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Question Number 44610 by rahul 19 last updated on 02/Oct/18 $${Let}\:{A}\:{be}\:\mathrm{2}×\mathrm{3}\:{matrix}\:,\:{whereas}\:{B}\:{be} \\ $$$$\mathrm{3}×\mathrm{2}\:{matrix}.\:{If}\:{determinant}\left({AB}\right)=\mathrm{4}, \\ $$$${then}\:{the}\:{value}\:{of}\:{determinant}\:\left({BA}\right)\:? \\ $$ Commented by rahul 19 last updated on 02/Oct/18…
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Question Number 175265 by oustmuchiya@gmail.com last updated on 25/Aug/22 $$\mathrm{Given}\:\mathrm{that}\:\mathrm{matrix}\: \\ $$$$\mathrm{B}=\begin{cases}{\:\:\sqrt{\mathrm{3}}}\\{-\sqrt{\mathrm{5}}}\end{cases}\left.\begin{matrix}{\:\:\:\:\sqrt{\mathrm{2}}}\\{\:\:\:\:\sqrt{\mathrm{7}}}\end{matrix}\right\}\:\mathrm{find}\: \\ $$$$\mathrm{B}^{-\mathrm{1}} \:\mathrm{using}\:\mathrm{row}\:\mathrm{operation} \\ $$ Answered by CElcedricjunior last updated on 25/Aug/22 $$\boldsymbol{{det}}\left(\boldsymbol{{B}}\right)=\sqrt{\mathrm{21}}+\sqrt{\mathrm{10}}…
Question Number 175266 by oustmuchiya@gmail.com last updated on 25/Aug/22 $$\mathrm{Find}\:\mathrm{matrix}\:\mid\mathrm{A}\mid\mathrm{A}^{-\mathrm{1}} \: \\ $$$$\mathrm{given}\:\mathrm{that}\:\mathrm{matrix}\: \\ $$$$\mathrm{A}=\begin{pmatrix}{\sqrt{\mathrm{2}}}&{-\mathrm{1}}&{\mathrm{1}}&{\:\mathrm{0}}\\{\mathrm{4}}&{\mathrm{3}}&{\mathrm{2}}&{-\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}}&{\mathrm{3}}&{\:\mathrm{1}}\\{\mathrm{1}}&{-\mathrm{1}}&{\mathrm{0}}&{\:\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{using}\:\mathrm{row}\:\mathrm{operations} \\ $$ Commented by kaivan.ahmadi last updated on…
Question Number 175110 by nadovic last updated on 19/Aug/22 $$\mathrm{Prove}\:\mathrm{that}\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{{a}}&{{b}}&{{c}}\\{{a}^{\mathrm{2}} }&{{b}^{\mathrm{2}} }&{{c}^{\mathrm{2}} }\end{vmatrix}=\:\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right) \\ $$ Commented by kaivan.ahmadi last updated on 19/Aug/22 $$\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{1}}\\{{a}\:\:\:\:\:{b}\:\:\:\:\:{c}}\\{{a}^{\mathrm{2}} \:\:\:{b}^{\mathrm{2}} \:\:\:{c}^{\mathrm{2}}…
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Question Number 109240 by bemath last updated on 22/Aug/20 $$\:\:\frac{\flat{emath}}{\bullet\bullet\bullet\bullet\bullet} \\ $$$${use}\:{cayley}\:−\:{hamilton}\:{theorem} \\ $$$${to}\:{calculate}\:{A}^{−\mathrm{1}} \:{for}\:{A}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{2}}\\{\mathrm{1}\:\:\:\:\:\:\mathrm{2}\:\:−\mathrm{1}}\\{−\mathrm{1}\:\:\mathrm{1}\:\:\:\:\mathrm{4}}\end{pmatrix} \\ $$ Answered by bobhans last updated on 22/Aug/20 $$\:\:\:\:\frac{\flat{o}\flat{hans}}{\leqslant\geqslant\equiv\bullet°#}…
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Question Number 42988 by layo last updated on 05/Sep/18 $${reduce}\:{this}\:{matrix}\begin{bmatrix}{\mathrm{2}}&{\mathrm{3}}&{\mathrm{4}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{7}}&{\mathrm{2}}&{\mathrm{3}}\\{−\mathrm{1}}&{\mathrm{4}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{0}}\end{bmatrix} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com