Question Number 191030 by cortano12 last updated on 16/Apr/23 Answered by Frix last updated on 16/Apr/23 $${t}+\frac{\mathrm{1}}{{t}}=\mathrm{3} \\ $$$$\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} =\mathrm{9}\:\Leftrightarrow\:{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }=\mathrm{7} \\ $$$$\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{3}} =\mathrm{27}\:\Leftrightarrow\:{t}^{\mathrm{3}}…
Question Number 190860 by SLVR last updated on 13/Apr/23 Commented by SLVR last updated on 13/Apr/23 $${kindly}\:{help}\:{me} \\ $$$${i}\:{got}\:\mathrm{24}\:{while}\:{repetetions}\:{not} \\ $$$${allowed}.{but}\:{it}\:{become}\:{heavy} \\ $$$${if}\:{repetetions}\:{allowed} \\ $$…
Question Number 59295 by pooja24 last updated on 07/May/19 $${Let}\:{A}\:{be}\mathrm{3}×\mathrm{3}\:{matrix}\:{with}\:{eigen}\:{values} \\ $$$$\mathrm{1},−\mathrm{1},\mathrm{0}.\:{Then}\:{determinant}\:{of}\:{I}+{A}^{\mathrm{100}\:} =?? \\ $$ Answered by alex041103 last updated on 10/May/19 $${basically} \\ $$$${A}={PDP}^{−\mathrm{1}}…
Question Number 190266 by alcohol last updated on 30/Mar/23 $${a}\:{ball}\:{is}\:{thrown}\:{vertically}\:{upward} \\ $$$${from}\:{a}\:{point}\:\mathrm{0}.\mathrm{5}{m}\:{above}\:{the}\:{ground}\:{with} \\ $$$${speed}\:{u}\:=\:\mathrm{7}{m}/{s} \\ $$$${find}\:{the}\:{height}\:{reached}\:{above}\:{ground} \\ $$$${g}\:=\:\mathrm{10}{m}/{s}^{\mathrm{2}} \\ $$ Answered by ARUNG_Brandon_MBU last updated…
Question Number 124707 by Mammadli last updated on 05/Dec/20 $$\mathrm{1}.\:\boldsymbol{{A}}=\left(\mathrm{3152}\right)\:,\:\:\boldsymbol{{A}}^{−\mathrm{1}} =? \\ $$$$\mathrm{2}.\:\boldsymbol{{Solve}}\:\boldsymbol{{the}}\:\boldsymbol{{equation}}: \\ $$$$\mid\mathrm{2}\boldsymbol{{x}}\mathrm{611}\mid=\mathrm{0} \\ $$$$\mathrm{3}.\:\boldsymbol{{Calculate}}\:\boldsymbol{{the}}\:\boldsymbol{{determinant}}: \\ $$$$\mid\mathrm{132}\:\mathrm{21}\:\mathrm{13}\:\mathrm{42}\mid \\ $$ Commented by MJS_new last…
Question Number 124170 by Pengu last updated on 01/Dec/20 $$ \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{learning}\:\mathrm{linear}\:\mathrm{algebra}\:\mathrm{and}\:\mathrm{have}\:\mathrm{a} \\ $$$$\mathrm{qustion}\:\mathrm{in}\:\mathrm{regards}\:\mathrm{to}\:\mathrm{nullspace}. \\ $$$$\: \\ $$$$\mathrm{So},\:\boldsymbol{{Ax}}=\mathrm{0}. \\ $$$$\mathrm{If}\:\boldsymbol{{A}}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{0}}\end{bmatrix},\:\mathrm{this}\:\mathrm{is}\:\mathrm{simply}\:\mathrm{solved}\:\mathrm{for} \\ $$$$\boldsymbol{{x}}=\begin{bmatrix}{{x}_{\mathrm{1}} }\\{{x}_{\mathrm{2}} }\end{bmatrix} \\…
Question Number 58480 by rahul 19 last updated on 23/Apr/19 Commented by kaivan.ahmadi last updated on 23/Apr/19 $${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} \\…