Question Number 180855 by srikanth2684 last updated on 18/Nov/22 $${Find}\:{number}\:{of}\:{skew}\:{symmetric} \\ $$$${matrices}\:{of}\:{order}\:\mathrm{3}×\mathrm{3}\:{in}\:{which} \\ $$$${all}\:{non}\:{diagonal}\:{elements}\:{are}\: \\ $$$${different}\:{and}\:{belong}\:{to}\:{the}\: \\ $$$${set}\:\left\{−\mathrm{9},−\mathrm{8},−\mathrm{7},…,\mathrm{7},\mathrm{8},\mathrm{9}\right\}. \\ $$ Terms of Service Privacy Policy…
Question Number 114879 by bobhans last updated on 21/Sep/20 $${If}\:{the}\:{product}\:{of}\:{the}\:{matrices}\: \\ $$$$\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{1}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{3}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{pmatrix}…\begin{pmatrix}{\mathrm{1}\:\:\:\:\:{k}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{378}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$${then}\:{k}\:=\: \\ $$ Answered by john santu last updated on 21/Sep/20 $$\left({i}\right)\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{1}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{3}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}…
Question Number 114567 by dw last updated on 19/Sep/20 Commented by dw last updated on 19/Sep/20 $${Determinant}\:{D}_{{n}} =? \\ $$ Terms of Service Privacy Policy…
Question Number 114554 by bemath last updated on 19/Sep/20 $${Given}\:{a}\:{matrix}\:{A}\:=\:\begin{pmatrix}{{a}\:\:\:{b}}\\{{c}\:\:\:{d}}\end{pmatrix}\:{satisfies} \\ $$$${the}\:{equation}\:{A}^{\mathrm{2}} +\lambda{A}+\mathrm{7}{I}\:=\:\mathrm{0} \\ $$$${where}\:{I}=\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}\:.\:{Find}\:{the}\:{value}\:{of}\: \\ $$$${trace}\:{of}\:{A}\: \\ $$ Answered by bobhans last updated on…
Question Number 48103 by wasim last updated on 19/Nov/18 $$\mathrm{6} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 112505 by bemath last updated on 08/Sep/20 $$\mathrm{Given}\:\mathrm{matrix}\:\mathrm{A}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{0}}\\{\mathrm{2}\:\:\:\mathrm{3}}\end{pmatrix}\:,\:\mathrm{B}=\begin{pmatrix}{−\mathrm{1}\:\:\:\:\:−\mathrm{3}}\\{\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{and}\:\mathrm{AX}\:+\mathrm{2B}\:=\:\mathrm{I}\:.\:\mathrm{what}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{det}\:\left(\mathrm{X}\right)\:? \\ $$ Answered by john santu last updated on 08/Sep/20 $$\Leftrightarrow\:{AX}\:=\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}\:−\begin{pmatrix}{−\mathrm{2}\:\:\:\:\:−\mathrm{6}}\\{\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}…
Question Number 176371 by SLVR last updated on 17/Sep/22 Commented by SLVR last updated on 17/Sep/22 $${kindly}\:{tell}\:{me}\:{plese} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 45211 by Tip Top last updated on 10/Oct/18 $$\mathrm{Let}\:\mathrm{A}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{3}}\\{\mathrm{2}\:\:\:\:\:\mathrm{5}}\end{pmatrix}.\:\mathrm{Find}\:\mathrm{an}\:\mathrm{expression}\: \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{enteries}\:\mathrm{of}\:\mathrm{A}^{\mathrm{n}} . \\ $$ Answered by kaivan.ahmadi last updated on 15/Dec/18 Terms of…
Question Number 44737 by rahul 19 last updated on 04/Oct/18 $${A}\:{and}\:{B}\:{are}\:{two}\:{non}−{singular}\:{matrices} \\ $$$${such}\:{that}\:{A}^{\mathrm{6}} ={I}\:{and}\:{AB}^{\mathrm{2}} ={BA}\left({B}\neq{I}\right). \\ $$$${Then}\:{value}\:{of}\:{K}\:{for}\:{which}\:{B}^{{K}} ={I}. \\ $$ Terms of Service Privacy Policy…
Question Number 44708 by rahul 19 last updated on 03/Oct/18 $${Let}\:{A},{B}\:{be}\:{two}\:{n}×{n}\:{matrices}\:{such} \\ $$$${that}\:{A}+{B}={AB}\:{then}\:{prove}\:: \\ $$$${AB}={BA}\:? \\ $$ Answered by arcana last updated on 03/Oct/18 $$\mathrm{B}=\mathrm{A}\left(\mathrm{B}−\mathrm{I}\right)…