Question Number 57997 by peter frank last updated on 15/Apr/19 $${find}\:{all}\:{second}\:{order}\:{partial} \\ $$$${derivatives} \\ $$$${f}\left({x},{y}\right)={xe}^{{x}^{{y}} } .{y}^{{x}} \\ $$ Commented by MJS last updated on…
Question Number 57805 by pete last updated on 12/Apr/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{image}\:\mathrm{of}\:\mathrm{y}=\mathrm{3x}+\mathrm{1}\:\mathrm{under}\:\mathrm{the} \\ $$$$\mathrm{mapping}\:\begin{pmatrix}{\mathrm{2}\:\:\:\mathrm{3}}\\{\mathrm{1}\:\:\:\mathrm{2}}\end{pmatrix}. \\ $$ Commented by pete last updated on 12/Apr/19 $$\mathrm{thanks}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir} \\ $$ Commented…
Question Number 57790 by pete last updated on 12/Apr/19 $$\mathrm{Given}\:\mathrm{N}=\begin{bmatrix}{\mathrm{5}\:\:\:\:\:\:\mathrm{3}}\\{\mathrm{6}\:\:\:\:\:\:\:\mathrm{4}}\end{bmatrix}\mathrm{and}\:\mathrm{P}=\begin{bmatrix}{\mathrm{4}\:\:\:\:\:−\mathrm{3}}\\{−\mathrm{6}\:\:\:\:\mathrm{5}}\end{bmatrix}, \\ $$$$\mathrm{find}\:\mathrm{NP}\:\mathrm{and}\:\mathrm{deduce}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{of}\:\mathrm{P}. \\ $$ Answered by math1967 last updated on 12/Apr/19 $$\begin{bmatrix}{\mathrm{5}}&{\mathrm{3}}\\{\mathrm{6}}&{\mathrm{4}}\end{bmatrix}\begin{bmatrix}{\mathrm{4}}&{−\mathrm{3}}\\{−\mathrm{6}}&{\mathrm{5}}\end{bmatrix}=\begin{bmatrix}{\mathrm{20}−\mathrm{18}}&{−\mathrm{15}+\mathrm{15}}\\{\mathrm{24}−\mathrm{24}}&{−\mathrm{18}+\mathrm{20}}\end{bmatrix} \\ $$$$\begin{bmatrix}{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{2}}\end{bmatrix}=\mathrm{2}{I} \\…
Question Number 57698 by rahul 19 last updated on 10/Apr/19 Commented by rahul 19 last updated on 10/Apr/19 $$\mathrm{3}{rd}\:{plane}\:{can}\:{be}\:{written}\:{as}\:: \\ $$$${P}_{\mathrm{1}} +\Upsilon{P}_{\mathrm{2}} \:,\:{how}\:{to}\:{find}\:\:\Upsilon? \\ $$$${here}\:\lambda=\mathrm{3},\mu=\mathrm{7}\:{but}\:{then}\:\Upsilon\:{is}\:{not}\:{unique}!…
Question Number 57633 by rahul 19 last updated on 09/Apr/19 Answered by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19 $${A}=\mid\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\mathrm{1}\mid \\ $$$$\:\:\:\:\:\:\:\:\mid\left(\mathrm{2}−{c}\right)\:\:\left({b}−{c}\right)\:\:\:\:{c}\mid \\ $$$$\:\:\:\:\:\:\:\:\mid\mathrm{4}−{c}^{\mathrm{2}} \:\:\:\:\:\:{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \:\:\:\:{c}^{\mathrm{2}}…
Question Number 57306 by Aditya789 last updated on 02/Apr/19 Commented by tanmay.chaudhury50@gmail.com last updated on 02/Apr/19 $${is}\:{it}\:{ab}\:\:{or}\:−{ab}\:\: \\ $$ Commented by Aditya789 last updated on…
Question Number 57044 by mr W last updated on 29/Mar/19 Commented by mr W last updated on 29/Mar/19 $${reposted} \\ $$$${Question}\:{from}\:{Tinkutara}\:{sir} \\ $$ Answered by…
Question Number 56510 by Tawa1 last updated on 17/Mar/19 Commented by MJS last updated on 17/Mar/19 $$\mathrm{well},\:\mathrm{the}\:\mathrm{determinant}\:\mathrm{is} \\ $$$$\mathrm{2}{a}^{\mathrm{4}} {bc}+\mathrm{2}{ab}^{\mathrm{4}} {c}+\mathrm{2}{abc}^{\mathrm{4}} +\mathrm{6}{a}^{\mathrm{3}} {b}^{\mathrm{2}} {c}+\mathrm{6}{a}^{\mathrm{3}} {bc}^{\mathrm{2}}…
Question Number 56339 by Tinkutara last updated on 14/Mar/19 Answered by tanmay.chaudhury50@gmail.com last updated on 15/Mar/19 $${x}={a}_{\mathrm{3}} {y}+{a}_{\mathrm{2}} \left({a}_{\mathrm{2}} {x}+{a}_{\mathrm{1}} {y}\right) \\ $$$${x}−{a}_{\mathrm{2}} ^{\mathrm{2}} {x}={y}\left({a}_{\mathrm{3}}…
Question Number 56090 by Tawa1 last updated on 10/Mar/19 Commented by math1967 last updated on 10/Mar/19 $${pls}\:{check}\:{I}\:{think}\:{it}\:{should}\:{be} \\ $$$$\begin{vmatrix}{{bc}}&{{ca}}&{{ab}}\\{{a}}&{{b}}&{{c}}\\{{a}^{\mathrm{2}} }&{{b}^{\mathrm{2}} }&{{c}^{\mathrm{2}} }\end{vmatrix} \\ $$ Commented…