: {#1}{#2}{#3}{\scriptscriptstyle }}{\MT_prescript_inner: {#1}{#2}{#3}{\scriptscriptstyle }}} \newcommand {\spreadlines }[1]{\setlength {\jot }{#1}\ignorespaces } \newcommand {\newgathered }[4]{\newenvironment {#1}{\def \MT_gathered_pre: {#2}\def \MT_gathered_post: {#3}\def \MT_gathered_env_end: {#4}\MT_gathered_env }{\endMT_gathered_env }} \newcommand {\renewgathered }[4]{\renewenvironment {#1}{\def \MT_gathered_pre: {#2}\def \MT_gathered_post: {#3}\def \MT_gathered_env_end: {#4}\MT_gathered_env }{\endMT_gathered_env }} \newcommand {\lgathered }[0]{\def \MT_gathered_pre: {}\def \MT_gathered_post: {\hfil }\def \MT_gathered_env_end: {}\MT_gathered_env } \newcommand {\rgathered }[0]{\def \MT_gathered_pre: {\hfil }\def \MT_gathered_post: {}\def \MT_gathered_env_end: {}\MT_gathered_env } \newcommand {\gathered }[0]{\def \MT_gathered_pre: {\hfil }\def \MT_gathered_post: {\hfil }\def \MT_gathered_env_end: {}\MT_gathered_env } \newcommand {\splitfrac }[2]{\genfrac {}{}{0pt}{1}{\textstyle #1\quad \hfill }{\textstyle \hfill \quad \mathstrut #2}} \newcommand {\splitdfrac }[2]{\genfrac {}{}{0pt}{0}{#1\quad \hfill }{\hfill \quad \mathstrut #2}} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\dblcolon }[0]{\vcentcolon \mathrel {\mkern -.9mu}\vcentcolon } \newcommand {\coloneqq }[0]{\vcentcolon \mathrel {\mkern -1.2mu}=} \newcommand {\Coloneqq }[0]{\dblcolon \mathrel {\mkern -1.2mu}=} \newcommand {\coloneq }[0]{\vcentcolon \mathrel {\mkern -1.2mu}\mathrel {-}} \newcommand {\Coloneq }[0]{\dblcolon \mathrel {\mkern -1.2mu}\mathrel {-}} \newcommand {\eqqcolon }[0]{=\mathrel {\mkern -1.2mu}\vcentcolon } \newcommand {\Eqqcolon }[0]{=\mathrel {\mkern -1.2mu}\dblcolon } \newcommand {\eqcolon }[0]{\mathrel {-}\mathrel {\mkern -1.2mu}\vcentcolon } \newcommand {\Eqcolon }[0]{\mathrel {-}\mathrel {\mkern -1.2mu}\dblcolon } \newcommand {\colonapprox }[0]{\vcentcolon \mathrel {\mkern -1.2mu}\approx } \newcommand {\Colonapprox }[0]{\dblcolon \mathrel {\mkern -1.2mu}\approx } \newcommand {\colonsim }[0]{\vcentcolon \mathrel {\mkern -1.2mu}\sim } \newcommand {\Colonsim }[0]{\dblcolon \mathrel {\mkern -1.2mu}\sim } \newcommand {\nuparrow }[0]{\MH_nuparrow: } \newcommand {\ndownarrow }[0]{\MH_ndownarrow: } \newcommand {\bigtimes }[0]{\MH_csym_bigtimes: }$

Additional Exercises for Chapter 7: Determinants

Find the determinants of the following matrices.

(a): $\left [ \begin {array}{rr} 1 & 3 \\ 0 & 2 \end {array} \right ]$
(b): $\left [ \begin {array}{rr} 0 & 3 \\ 0 & 2 \end {array} \right ]$
(c): $\left [ \begin {array}{rr} 4 & 3 \\ 6 & 2 \end {array} \right ]$

Let $A = \left [ \begin {array}{rrr} 1 & 2 & 4 \\ 0 & 1 & 3 \\ -2 & 5 & 1 \end {array} \right ]$ . When doing cofactor expansion along the top row, we encounter three minor matrices. What are they?

Click the arrow to see answer.

$\begin {bmatrix}1&3\\5&1\end {bmatrix},\quad \begin {bmatrix}0&3\\-2&1\end {bmatrix},\quad \begin {bmatrix}0&1\\-2&5\end {bmatrix}$

Find the determinants of the following matrices.

(a): $\left [ \begin {array}{rrr} 1 & 2 & 3 \\ 3 & 2 & 2 \\ 0 & 9 & 8 \end {array} \right ]$
(b): $\left [ \begin {array}{rrr} 4 & 3 & 2 \\ 1 & 7 & 8 \\ 3 & -9 & 3 \end {array} \right ]$
(c): $\left [ \begin {array}{rrrr} 1 & 2 & 3 & 2 \\ 1 & 3 & 2 & 3 \\ 4 & 1 & 5 & 0 \\ 1 & 2 & 1 & 2 \end {array} \right ]$

Click the arrow to see answer.

(a): The answer is $31$ .
(b): The answer is $375$ .
(c): The answer is $-2$ .

Find the following determinant by (a) expanding along the first row, (b) second column. $\begin{equation*} \left| \begin{array}{rrr} 1 & 2 & 1 \\ 2 & 1 & 3 \\ 2 & 1 & 1 \end{array} \right| \end{equation*}$

Click the arrow to see answer.

$\left | \begin {array}{ccc} 1 & 2 & 1 \\ 2 & 1 & 3 \\ 2 & 1 & 1 \end {array} \right | = 6$

Find the following determinant by (a) expanding along the first column, (b) third row. $\begin{equation*} \left| \begin{array}{rrr} 1 & 2 & 1 \\ 1 & 0 & 1 \\ 2 & 1 & 1 \end{array} \right| \end{equation*}$

Click the arrow to see answer.

$\left | \begin {array}{ccc} 1 & 2 & 1 \\ 1 & 0 & 1 \\ 2 & 1 & 1 \end {array} \right | = 2$

Find the following determinant by expanding (a) along the second row, (b) first column. $\begin{equation*} \left| \begin{array}{rrr} 1 & 2 & 1 \\ 2 & 1 & 3 \\ 2 & 1 & 1 \end{array} \right| \end{equation*}$

Click the arrow to see answer.

$\left | \begin {array}{ccc} 1 & 2 & 1 \\ 2 & 1 & 3 \\ 2 & 1 & 1 \end {array} \right | = 6$

Compute the determinant by cofactor expansion. Pick the easiest row or column to use. $\begin{equation*} \left| \begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 2 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 \\ 2 & 1 & 3 & 1 \end{array} \right| \end{equation*}$

Click the arrow to see answer.

$\left | \begin {array}{cccc} 1 & 0 & 0 & 1 \\ 2 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 \\ 2 & 1 & 3 & 1 \end {array} \right | = -4$

Find the determinants of the following matrices.

(a): $A = \left [ \begin {array}{rr} 1 & -34 \\ 0 & 2 \end {array} \right ]$
(b): $A = \left [ \begin {array}{rrr} 4 & 3 & 14 \\ 0 & -2 & 0 \\ 0 & 0 & 5 \end {array} \right ]$
(c): $A = \left [ \begin {array}{rrrr} 2 & 3 & 15 & 0 \\ 0 & 4 & 1 & 7 \\ 0 & 0 & -3 & 5 \\ 0 & 0 & 0 & 1 \end {array} \right ]$

An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant. $\begin{equation*} \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \rightarrow \cdots \rightarrow \left[ \begin{array}{cc} a & c \\ b & d \end{array} \right] \end{equation*}$

Click the arrow to see answer.

An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant. $\begin{equation*} \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \rightarrow \cdots \rightarrow \left[ \begin{array}{cc} c & d \\ a & b \end{array} \right] \end{equation*}$

Click the arrow to see answer.

In this case two rows were switched and so the resulting determinant is $-1$ times the first.

Click the arrow to see answer.

An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant. $\begin{equation*} \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \rightarrow \cdots \rightarrow \left[ \begin{array}{cc} b & a \\ d & c \end{array} \right] \end{equation*}$

Click the arrow to see answer.

In this case the two columns were switched so the determinant of the second is $-1$ times the determinant of the first.

Let $A$ be an $r\times r$ matrix and suppose there are $r-1$ rows (columns) such that all rows (columns) are linear combinations of these $r-1$ rows (columns). Show $\det \left ( A\right ) =0.$

If the determinant is nonzero, then it will remain nonzero with row operations applied to the matrix. In this case, you can obtain a row of zeros by doing row operations. Thus the determinant must be zero.

Show $\det \left ( aA\right ) =a^{n}\det \left ( A\right )$ for an $n \times n$ matrix $A$ and scalar $a$ .

Click the arrow to see answer.

$\det \left ( aA\right ) =\det \left ( aIA\right ) =\det \left ( aI\right ) \det \left ( A\right ) =a^{n}\det \left ( A\right ) .$ The matrix which has $a$ down the main diagonal has determinant equal to $a^{n}$ .

Construct $2\times 2$ matrices $A$ and $B$ to show that the $\det A \det B = \det (AB)$ .

Click the arrow to see answer.

$\det \left ( \left [ \begin {array}{cc} 1 & 2 \\ 3 & 4 \end {array} \right ] \left [ \begin {array}{rr} -1 & 2 \\ -5 & 6 \end {array} \right ] \right ) = -8$ $\det \left [ \begin {array}{cc} 1 & 2 \\ 3 & 4 \end {array} \right ] \det \left [ \begin {array}{rr} -1 & 2 \\ -5 & 6 \end {array} \right ] = -2 \times 4 = -8$

Is it true that $\det \left ( A+B\right ) =\det \left ( A\right ) +\det \left ( B\right ) ?$ If this is so, explain why. If it is not so, give a counter example.

Click the arrow to see answer.

This is not true at all. Consider $A=\left [ \begin {array}{cc} 1 & 0 \\ 0 & 1 \end {array} \right ] ,B=\left [ \begin {array}{rr} -1 & 0 \\ 0 & -1 \end {array} \right ] .$

An $n\times n$ matrix is called nilpotent if for some positive integer, $k$ it follows $A^{k}=O.$ If $A$ is a nilpotent matrix and $k$ is the smallest possible integer such that $A^{k}=O,$ what are the possible values of $\det \left ( A\right )$ ?

Click the arrow to see answer.

It must be $0$ because $0=\det \left ( 0\right ) =\det \left ( A^{k}\right ) =\left ( \det \left ( A\right ) \right ) ^{k}.$

A matrix is said to be orthogonal if $A^{T}A=I.$ Thus the inverse of an orthogonal matrix is just its transpose. What are the possible values of $\det \left ( A\right )$ if $A$ is an orthogonal matrix?

Click the arrow to see answer.

You would need $\det \left ( AA^{T}\right ) =\det \left ( A\right ) \det \left ( A^{T}\right ) =\det \left ( A\right ) ^{2}=1$ and so $\det \left ( A\right ) =1,$ or $-1$ .

Let $A$ and $B$ be two $n\times n$ matrices. $A\sim B$ ( $A$ is similar to $B$ ) means there exists an invertible matrix $P$ such that $A=P^{-1}BP.$ Show that if $A\sim B,$ then $\det \left ( A\right ) =\det \left ( B\right )$ .

Click the arrow to see answer.

$\det \left ( A\right ) =\det \left ( S^{-1}BS\right ) =\det \left ( S^{-1}\right ) \det \left ( B\right ) \det \left ( S\right ) =\det \left ( B\right ) \det \left ( S^{-1}S\right ) =\det \left ( B\right )$ .

Tell whether each statement is true or false. If true, provide a proof. If false, provide a counter example.

(a): If $A$ is a $3\times 3$ matrix with a zero determinant, then one column must be a multiple of some other column.
(b): If any two columns of a square matrix are equal, then the determinant of the matrix equals zero.
(c): For two $n\times n$ matrices $A$ and $B$ , $\det \left ( A+B\right ) =\det \left ( A\right ) +\det \left ( B\right ) .$
(d): For an $n\times n$ matrix $A$ , $\det \left ( 3A\right ) =3\det \left ( A\right )$
(e): If $A^{-1}$ exists then $\det \left ( A^{-1}\right ) =\det \left ( A\right ) ^{-1}.$
(f): If $B$ is obtained by multiplying a single row of $A$ by $4$ then $\det \left ( B\right ) =4\det \left ( A\right ) .$
(g): For $A$ an $n\times n$ matrix, $\det \left ( -A\right ) =\left ( -1\right ) ^{n}\det \left ( A\right ) .$
(h): If $A$ is a real $n\times n$ matrix, then $\det \left ( A^{T}A\right ) \geq 0.$
(i): If $A^{k}=O$ for some positive integer $k,$ then $\det \left ( A\right ) =0.$
(j): If $AX=0$ for some $X \neq 0,$ then $\det \left ( A\right ) =0.$

Click the arrow to see answer.

(a): False. Consider $\left [ \begin {array}{rrr} 1 & 1 & 2 \\ -1 & 5 & 4 \\ 0 & 3 & 3 \end {array} \right ]$
(b): True.
(c): False.
(d): False.
(e): True.
(f): True.
(g): True.
(h): True.
(i): True.
(j): True.

Find the determinant using row operations to first simplify. $\begin{equation*} \left| \begin{array}{rrr} 1 & 2 & 1 \\ 2 & 3 & 2 \\ -4 & 1 & 2 \end{array} \right| \end{equation*}$

Click the arrow to see answer.

$\left | \begin {array}{rrr} 1 & 2 & 1 \\ 2 & 3 & 2 \\ -4 & 1 & 2 \end {array} \right | = -6$

Find the determinant using row operations to first simplify. $\begin{equation*} \left| \begin{array}{rrr} 2 & 1 & 3 \\ 2 & 4 & 2 \\ 1 & 4 & -5 \end{array} \right| \end{equation*}$

Click the arrow to see answer.

$\left | \begin {array}{rrr} 2 & 1 & 3 \\ 2 & 4 & 2 \\ 1 & 4 & -5 \end {array} \right | = -32$

Find the determinant using row operations to first simplify. $\begin{equation*} \left| \begin{array}{rrrr} 1 & 2 & 1 & 2 \\ 3 & 1 & -2 & 3 \\ -1 & 0 & 3 & 1 \\ 2 & 3 & 2 & -2 \end{array} \right| \end{equation*}$

Click the arrow to see answer.

One can row reduce this using only row operation 3 to $\left [ \begin {array}{rrrr} 1 & 2 & 1 & 2 \\ 0 & -5 & -5 & -3 \\ 0 & 0 & 2 & \frac {9}{5} \\ 0 & 0 & 0 & -\frac {63}{10} \end {array} \right ]$ and therefore, the determinant is $-63.$ $\left | \begin {array}{rrrr} 1 & 2 & 1 & 2 \\ 3 & 1 & -2 & 3 \\ -1 & 0 & 3 & 1 \\ 2 & 3 & 2 & -2 \end {array} \right | = 63$

Find the determinant using row operations to first simplify. $\begin{equation*} \left| \begin{array}{rrrr} 1 & 4 & 1 & 2 \\ 3 & 2 & -2 & 3 \\ -1 & 0 & 3 & 3 \\ 2 & 1 & 2 & -2 \end{array} \right| \end{equation*}$

Click the arrow to see answer.

$\left [ \begin {array}{rrrr} 1 & 4 & 1 & 2 \\ 0 & -10 & -5 & -3 \\ 0 & 0 & 2 & \frac {19}{5} \\ 0 & 0 & 0 & -\frac {211}{20} \end {array} \right ]$ Thus the determinant is given by $\left | \begin {array}{rrrr} 1 & 4 & 1 & 2 \\ 3 & 2 & -2 & 3 \\ -1 & 0 & 3 & 3 \\ 2 & 1 & 2 & -2 \end {array} \right | = 211$

Let $\begin{equation*} A= \left[ \begin{array}{rrr} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 3 & 1 & 0 \end{array} \right] \end{equation*}$ Determine whether the matrix $A$ has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse which involves the cofactor matrix.

Click the arrow to see answer.

$\det \left [ \begin {array}{ccc} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 3 & 1 & 0 \end {array} \right ] = -13$ and so it has an inverse. This inverse is

$\begin{eqnarray*} \frac{1}{-13}\left[ \begin{array}{rrr} \left\vert \begin{array}{cc} 2 & 1 \\ 1 & 0 \end{array} \right\vert & -\left\vert \begin{array}{cc} 0 & 1 \\ 3 & 0 \end{array} \right\vert & \left\vert \begin{array}{cc} 0 & 2 \\ 3 & 1 \end{array} \right\vert \\ -\left\vert \begin{array}{cc} 2 & 3 \\ 1 & 0 \end{array} \right\vert & \left\vert \begin{array}{cc} 1 & 3 \\ 3 & 0 \end{array} \right\vert & -\left\vert \begin{array}{cc} 1 & 2 \\ 3 & 1 \end{array} \right\vert \\ \left\vert \begin{array}{cc} 2 & 3 \\ 2 & 1 \end{array} \right\vert & -\left\vert \begin{array}{cc} 1 & 3 \\ 0 & 1 \end{array} \right\vert & \left\vert \begin{array}{cc} 1 & 2 \\ 0 & 2 \end{array} \right\vert \end{array} \right] ^{T} &=&\frac{1}{-13}\left[ \begin{array}{rrr} -1 & 3 & -6 \\ 3 & -9 & 5 \\ -4 & -1 & 2 \end{array} \right] ^{T} \\ &=& \left[ \begin{array}{rrr} \frac{1}{13} & -\frac{3}{13} & \frac{4}{13} \\ -\frac{3}{13} & \frac{9}{13} & \frac{1}{13} \\ \frac{6}{13} & -\frac{5}{13} & -\frac{2}{13} \end{array} \right] \end{eqnarray*}$

Let $\begin{equation*} A= \left[ \begin{array}{rrr} 1 & 2 & 0 \\ 0 & 2 & 1 \\ 3 & 1 & 1 \end{array} \right] \end{equation*}$ Determine whether the matrix $A$ has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse.

Click the arrow to see answer.

$\det \left [ \begin {array}{ccc} 1 & 2 & 0 \\ 0 & 2 & 1 \\ 3 & 1 & 1 \end {array} \right ] = 7$ so it has an inverse. This inverse is $\frac {1}{7} \left [ \begin {array}{rrr} 1 & 3 & -6 \\ -2 & 1 & 5 \\ 2 & -1 & 2 \end {array} \right ]^{T} = \left [ \begin {array}{rrr} \frac {1}{7} & -\frac {2}{7} & \frac {2}{7} \\ \frac {3}{7} & \frac {1}{7} & -\frac {1}{7} \\ -\frac {6}{7} & \frac {5}{7} & \frac {2}{7} \end {array} \right ]$

Let $\begin{equation*} A= \left[ \begin{array}{rrr} 1 & 3 & 3 \\ 2 & 4 & 1 \\ 0 & 1 & 1 \end{array} \right] \end{equation*}$ Determine whether the matrix $A$ has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse.

Click the arrow to see answer.

$\det \left [ \begin {array}{ccc} 1 & 3 & 3 \\ 2 & 4 & 1 \\ 0 & 1 & 1 \end {array} \right ] = 3$ so it has an inverse which is $\left [ \begin {array}{rrr} 1 & 0 & -3 \\ -\frac {2}{3} & \frac {1}{3} & \frac {5}{3} \\ \frac {2}{3} & -\frac {1}{3} & -\frac {2}{3} \end {array} \right ]$

Let $\begin{equation*} A = \left[ \begin{array}{rrr} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 2 & 6 & 7 \end{array} \right] \end{equation*}$ Determine whether the matrix $A$ has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse.

Let $\begin{equation*} A = \left[ \begin{array}{rrr} 1 & 0 & 3 \\ 1 & 0 & 1 \\ 3 & 1 & 0 \end{array} \right] \end{equation*}$ Determine whether the matrix $A$ has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse.

Click the arrow to see answer.

$\det \left [ \begin {array}{rrr} 1 & 0 & 3 \\ 1 & 0 & 1 \\ 3 & 1 & 0 \end {array} \right ] = 2$ and so it has an inverse. The inverse turns out to equal $\left [ \begin {array}{rrr} -\frac {1}{2} & \frac {3}{2} & 0 \\ \frac {3}{2} & -\frac {9}{2} & 1 \\ \frac {1}{2} & -\frac {1}{2} & 0 \end {array} \right ]$

For the following matrices, determine if they are invertible. If so, use the formula for the inverse in terms of the cofactor matrix to find each inverse. If the inverse does not exist, explain why.

(a): $\left [ \begin {array}{rr} 1 & 1 \\ 1 & 2 \end {array} \right ]$
(b): $\left [ \begin {array}{rrr} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 4 & 1 & 1 \end {array} \right ]$
(c): $\left [ \begin {array}{rrr} 1 & 2 & 1 \\ 2 & 3 & 0 \\ 0 & 1 & 2 \end {array} \right ]$

(a): $\left \vert \begin {array}{cc} 1 & 1 \\ 1 & 2 \end {array} \right \vert = 1$
(b): $\left \vert \begin {array}{ccc} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 4 & 1 & 1\end {array} \right \vert = -15$
(c): $\left \vert \begin {array}{ccc} 1 & 2 & 1 \\ 2 & 3 & 0 \\ 0 & 1 & 2 \end {array} \right \vert = 0$

Challenge Exercises

Consider the matrix $\begin{equation*} A = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos t & -\sin t \\ 0 & \sin t & \cos t \end{array} \right] \end{equation*}$ Does there exist a value of $t$ for which this matrix fails to have an inverse? Explain.

Click the arrow to see answer.

No. It has a nonzero determinant for all $t$

Consider the matrix $\begin{equation*} A = \left[ \begin{array}{rrr} 1 & t & t^{2} \\ 0 & 1 & 2t \\ t & 0 & 2 \end{array} \right] \end{equation*}$ Does there exist a value of $t$ for which this matrix fails to have an inverse? Explain.

Click the arrow to see answer.

$\det \left [ \begin {array}{ccc} 1 & t & t^{2} \\ 0 & 1 & 2t \\ t & 0 & 2 \end {array} \right ] = t^{3}+2$ and so it has no inverse when $t=-\sqrt [3]{2}$

Consider the matrix $\begin{equation*} A = \left[ \begin{array}{ccc} e^{t} & \cosh t & \sinh t \\ e^{t} & \sinh t & \cosh t \\ e^{t} & \cosh t & \sinh t \end{array} \right] \end{equation*}$ Does there exist a value of $t$ for which this matrix fails to have an inverse? Explain.

Click the arrow to see answer.

$\det \left [ \begin {array}{ccc} e^{t} & \cosh t & \sinh t \\ e^{t} & \sinh t & \cosh t \\ e^{t} & \cosh t & \sinh t \end {array} \right ] = 0$ and so this matrix fails to have a nonzero determinant at any value of $t$ .

Consider the matrix $\begin{equation*} A = \left[ \begin{array}{ccc} e^{t} & e^{-t}\cos t & e^{-t}\sin t \\ e^{t} & -e^{-t}\cos t-e^{-t}\sin t & -e^{-t}\sin t+e^{-t}\cos t \\ e^{t} & 2e^{-t}\sin t & -2e^{-t}\cos t \end{array} \right] \end{equation*}$ Does there exist a value of $t$ for which this matrix fails to have an inverse? Explain.

Click the arrow to see answer.

$\det \left [ \begin {array}{ccc} e^{t} & e^{-t}\cos t & e^{-t}\sin t \\ e^{t} & -e^{-t}\cos t-e^{-t}\sin t & -e^{-t}\sin t+e^{-t}\cos t \\ e^{t} & 2e^{-t}\sin t & -2e^{-t}\cos t\end {array} \right ] = 5e^{-t} \neq 0$ and so this matrix is always invertible.

Show that if $\det \left ( A\right ) \neq 0$ for $A$ an $n\times n$ matrix, it follows that if $AX=0,$ then $X=0$ .

If $\det \left ( A\right ) \neq 0,$ then $A^{-1}$ exists and so you could multiply on both sides on the left by $A^{-1}$ and obtain that $X=0$ .

Suppose $A,B$ are $n\times n$ matrices and that $AB=I.$ Show that then $BA=I.$

First explain why $\det \left ( A\right ) ,\det \left ( B\right )$ are both nonzero. Then $\left ( AB\right ) A=A$ and then show $BA\left ( BA-I\right ) =0.$ From this use what is given to conclude $A\left ( BA-I\right ) =0.$ Then use Problem prb:7.36.

Click the arrow to see answer.

You have $1=\det \left ( A\right ) \det \left ( B\right )$ . Hence both $A$ and $B$ have inverses. Letting $X$ be given, $A\left ( BA-I\right ) X=\left ( AB\right ) AX-AX=AX-AX = 0$ and so it follows from the above problem that $\left ( BA-I\right )X=0.$ Since $X$ is arbitrary, it follows that $BA=I.$

Use the formula for the inverse in terms of the cofactor matrix to find the inverse of the matrix $\begin{equation*} A=\left[ \begin{array}{ccc} e^{t} & 0 & 0 \\ 0 & e^{t}\cos t & e^{t}\sin t \\ 0 & e^{t}\cos t-e^{t}\sin t & e^{t}\cos t+e^{t}\sin t \end{array} \right] \end{equation*}$

Click the arrow to see answer.

$\det \left [ \begin {array}{ccc} e^{t} & 0 & 0 \\ 0 & e^{t}\cos t & e^{t}\sin t \\ 0 & e^{t}\cos t-e^{t}\sin t & e^{t}\cos t+e^{t}\sin t \end {array} \right ] = e^{3t}.$ Hence the inverse is

$\begin{eqnarray*} &&e^{-3t}\left[ \begin{array}{ccc} e^{2t} & 0 & 0 \\ 0 & e^{2t}\cos t+e^{2t}\sin t & -\left( e^{2t}\cos t-e^{2t}\sin \right) t \\ 0 & -e^{2t}\sin t & e^{2t}\cos \left( t\right) \end{array} \right] ^{T} \\ &=& \left[ \begin{array}{ccc} e^{-t} & 0 & 0 \\ 0 & e^{-t}\left( \cos t+\sin t\right) & -\left( \sin t\right) e^{-t} \\ 0 & -e^{-t}\left( \cos t-\sin t\right) & \left( \cos t\right) e^{-t} \end{array} \right] \end{eqnarray*}$

Find the inverse, if it exists, of the matrix $\begin{equation*} A = \left[ \begin{array}{ccc} e^{t} & \cos t & \sin t \\ e^{t} & -\sin t & \cos t \\ e^{t} & -\cos t & -\sin t \end{array} \right] \end{equation*}$

Click the arrow to see answer.

$\begin{eqnarray*} &&\left[ \begin{array}{ccc} e^{t} & \cos t & \sin t \\ e^{t} & -\sin t & \cos t \\ e^{t} & -\cos t & -\sin t \end{array} \right] ^{-1} \\ &=&\left[ \begin{array}{ccc} \frac{1}{2}e^{-t} & 0 & \frac{1}{2}e^{-t} \\ \frac{1}{2}\cos t+\frac{1}{2}\sin t & -\sin t & \frac{1}{2}\sin t-\frac{1}{2} \cos t \\ \frac{1}{2}\sin t-\frac{1}{2}\cos t & \cos t & -\frac{1}{2}\cos t-\frac{1}{2} \sin t \end{array} \right] \end{eqnarray*}$

Suppose $A$ is an upper triangular matrix. Show that $A^{-1}$ exists if and only if all elements of the main diagonal are non zero. Is it true that $A^{-1}$ will also be upper triangular? Explain. Could the same be concluded for lower triangular matrices?

The given condition is what it takes for the determinant to be non zero. Recall that the determinant of an upper triangular matrix is just the product of the entries on the main diagonal.

If $A,B,$ and $C$ are each $n\times n$ matrices and $ABC$ is invertible, show why each of $A,B,$ and $C$ are invertible.

Click the arrow to see answer.

This follows because $\det \left ( ABC\right ) =\det \left ( A\right ) \det \left ( B\right ) \det \left ( C\right )$ and if this product is nonzero, then each determinant in the product is nonzero and so each of these matrices is invertible.

Practice Problem Source

These problems come from Chapter 3 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)

Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, pp. 272–315.

Press...	...to do
left/right arrows	Move cursor
shift+left/right arrows	Select region
ctrl+a	Select all
ctrl+x/c/v	Cut/copy/paste
ctrl+z/y	Undo/redo
ctrl+left/right	Add entry to list or column to matrix
shift+ctrl+left/right	Add copy of current entry/column to to list/matrix
ctrl+up/down	Add row to matrix
shift+ctrl+up/down	Add copy of current row to matrix
ctrl+backspace	Delete current entry in list or column in matrix
ctrl+shift+backspace	Delete current row in matrix

Type...	...to get
norm	$\|\|\blue{[?]}\|\|$
text	$\text{\blue{[?]}}$
sym_name	$\backslash\texttt{\blue{[?]}}$
abs	$\left\|\blue{[?]}\right\|$
sqrt	$\sqrt{\blue{[?]}}$
paren	$\left(\blue{[?]}\right)$
floor	$\lfloor \blue{[?]} \rfloor$
factorial	$\blue{[?]}!$
exp	${\blue{[?]}}^{\blue{[?]}}$
sub	${\blue{[?]}}_{\blue{[?]}}$
frac	$\dfrac{\blue{[?]}}{\blue{[?]}}$
int	$\displaystyle\int{\blue{[?]}}d\blue{[?]}$
defi	$\displaystyle\int_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}d\blue{[?]}$
deriv	$\displaystyle\frac{d}{d\blue{[?]}}\blue{[?]}$
sum	$\displaystyle\sum_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}$
prod	$\displaystyle\prod_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}$
root	$\sqrt[\blue{[?]}]{\blue{[?]}}$
vec	$\left\langle \blue{[?]} \right\rangle$
mat	$\left(\begin{matrix} \blue{[?]} \end{matrix}\right)$
*	$\cdot$
infinity	$\infty$
arcsin	$\arcsin\left(\blue{[?]}\right)$
arccos	$\arccos\left(\blue{[?]}\right)$
arctan	$\arctan\left(\blue{[?]}\right)$
sin	$\sin\left(\blue{[?]}\right)$
cos	$\cos\left(\blue{[?]}\right)$
tan	$\tan\left(\blue{[?]}\right)$
sec	$\sec\left(\blue{[?]}\right)$
csc	$\csc\left(\blue{[?]}\right)$
cot	$\cot\left(\blue{[?]}\right)$
log	$\log\left(\blue{[?]}\right)$
ln	$\ln\left(\blue{[?]}\right)$
alpha	$\alpha$
beta	$\beta$
gamma	$\gamma$
delta	$\delta$
epsilon	$\epsilon$
zeta	$\zeta$
eta	$\eta$
theta	$\theta$
iota	$\iota$
kappa	$\kappa$
lambda	$\lambda$
mu	$\mu$
nu	$\nu$
xi	$\xi$
omicron	$\omicron$
pi	$\pi$
rho	$\rho$
sigma	$\sigma$
tau	$\tau$
upsilon	$\upsilon$
phi	$\phi$
chi	$\chi$
psi	$\psi$
omega	$\omega$
Gamma	$\Gamma$
Delta	$\Delta$
Theta	$\Theta$
Lambda	$\Lambda$
Xi	$\Xi$
Pi	$\Pi$
Sigma	$\Sigma$
Phi	$\Phi$
Psi	$\Psi$
Omega	$\Omega$

Additional Exercises for Chapter 7: Determinants

Challenge Exercises

Practice Problem Source

Controls

Symbols

Settings