Question Number 15 by user1 last updated on 25/Jan/15 $$\mathrm{If}\:{A}=\begin{bmatrix}{\:\:\mathrm{3}}&{−\mathrm{5}}\\{−\mathrm{4}}&{\:\:\:\mathrm{2}}\end{bmatrix}, \\ $$$$\mathrm{show}\:\mathrm{that}\:{A}^{\mathrm{2}} −\mathrm{5}{A}−\mathrm{14}{I}=\mathrm{0} \\ $$ Answered by user1 last updated on 30/Oct/14 $$\mathrm{We}\:\mathrm{have}: \\ $$$${A}^{\mathrm{2}}…
Question Number 11 by user1 last updated on 25/Jan/15 $$\mathrm{Evaluate}:\:\: \\ $$$$\:\begin{vmatrix}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}&{{x}−\mathrm{1}}\\{\:\:\:\:\:{x}+\mathrm{1}}&{{x}+\mathrm{1}}\end{vmatrix} \\ $$ Answered by user1 last updated on 30/Oct/14 $$=\left[\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\left({x}+\mathrm{1}\right)−\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)\right] \\…
Question Number 13 by user1 last updated on 25/Jan/15 $$\mathrm{Expand}\:\mathrm{the}\:\mathrm{determnent}\: \\ $$$$\:\:\bigtriangleup=\begin{vmatrix}{{a}}&{{h}}&{{g}}\\{{h}}&{{b}}&{{f}}\\{{g}}&{{f}}&{{c}}\end{vmatrix} \\ $$ Answered by user1 last updated on 30/Oct/14 $$\mathrm{Expanding}\:\mathrm{by}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{row},\:\mathrm{we}\:\mathrm{have}: \\ $$$$\bigtriangleup={a}\centerdot\begin{vmatrix}{{b}}&{{f}}\\{{f}}&{{c}}\end{vmatrix}−{h}\centerdot\begin{vmatrix}{{h}}&{{f}}\\{{g}}&{{c}}\end{vmatrix}+{g}\centerdot\begin{vmatrix}{{h}}&{{b}}\\{{g}}&{{f}}\end{vmatrix}…
Question Number 7 by user1 last updated on 25/Jan/15 $$\mathrm{If}\:\:\:{A}=\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{4}}&{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}{and}\:\:{B}=\begin{bmatrix}{\mathrm{1}}&{−\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}}\\{\mathrm{5}}&{\mathrm{0}}\end{bmatrix}, \\ $$$$\mathrm{verify}\:\mathrm{that}\:\:\left(\mathrm{A}{B}\right)'={B}'{A}' \\ $$ Answered by user1 last updated on 30/Oct/14 $$\mathrm{We}\:\mathrm{have}\: \\ $$$${AB}=\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{4}}&{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}\centerdot\begin{bmatrix}{\mathrm{1}}&{−\mathrm{1}}\\{\mathrm{0}}&{\:\:\:\:\mathrm{2}}\\{\mathrm{5}}&{\:\:\:\:\mathrm{0}}\end{bmatrix} \\…
Question Number 9 by user1 last updated on 25/Jan/15 $$\mathrm{Let}\:{A}=\begin{bmatrix}{\:\:\:\mathrm{0}}&{−\mathrm{tan}\frac{\mathrm{x}}{\mathrm{2}}}\\{\mathrm{tan}\frac{\mathrm{x}}{\mathrm{2}}}&{\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{bmatrix}\:\mathrm{and}\:{I}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{identity}\:\mathrm{matrix}\:\mathrm{of}\:\mathrm{order}\:\mathrm{2}.\:\mathrm{Show}\:\mathrm{that}\: \\ $$$$\left({I}+{A}\right)=\left({I}−{A}\right)\centerdot\begin{bmatrix}{\mathrm{cos}\:{x}}&{−\mathrm{sin}\:{x}}\\{\mathrm{sin}\:{x}}&{\:\:\:\:\mathrm{cos}\:{x}}\end{bmatrix}. \\ $$ Answered by user1 last updated on 30/Oct/14 $$\mathrm{Let}\:\:\:\:\mathrm{tan}\:\frac{{x}}{\mathrm{2}}={t} \\…
Question Number 143270 by SLVR last updated on 12/Jun/21 Answered by Olaf_Thorendsen last updated on 12/Jun/21 $$\mathrm{B}\left({i},{j}\right)\:\geqslant\:\mathrm{1} \\ $$$$\mathrm{Let}\:{c}_{{ij}} \:=\:\mathrm{B}^{\mathrm{2}} \left({i},{j}\right)\:=\:\underset{{p}=\mathrm{1}} {\overset{\lambda} {\sum}}\underset{{q}=\mathrm{1}} {\overset{\lambda} {\sum}}{b}_{{ip}}…
Question Number 143200 by Guddone last updated on 11/Jun/21 Answered by Dwaipayan Shikari last updated on 11/Jun/21 $$\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}…