Question Number 117973 by bemath last updated on 14/Oct/20 $$\mathrm{consider}\:\mathrm{a}\:\mathrm{non}−\mathrm{singular}\:\mathrm{2}×\mathrm{2}\: \\ $$$$\mathrm{square}\:\mathrm{matrix}\:\mathrm{T}.\:\mathrm{If}\:\mathrm{trace}\:\left(\mathrm{T}\right)\:=\mathrm{4} \\ $$$$\mathrm{and}\:\mathrm{trace}\:\left(\mathrm{T}^{\mathrm{2}} \right)=\mathrm{5}\:\mathrm{what}\:\mathrm{is}\:\mathrm{determinant} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{matrix}\:\mathrm{T}\:? \\ $$ Answered by bobhans last updated on…
Question Number 183425 by ali009 last updated on 25/Dec/22 $${find}\:{the}\:{rank}\:{of}\:{the}\:{matrix}\:{A}\:{and}\:{B}\:{by} \\ $$$$\:{following}\:{row}\:{operation}: \\ $$$${A}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}&{−\mathrm{1}}\\{−\mathrm{2}}&{−\mathrm{1}}&{−\mathrm{3}}&{−\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{1}}&{\:\:\:\:\mathrm{1}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{1}}&{−\mathrm{1}}\end{bmatrix} \\ $$$${B}=\begin{bmatrix}{\:\:\:\:\mathrm{1}}&{\:\:\:\:\mathrm{2}}&{−\mathrm{1}}&{\:\:\:\:\:\mathrm{4}}\\{\:\:\:\:\mathrm{2}}&{\:\:\:\:\mathrm{4}}&{\:\:\:\:\:\mathrm{3}}&{\:\:\:\:\:\mathrm{5}}\\{−\mathrm{1}}&{−\mathrm{2}}&{\:\:\:\:\:\mathrm{6}}&{−\mathrm{7}}\end{bmatrix} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 117844 by fitria last updated on 14/Oct/20 Commented by fitria last updated on 14/Oct/20 $${Help} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 183240 by ali009 last updated on 23/Dec/22 $${find}\:{the}\:{value}\:{of}\:{cofficent}\:\mu\:{in}\:{the}\:{following} \\ $$$${system}\:{from}\:{the}\:{determinat}: \\ $$$$\mathrm{2}{x}_{\mathrm{1}} +\mu{x}_{\mathrm{2}} +{x}_{\mathrm{3}} =\mathrm{0} \\ $$$$\left(\mu−\mathrm{1}\right){x}_{\mathrm{1}} −{x}_{\mathrm{2}} +\mathrm{2}{x}_{\mathrm{3}} =\mathrm{0} \\ $$$$\mathrm{4}{x}_{\mathrm{1}} +{x}^{\mathrm{2}}…
Question Number 183239 by ali009 last updated on 24/Dec/22 $${determine}\:{eigenvalues}\:{and}\:{digonalize} \\ $$$${by}\:{row}\:{operation} \\ $$$$\begin{bmatrix}{\mathrm{4}}&{−\mathrm{9}}&{\mathrm{6}}&{\mathrm{12}}\\{\mathrm{9}}&{−\mathrm{1}}&{\mathrm{4}}&{\mathrm{6}}\\{\mathrm{2}}&{−\mathrm{11}}&{\mathrm{8}}&{\mathrm{16}}\\{−\mathrm{1}}&{\:\:\:\:\mathrm{3}}&{\mathrm{0}}&{−\mathrm{1}}\end{bmatrix} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 116746 by bemath last updated on 06/Oct/20 $$\mathrm{Find}\:\mathrm{an}\:\mathrm{orthogonal}\:\mathrm{matrix}\:\mathrm{A}\:\mathrm{whose} \\ $$$$\mathrm{first}\:\mathrm{row}\:\mathrm{is}\:\mathrm{u}_{\mathrm{1}} =\:\left(\frac{\mathrm{1}}{\mathrm{3}},\:\frac{\mathrm{2}}{\mathrm{3}},\:\frac{\mathrm{2}}{\mathrm{3}}\right). \\ $$ Answered by john santu last updated on 06/Oct/20 $${First}\:{step}\:{find}\:{a}\:{nonzero}\:{vector}\: \\…
Question Number 116695 by bemath last updated on 06/Oct/20 $$\mathrm{Solving}\:\mathrm{by}\:\mathrm{Gaussian}\:\mathrm{elimination} \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{following}\:\mathrm{system}\:\mathrm{of} \\ $$$$\mathrm{linear}\:\mathrm{equation}\:\begin{cases}{\mathrm{x}−\mathrm{3y}−\mathrm{2z}=\mathrm{6}}\\{\mathrm{2x}−\mathrm{4y}−\mathrm{3z}=\mathrm{8}}\\{−\mathrm{3x}+\mathrm{6y}+\mathrm{8z}=−\mathrm{5}}\end{cases} \\ $$ Answered by bobhans last updated on 06/Oct/20 $$\:\mathrm{Solving}\:\mathrm{by}\:\mathrm{Gaussian}\:\mathrm{elimination}\:\mathrm{using} \\…
Question Number 116669 by Eric002 last updated on 05/Oct/20 Commented by Eric002 last updated on 05/Oct/20 $${d}\acute {{e}montrer}\:{que}\:\:{les}\:{triangles}\:{ABC}\:{et}\:{IJK} \\ $$$${ont}\:{le}\:{m}\hat {{e}me}\:{centre}\:{de}\:{gravit}\acute {{e}} \\ $$ Answered…
Question Number 116578 by bobhans last updated on 05/Oct/20 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}\: \\ $$$$\:\begin{vmatrix}{\mathrm{1}\:\:\:\mathrm{x}\:\:\:\:\mathrm{x}^{\mathrm{2}} \:\:\:\:\mathrm{x}^{\mathrm{3}} }\\{\mathrm{1}\:\:\:\mathrm{2}\:\:\:\:\mathrm{2}^{\mathrm{2}} \:\:\:\:\mathrm{2}^{\mathrm{3}} }\\{\mathrm{1}\:\:\:\mathrm{3}\:\:\:\:\mathrm{3}^{\mathrm{2}} \:\:\:\:\mathrm{3}^{\mathrm{3}} }\\{\mathrm{1}\:\:\:\mathrm{4}\:\:\:\:\mathrm{4}^{\mathrm{2}} \:\:\:\:\mathrm{4}^{\mathrm{3}} }\end{vmatrix}=\:\mathrm{0} \\ $$ Answered by bemath…
Question Number 182050 by ali009 last updated on 03/Dec/22 $${A}=\begin{bmatrix}{{a}}&{{b}}&{{c}}\\{−\mathrm{2}}&{\mathrm{3}}&{\mathrm{6}}\\{\mathrm{0}}&{−\mathrm{2}}&{\mathrm{5}}\end{bmatrix}{and}\:{B}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{4}}\\{\mathrm{0}}&{\mathrm{3}}&{\mathrm{9}}\\{−\mathrm{1}}&{\mathrm{2}}&{\mathrm{2}}\end{bmatrix} \\ $$$${A}×{B}=\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{−\mathrm{1}}\\{−\mathrm{8}}&{{d}}&{\mathrm{31}}\\{−\mathrm{5}}&{\mathrm{4}}&{{e}}\end{bmatrix}{find}\:{the}\:{missing}\:{value} \\ $$ Answered by cortano1 last updated on 04/Dec/22 $$\:\begin{bmatrix}{\:\:\:\mathrm{a}\:\:\:\:\:\mathrm{b}\:\:\:\:\:\:\mathrm{c}}\\{−\mathrm{2}\:\:\:\:\mathrm{3}\:\:\:\:\:\:\mathrm{6}}\\{\:\:\:\mathrm{0}\:\:−\mathrm{2}\:\:\:\:\:\mathrm{5}}\end{bmatrix}\begin{bmatrix}{\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\mathrm{4}}\\{\:\:\:\mathrm{0}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\mathrm{9}}\\{−\mathrm{1}\:\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{2}}\end{bmatrix}= \\ $$$$\:\:\begin{bmatrix}{\mathrm{a}−\mathrm{c}\:\:\:\:\:\:\mathrm{2a}+\mathrm{3b}+\mathrm{2c}\:\:\:\:\:\:\:\mathrm{4a}+\mathrm{9b}+\mathrm{2c}}\\{−\mathrm{8}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{17}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{31}}\\{−\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{8}}\end{bmatrix}= \\…