Question Number 54903 by Meritguide1234 last updated on 14/Feb/19 Answered by kaivan.ahmadi last updated on 14/Feb/19 $$\mathrm{det}\left(\mathrm{ABA}^{\mathrm{T}} \right)=\mathrm{det}\left(\mathrm{A}^{\mathrm{2}} \mathrm{B}\right)=\mathrm{detA}^{\mathrm{2}} \mathrm{detB}=\mathrm{8}\:\:\left(\mathrm{i}\right) \\ $$$$\mathrm{det}\left(\mathrm{AB}^{−\mathrm{1}} \right)=\frac{\mathrm{detA}}{\mathrm{detB}}=\mathrm{8}\Rightarrow\mathrm{detA}=\mathrm{8detB}\:\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{replace}\:\left(\mathrm{ii}\right)\:\mathrm{in}\:\left(\mathrm{i}\right)\:\mathrm{deduce}…
Question Number 54741 by gunawan last updated on 10/Feb/19 $$\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{1}.\begin{pmatrix}{\mathrm{n}}\\{\mathrm{0}}\end{pmatrix}^{\mathrm{2}} +\begin{pmatrix}{\mathrm{n}}\\{\mathrm{1}}\end{pmatrix}^{\mathrm{2}} +…+\begin{pmatrix}{\mathrm{n}}\\{\mathrm{n}}\end{pmatrix}^{\mathrm{2}} =\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}.\:\begin{pmatrix}{\mathrm{n}}\\{\mathrm{0}}\end{pmatrix}+\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{1}}\end{pmatrix}+…+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{n}}\end{pmatrix}^{\mathrm{2}} =\frac{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}} \\ $$ Commented by Abdo…
Question Number 54644 by gunawan last updated on 08/Feb/19 $$\mathrm{A}=\begin{bmatrix}{\mathrm{3}}&{\mathrm{7}}\\{−\mathrm{1}}&{−\mathrm{2}}\end{bmatrix} \\ $$$${A}^{\mathrm{27}} +{A}^{\mathrm{31}} +{A}^{\mathrm{40}} =… \\ $$ Answered by kaivan.ahmadi last updated on 10/Feb/19 $$\mathrm{A}^{\mathrm{2}}…
Question Number 119124 by abdullahquwatan last updated on 22/Oct/20 $$\mathrm{if}\:\mathrm{matrix}\:\mathrm{A}^{\mathrm{2}} \:=\:\begin{pmatrix}{\mathrm{1}\:\:\mathrm{3}}\\{\mathrm{0}\:\:\mathrm{1}}\end{pmatrix}\:\:\mathrm{then}\:\mathrm{matrix}\:\mathrm{A}=… \\ $$ Answered by benjo_mathlover last updated on 22/Oct/20 $$\Leftrightarrow\:{A}=\:\begin{pmatrix}{\mathrm{1}\:\:\:\frac{\mathrm{3}}{\mathrm{2}}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$${A}^{\mathrm{2}} \:=\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\frac{\mathrm{3}}{\mathrm{2}}}\\{\mathrm{0}\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\frac{\mathrm{3}}{\mathrm{2}}}\\{\mathrm{0}\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\frac{\mathrm{6}}{\mathrm{2}}}\\{\mathrm{0}\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\…
Question Number 118640 by 281981 last updated on 18/Oct/20 Answered by MJS_new last updated on 19/Oct/20 $$\left.\mathrm{1}\right) \\ $$$${x}=−\mathrm{2}\wedge{y}=\mathrm{0}\:\mathrm{solves}\:\mathrm{all}\:\mathrm{given}\:\mathrm{equations} \\ $$ Answered by 1549442205PVT last…
Question Number 183965 by nadovic last updated on 01/Jan/23 $$\mathrm{A}\:\mathrm{linear}\:\mathrm{transformation}\:\boldsymbol{{E}},\:\mathrm{of}\:\mathrm{the} \\ $$$${x}−{y}\:\mathrm{plane}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{as} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{E}}:\left({x},\:{y}\right)\:\rightarrow\:\left(\mathrm{2}{x}+{y},\:\mathrm{2}{x}+\mathrm{3}{y}\right) \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line}\:\mathrm{that} \\ $$$$\mathrm{remains}\:\mathrm{invariant}\:\mathrm{under}\:\mathrm{the} \\ $$$$\mathrm{transformation}. \\ $$ Answered by mr…
Question Number 183863 by Michaelfaraday last updated on 31/Dec/22 Answered by qaz last updated on 31/Dec/22 $$\frac{{d}}{{d}\theta}\left\{\begin{vmatrix}{\begin{pmatrix}{{A}}&{{B}}\\{{C}}&{{D}}\end{pmatrix}}&{\begin{pmatrix}{{E}}&{{F}}\\{{G}}&{{H}}\end{pmatrix}}\end{vmatrix}\right\} \\ $$$$=\frac{{d}}{{d}\theta}\left\{\begin{vmatrix}{{A}}&{{B}}\\{{C}}&{{D}}\end{vmatrix}\centerdot\begin{vmatrix}{{E}}&{{F}}\\{{G}}&{{H}}\end{vmatrix}\right\} \\ $$$$=\begin{vmatrix}{{A}}&{{B}}\\{{C}}&{{D}}\end{vmatrix}^{'} \centerdot\begin{vmatrix}{{E}}&{{F}}\\{{G}}&{{H}}\end{vmatrix}+\begin{vmatrix}{{A}}&{{B}}\\{{C}}&{{D}}\end{vmatrix}\centerdot\begin{vmatrix}{{E}}&{{F}}\\{{G}}&{{H}}\end{vmatrix}^{'} \\ $$$$=\left(\begin{vmatrix}{{A}'}&{{B}'}\\{{C}}&{{D}}\end{vmatrix}+\begin{vmatrix}{{A}}&{{B}}\\{{C}'}&{{D}'}\end{vmatrix}\right)\centerdot\begin{vmatrix}{{E}}&{{F}}\\{{G}}&{{H}}\end{vmatrix}+\begin{vmatrix}{{A}}&{{B}}\\{{C}}&{{D}}\end{vmatrix}\centerdot\left(\begin{vmatrix}{{E}'}&{{F}'}\\{{G}}&{{H}}\end{vmatrix}+\begin{vmatrix}{{E}}&{{F}}\\{{G}'}&{{H}'}\end{vmatrix}\right) \\…
Question Number 183814 by ali009 last updated on 30/Dec/22 $${determine}\:{eigen}\:{values}\:{and}\:{eigen}\:{vectors}\:{for} \\ $$$${each}\:\lambda\:.\:{and}\:{verify}\:{Ax}=\lambda{x} \\ $$$${A}=\begin{bmatrix}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}&{−\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{\mathrm{2}}}&{\:\:\:\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\end{bmatrix} \\ $$ Answered by TheSupreme last updated on 30/Dec/22 $$\left\{{x}\right\}'={A}\left\{{x}\right\}\:{is}\:{a}\:{rotation}\:{of}\:\frac{\pi}{\mathrm{6}}\:{rads}\:{counterclockwise} \\…
Question Number 118231 by bemath last updated on 16/Oct/20 $${Given}\:{a}\:{matrix}\:{A}=\:\begin{pmatrix}{−\mathrm{1}\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{2}}\\{\:\:\:\mathrm{0}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{4}}\\{−\mathrm{2}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{2}}\end{pmatrix} \\ $$$${and}\:{A}^{−\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{10}}\left({kA}+\mathrm{9}{I}−{A}^{\mathrm{2}} \right). \\ $$$${find}\:{k}. \\ $$ Answered by bobhans last updated on 16/Oct/20…
Question Number 118090 by bemath last updated on 15/Oct/20 $$\mathrm{find}\:\begin{vmatrix}{\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{c}}\:\:\:\:\:\:\:\:\mathrm{c}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}}\\{\:\:\:\:\:\mathrm{a}\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }{\mathrm{a}}\:\:\:\:\:\:\:\mathrm{a}}\\{\:\:\:\:\:\mathrm{b}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:\:\:\:\:\:\:\:\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }{\mathrm{b}}}\end{vmatrix}=?\: \\ $$ Commented by bemath last updated on 15/Oct/20…