Question Number 129886 by zarawan last updated on 23/Jan/21 $${Find}\:{Null}\:{space}\:{of}\:{the}\:{following}\:{matrix}\:{and}\:{also}\:{find}\:{basis}\:{for}\:{the}\:{null}\:{space}. \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{−\mathrm{2}}&{\mathrm{0}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{−\mathrm{2}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{2}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{4}}&{\mathrm{1}}\end{bmatrix} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 64124 by ANTARES VY last updated on 13/Jul/19 $$\sqrt{\mathrm{2014}}\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\mathrm{4029}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{x}}_{\mathrm{1}} <\boldsymbol{\mathrm{x}}_{\mathrm{2}} <\boldsymbol{\mathrm{x}}_{\mathrm{3}} \\ $$$$\boldsymbol{\mathrm{x}}_{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}_{\mathrm{1}} +\boldsymbol{\mathrm{x}}_{\mathrm{3}} \right)=? \\ $$ Terms…
Question Number 64012 by aditya@345 last updated on 12/Jul/19 Answered by MJS last updated on 12/Jul/19 $$\begin{vmatrix}{\mathrm{3}}&{−\mathrm{2}}&{\mathrm{sin}\:\mathrm{3}\theta}\\{−\mathrm{7}}&{\mathrm{8}}&{\mathrm{cos}\:\mathrm{2}\theta}\\{−\mathrm{11}}&{\mathrm{14}}&{\mathrm{2}}\end{vmatrix}=\mathrm{0} \\ $$$$\mathrm{20}−\mathrm{20cos}\:\mathrm{2}\theta\:−\mathrm{10sin}\:\mathrm{3}\theta\:=\mathrm{0} \\ $$$$\mathrm{2cos}\:\mathrm{2}\theta\:+\mathrm{sin}\:\mathrm{3}\theta\:−\mathrm{2}=\mathrm{0} \\ $$$$\left(\mathrm{1}−\mathrm{2sin}\:\theta\right)\left(\mathrm{3}+\mathrm{2sin}\:\theta\right)\mathrm{sin}\:\theta\:=\mathrm{0} \\ $$$$\mathrm{sin}\:\theta\:=−\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow\:\theta\notin\mathbb{R}…
Question Number 63957 by meme last updated on 11/Jul/19 $${f}\mathrm{0}{f}=\theta \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 127541 by Study last updated on 30/Dec/20 $${can}\:{we}\:{find}\:{the}\:\mathrm{4}×\mathrm{4}\:{matrics}\:{inevers}\: \\ $$$${multiplicative}\:{matrics}??? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 127143 by peter frank last updated on 27/Dec/20 Commented by Dwaipayan Shikari last updated on 27/Dec/20 $$\mathrm{042} \\ $$ Answered by Ar Brandon…
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}…