Question Number 58480 by rahul 19 last updated on 23/Apr/19 Commented by kaivan.ahmadi last updated on 23/Apr/19 $${Let}\:{P}=\begin{bmatrix}{{a}\:\:\:{b}}\\{{c}\:\:\:\:{d}}\end{bmatrix} \\ $$$$\left[\mathrm{1}\:\:\mathrm{0}\right]\begin{bmatrix}{{a}\:\:\:\:{b}}\\{{c}\:\:\:\:{d}}\end{bmatrix}=\left[{a}\:\:{b}\right]\Rightarrow{a}={b}=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$${and} \\ $$$$\left[\mathrm{0}\:\:\mathrm{1}\right]\begin{bmatrix}{{a}\:\:\:{b}}\\{{c}\:\:\:\:{d}}\end{bmatrix}=\left[{c}\:\:\:{d}\right]\Rightarrow{c}=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:,\:{d}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\…
Question Number 58479 by rahul 19 last updated on 23/Apr/19 Commented by rahul 19 last updated on 24/Apr/19 $$?????? \\ $$ Terms of Service Privacy…
Question Number 58422 by peter frank last updated on 22/Apr/19 Answered by 2pac last updated on 23/Apr/19 $${DB}=\frac{{y}}{{sin}\left(\beta\right)} \\ $$$$<{ADB}=\mathrm{180}−\alpha−\beta \\ $$$${we}\:{have}\:\frac{{DB}}{{sin}\left(\alpha\right)}=\frac{{AB}}{{sin}\left(<{ADB}\right)}==>\frac{{y}}{{sin}\left(\alpha\right)}=\frac{{AB}}{{sin}\left(\mathrm{180}−\alpha−\beta\right)} \\ $$$${sin}\left(\mathrm{180}−{z}\right)={sinz}\:{so}\: \\…
Question Number 123868 by benjo_mathlover last updated on 28/Nov/20 Answered by liberty last updated on 29/Nov/20 $${we}\:{begin}\:{from}\:{A}^{\mathrm{3}} ={A}.{A}^{\mathrm{2}} \:{and}\:{A}^{\mathrm{2}} =\mathrm{8}{A}−\mathrm{13}{I} \\ $$$${gives}\:{A}^{\mathrm{3}} ={A}\left(\mathrm{8}{A}−\mathrm{13}{I}\right)=\mathrm{8}{A}^{\mathrm{2}} −\mathrm{13}{A} \\…
Question Number 57997 by peter frank last updated on 15/Apr/19 $${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…