Question Number 11958 by 7991 last updated on 07/Apr/17 $${A}\in{M}_{\mathrm{2016}×\mathrm{2016}} \: \\ $$$${with}\:{the}\:{entries}\:{a}_{{ij}} \left\{_{\mathrm{0},\:{if}\:{i}+{j}\neq\mathrm{2016}} ^{\mathrm{1},\:{if}\:{i}+{j}=\mathrm{2016}} \right. \\ $$$${find}\:{the}\:{determinant}?? \\ $$ Commented by prakash jain last…
Question Number 11353 by tawa last updated on 21/Mar/17 $$\mathrm{reduce}\:\mathrm{the}\:\mathrm{matrix}\:\mathrm{below}\:\mathrm{to}\:\mathrm{echelon}\:\mathrm{form}\:\mathrm{and}\:\mathrm{then}\:\mathrm{to}\:\mathrm{row}\:\mathrm{canonical}\:\mathrm{form} \\ $$$$\mathrm{A}\:=\:\begin{bmatrix}{\mathrm{2}\:\:\mathrm{4}\:\:\mathrm{2}\:\:−\mathrm{2}\:\:\mathrm{5}\:\:\mathrm{1}}\\{\mathrm{3}\:\:\mathrm{6}\:\:\mathrm{2}\:\:\:\:\:\mathrm{2}\:\:\mathrm{0}\:\:\mathrm{4}}\\{\mathrm{4}\:\:\mathrm{8}\:\:\mathrm{2}\:\:\:\mathrm{6}\:\:−\mathrm{5}\:\mathrm{7}}\end{bmatrix} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 76812 by Rio Michael last updated on 30/Dec/19 $$\mathrm{Given}\:\mathrm{that}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{3}\:×\:\mathrm{3}\:\mathrm{invertible}\:\mathrm{matrices}, \\ $$$$\:\mathrm{then}\:\left(\mathrm{A}^{−\mathrm{1}} \mathrm{B}\right)^{−\mathrm{1}} \:= \\ $$$$\mathrm{A}.\:\mathrm{AB}^{−\mathrm{1}} \\ $$$$\mathrm{B}.\mathrm{B}^{−\mathrm{1}} \mathrm{A} \\ $$$$\mathrm{C}.\:\mathrm{B}^{−\mathrm{1}} \mathrm{A}^{−\mathrm{1}} \\ $$$$\mathrm{D}.\:\mathrm{BA}^{−\mathrm{1}}…
Question Number 11035 by suci last updated on 08/Mar/17 $${A}=\begin{bmatrix}{\mathrm{1}}&{−\mathrm{1}}\\{−\mathrm{4}}&{−\mathrm{2}}\end{bmatrix} \\ $$$${A}^{\mathrm{2016}} =…..? \\ $$ Answered by diofanto last updated on 19/Jul/17 $$\mathrm{Finding}\:\mathrm{the}\:\mathrm{eigenvalues}\:\mathrm{of}\:{A}: \\ $$$$\begin{vmatrix}{\mathrm{1}−\lambda}&{−\mathrm{1}}\\{−\mathrm{4}}&{−\mathrm{2}−\lambda}\end{vmatrix}\:=\:\mathrm{0}…
Question Number 11034 by suci last updated on 08/Mar/17 $${u}=\begin{bmatrix}{\mathrm{1}}&{−\mathrm{1}}&{\mathrm{2}}\end{bmatrix} \\ $$$${A}=\begin{bmatrix}{\mathrm{3}}&{\mathrm{2}}\\{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$${v}=\begin{bmatrix}{\mathrm{2}}&{−\mathrm{1}}\end{bmatrix} \\ $$$${uAv}^{{t}} =….? \\ $$ Commented by bahmanfeshki last updated on…
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Question Number 76326 by Master last updated on 26/Dec/19 Answered by benjo last updated on 26/Dec/19 $$\mathrm{using}\:\mathrm{determinan}\:\mathrm{Vandermon}\:\mathrm{de}\:\mathrm{method}\: \\ $$ Commented by Master last updated on…
Question Number 141594 by naz last updated on 20/May/21 Answered by MJS_new last updated on 20/May/21 $$\begin{vmatrix}{{a}}&{{b}}&{{c}}\\{{d}}&{{e}}&{{f}}\\{{g}}&{{h}}&{{i}}\end{vmatrix}={aei}+{bfg}+{cdh}−\left({afh}+{bdi}+{ceg}\right) \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 141400 by Raffaqet last updated on 18/May/21 $${fjnd}\:{inverse}\:{of}\:{matrix}\begin{bmatrix}{\mathrm{2}}&{−\mathrm{5}}\\{\mathrm{1}}&{\mathrm{3}}\end{bmatrix} \\ $$ Answered by iloveisrael last updated on 18/May/21 $$\:{A}^{−\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{6}−\left(−\mathrm{5}\right)}\:\begin{bmatrix}{\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{5}}\\{−\mathrm{1}\:\:\:\:\:\mathrm{2}}\end{bmatrix} \\ $$$${A}^{−\mathrm{1}} \:=\:\begin{bmatrix}{\mathrm{3}/\mathrm{11}\:\:\:\:\:\:\:\:\:\mathrm{5}/\mathrm{11}}\\{−\mathrm{1}/\mathrm{11}\:\:\:\:\:\mathrm{2}/\mathrm{11}}\end{bmatrix} \\…