Question Number 192464 by Engr_Jidda last updated on 18/May/23 Commented by a.lgnaoui last updated on 19/May/23 $${X}'? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 60775 by Pathen Vincent last updated on 25/May/19 Commented by Prithwish sen last updated on 25/May/19 $$\mathrm{Divide}\:\mathrm{all}\:\mathrm{3}\:\mathrm{eqiations}\:\mathrm{by}\:\mathrm{xyz}\:\mathrm{and}\:\mathrm{then}\:\mathrm{apply} \\ $$$$\mathrm{crammers}\:\mathrm{rule}\:\mathrm{for}\:\frac{\mathrm{1}}{\mathrm{x}},\:\frac{\mathrm{1}}{\mathrm{y}}\:,\frac{\mathrm{1}}{\mathrm{z}} \\ $$ Terms of…
Question Number 60303 by Cheyboy last updated on 19/May/19 Answered by tw000001 last updated on 18/Oct/19 $$\mathrm{det}\begin{vmatrix}{\mathrm{2}}&{−\mathrm{1}}&{\mathrm{5}}\\{\mathrm{3}}&{\mathrm{4}}&{\mathrm{0}}\\{−\mathrm{2}}&{\mathrm{1}}&{\mathrm{9}}\end{vmatrix}=\mathrm{154} \\ $$$$\rightarrow\begin{bmatrix}{\mathrm{2}}&{−\mathrm{1}}&{\mathrm{5}}\\{\mathrm{3}}&{\mathrm{4}}&{\mathrm{0}}\\{−\mathrm{2}}&{\mathrm{1}}&{\mathrm{9}}\end{bmatrix}^{−\mathrm{1}} =\begin{bmatrix}{\frac{\mathrm{18}}{\mathrm{77}}}&{\frac{\mathrm{1}}{\mathrm{11}}}&{−\frac{\mathrm{10}}{\mathrm{77}}}\\{−\frac{\mathrm{27}}{\mathrm{154}}}&{\frac{\mathrm{2}}{\mathrm{11}}}&{\frac{\mathrm{15}}{\mathrm{154}}}\\{\frac{\mathrm{1}}{\mathrm{14}}}&{\mathrm{0}}&{\frac{\mathrm{1}}{\mathrm{14}}}\end{bmatrix} \\ $$$$\mathrm{det}\begin{vmatrix}{\mathrm{9}}&{\mathrm{7}}&{\mathrm{5}}\\{\mathrm{1}}&{\mathrm{4}}&{\mathrm{3}}\\{\mathrm{7}}&{\mathrm{5}}&{−\mathrm{2}}\end{vmatrix}=−\mathrm{161} \\ $$$$\rightarrow\begin{bmatrix}{\mathrm{9}}&{\mathrm{7}}&{\mathrm{5}}\\{\mathrm{1}}&{\mathrm{4}}&{\mathrm{3}}\\{\mathrm{7}}&{\mathrm{5}}&{−\mathrm{2}}\end{bmatrix}^{−\mathrm{1}} =\begin{bmatrix}{\frac{\mathrm{1}}{\mathrm{7}}}&{−\frac{\mathrm{39}}{\mathrm{161}}}&{−\frac{\mathrm{1}}{\mathrm{161}}}\\{−\frac{\mathrm{1}}{\mathrm{7}}}&{\frac{\mathrm{53}}{\mathrm{161}}}&{\frac{\mathrm{22}}{\mathrm{161}}}\\{\frac{\mathrm{1}}{\mathrm{7}}}&{−\frac{\mathrm{4}}{\mathrm{161}}}&{−\frac{\mathrm{29}}{\mathrm{161}}}\end{bmatrix}…
Question Number 125768 by Don08q last updated on 13/Dec/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{2}×\mathrm{2}\:\mathrm{matrix}\:{A}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\begin{pmatrix}{−\mathrm{4}}&{\:\:\:\:\:\mathrm{0}}\\{\:\:\:\:\mathrm{0}}&{−\mathrm{4}}\end{pmatrix}\:\:\:−\:\:{A}\:\:=\:\:{A}^{−\mathrm{1}} \begin{pmatrix}{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}\end{pmatrix} \\ $$ Answered by 676597498 last updated on 13/Dec/20 $${A}=\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix}\: \\ $$$${det}\left({A}\right)={ad}−{cd}…
Question Number 191030 by cortano12 last updated on 16/Apr/23 Answered by Frix last updated on 16/Apr/23 $${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} \\…