Question Number 143200 by Guddone last updated on 11/Jun/21 Answered by Dwaipayan Shikari last updated on 11/Jun/21 $$\mathrm{2}{x}+\mathrm{3}{y}=\begin{bmatrix}{\mathrm{2}\:\:\mathrm{3}}\\{\mathrm{4}\:\:\mathrm{0}}\end{bmatrix}\Rightarrow\mathrm{6}{x}+\mathrm{9}{y}=\begin{bmatrix}{\:\mathrm{6}\:\:\:\:\mathrm{9}}\\{\:\mathrm{12}\:\:\mathrm{0}}\end{bmatrix}\rightarrow\left({a}\right) \\ $$$$\mathrm{3}{x}+\mathrm{2}{y}=\begin{bmatrix}{\mathrm{2}\:\:−\mathrm{2}}\\{−\mathrm{1}\:\:\mathrm{5}}\end{bmatrix}\Rightarrow\mathrm{6}{x}+\mathrm{4}{y}=\begin{bmatrix}{\:\mathrm{4}\:\:−\mathrm{4}}\\{\:−\mathrm{2}\:\:\mathrm{10}}\end{bmatrix}\rightarrow\left({b}\right) \\ $$$${a}−{b}\:\Rightarrow\mathrm{5}{y}=\begin{bmatrix}{\mathrm{2}\:\:\:\:\:\:\mathrm{13}\:\:}\\{\mathrm{14}\:−\mathrm{10}}\end{bmatrix}\Rightarrow{y}=\begin{bmatrix}{\frac{\mathrm{2}}{\mathrm{5}}\:\:\frac{\mathrm{13}}{\mathrm{5}}}\\{\frac{\mathrm{14}}{\mathrm{5}}\:−\mathrm{2}}\end{bmatrix} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\begin{bmatrix}{\mathrm{2}\:\mathrm{3}}\\{\mathrm{4}\:\mathrm{0}}\end{bmatrix}−\begin{bmatrix}{\frac{\mathrm{6}}{\mathrm{5}}\:\frac{\mathrm{39}}{\mathrm{5}}}\\{\:\frac{\mathrm{42}}{\mathrm{5}}\:\:−\mathrm{6}}\end{bmatrix}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\begin{bmatrix}{\frac{\mathrm{4}}{\mathrm{5}}\:\frac{−\mathrm{24}}{\mathrm{5}}}\\{\frac{−\mathrm{22}}{\mathrm{5}}\:\:\mathrm{6}}\end{bmatrix}\right)=\begin{bmatrix}{\frac{\mathrm{2}}{\mathrm{5}}\:−\frac{\mathrm{12}}{\mathrm{5}}}\\{−\frac{\mathrm{11}}{\mathrm{5}}\:\:\:\mathrm{3}}\end{bmatrix} \\…
Question Number 12046 by 7991 last updated on 10/Apr/17 $${how}\:{much}\:{matrices}\:{of}\:{integers}\:{number} \\ $$$${A}=\begin{bmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{bmatrix}{if}\:{A}^{\mathrm{2}} +{A}=\mathrm{2}{I},\:{c}=\mathrm{0},\:{det}\left({A}\right)=\mathrm{4} \\ $$ Answered by sma3l2996 last updated on 10/Apr/17 $${A}^{\mathrm{2}} =\begin{bmatrix}{{a}}&{{b}}\\{\mathrm{0}}&{{d}}\end{bmatrix}\begin{bmatrix}{{a}}&{{b}}\\{\mathrm{0}}&{{d}}\end{bmatrix}=\begin{bmatrix}{{a}^{\mathrm{2}} }&{{ab}+{db}}\\{\mathrm{0}}&{{d}^{\mathrm{2}}…
Question Number 12044 by 7991 last updated on 10/Apr/17 $${A}\in{M}_{{n}×{n}} \\ $$$${A}^{\mathrm{2}} ={A} \\ $$$$\left({I}+{A}\right)^{−\mathrm{1}} =….??? \\ $$ Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Apr/17…
Question Number 77565 by TawaTawa last updated on 07/Jan/20 Commented by MJS last updated on 07/Jan/20 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{that}\:{A}^{{B}} \:\mathrm{for}\:\mathrm{matrices}\:{A},\:{B}\:\mathrm{is} \\ $$$$\mathrm{defined}.\:\mathrm{I}\:\mathrm{only}\:\mathrm{know}\:{A}^{{n}} \:\mathrm{with}\:{n}\in\mathbb{N};\:{A}^{−{n}} \:\mathrm{is} \\ $$$$\mathrm{only}\:\mathrm{possible}\:\mathrm{if}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{matrix}\:{A}^{−\mathrm{1}} \:\mathrm{exists}…
Question Number 11958 by 7991 last updated on 07/Apr/17 $${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…
Question Number 141974 by 676597498 last updated on 25/May/21 Terms of Service Privacy Policy Contact: info@tinkutara.com