Question Number 206609 by universe last updated on 20/Apr/24 $$\:\:\:\mathrm{let}\:\mathrm{matrix}\:\:\mathrm{A}_{\mathrm{n}×\mathrm{n}} \:\mathrm{and}\:\mathrm{B}_{\mathrm{n}×\mathrm{n}} \:\mathrm{satisfying} \\ $$$$\:\:\:\mathrm{A}^{\mathrm{2}} =\:\mathrm{A}\:\:\&\:\:\mathrm{B}^{\mathrm{2}} =\:\mathrm{B}\:\:\mathrm{then}\:\mathrm{prove} \\ $$$$\:\:\:\rho\left(\mathrm{A}−\mathrm{B}\right)\:=\:\rho\left(\mathrm{A}−\mathrm{AB}\right)+\:\:\rho\left(\mathrm{B}−\mathrm{AB}\right) \\ $$$$\:\:\:\mathrm{here}\:\rho\:=\:\mathrm{rank} \\ $$ Terms of Service…
Question Number 206600 by naka3546 last updated on 20/Apr/24 Answered by aleks041103 last updated on 22/Apr/24 $$\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{1}}&{−\mathrm{3}}&{−\mathrm{2}}\end{bmatrix}\overset{\mathrm{3}{rd}\:{row}\:+\:\mathrm{1}{st}\:{row}} {\rightarrow}\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$$\Rightarrow\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{1}}&{−\mathrm{3}}&{−\mathrm{2}}\end{bmatrix}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{−\mathrm{1}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$$\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}\overset{\mathrm{1}{st}\:{row}\:+\:\mathrm{2}{nd}\:{row}} {\rightarrow}\begin{bmatrix}{\mathrm{0}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$$\Rightarrow\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\mathrm{1}}&{−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}\begin{bmatrix}{\mathrm{0}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}…
Question Number 206629 by aras18 last updated on 20/Apr/24 Answered by Frix last updated on 20/Apr/24 $${a}+{b}−{c}=\mathrm{170} \\ $$$${a}−{b}+{c}=\mathrm{130} \\ $$$$======= \\ $$$$\mathrm{2}{a}\:\:\:\:\:\:\:\:\:\:\:=\mathrm{300} \\ $$$$\:\:\:{a}\:\:\:\:\:\:\:\:\:\:\:=\mathrm{150}…
Question Number 206591 by necx122 last updated on 19/Apr/24 $${I}\:{have}\:{a}\:{background}\:{in}\:{building}\:{design} \\ $$$${but}\:{I}\:{have}\:{great}\:{interest}\:{in}\:{Mathematics}. \\ $$$${That}'{s}\:{one}\:{major}\:{reason}\:{why}\:{I}\:{joined} \\ $$$${this}\:{platform}\:{cause}\:{I}\:{did}\:{little}\:{or}\:{no} \\ $$$${maths}\:{in}\:{my}\:{undergraduate}\:{days}\:{as}\:{a} \\ $$$${building}\:{design}\:{student}. \\ $$$$ \\ $$$${This}\:{forum}\:{has}\:{contributed}\:{greatly}\:{to} \\…
Question Number 206592 by NasaSara last updated on 19/Apr/24 Answered by namphamduc last updated on 20/Apr/24 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}\right)}{\mathrm{1}−{x}}{dx}=\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}\right)}{\mathrm{1}−{x}}{dx}−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}\right)}{\mathrm{1}−{x}}{dx} \\ $$$${x}\rightarrow{x}^{\mathrm{2}}…
Question Number 206579 by York12 last updated on 19/Apr/24 $$\mathrm{Evaluate}\::\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}}{d}\left({x}\right). \\ $$ Answered by Berbere last updated on 19/Apr/24 $$=\left[{ln}\left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 206557 by luciferit last updated on 18/Apr/24 Answered by Frix last updated on 18/Apr/24 $$\mathrm{Without}\:\mathrm{substitution}: \\ $$$$\int\frac{\mathrm{3}+\mathrm{2sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}}{dx}=\int\left(\frac{\mathrm{5}}{\mathrm{cos}^{\mathrm{2}} \:{x}}−\mathrm{2}\right){dx}= \\ $$$$=\mathrm{5tan}\:{x}\:−\mathrm{2}{x}\:+{C} \\…
Question Number 206558 by luciferit last updated on 18/Apr/24 Answered by Frix last updated on 18/Apr/24 $$\int\sqrt{\mathrm{3}+\mathrm{5}\sqrt{{x}}}{dx}\:\overset{{t}=\sqrt{\mathrm{3}+\mathrm{5}\sqrt{{x}}}} {=}\: \\ $$$$=\frac{\mathrm{4}}{\mathrm{25}}\int\left({t}^{\mathrm{4}} −\mathrm{3}{t}^{\mathrm{2}} \right){dt}=\frac{\mathrm{4}}{\mathrm{25}}\left(\frac{{t}^{\mathrm{5}} }{\mathrm{5}}−{t}^{\mathrm{3}} \right)= \\…
Question Number 206554 by cortano21 last updated on 18/Apr/24 Answered by starsouf last updated on 18/Apr/24 $$ \\ $$ Answered by mr W last updated…
Question Number 206549 by Shrodinger last updated on 18/Apr/24 $$\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx} \\ $$ Commented by Frix last updated on 18/Apr/24…