Question Number 40910 by rahul 19 last updated on 29/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jul/18 $${let}\:{the}\:{required}\:{eqn}\:{is} \\ $$$$\left({x}−{A}\right)\left({x}−{B}\right)\left({x}−{C}\right)=\mathrm{0} \\ $$$${A}=\alpha+\beta+\gamma−\mathrm{2}\gamma \\ $$$$\:\:{A}=\frac{−{b}}{{a}}−\mathrm{2}\gamma\:\:\:{B}=\frac{−{b}}{{a}}−\mathrm{2}\alpha\:\:\:{C}=\frac{−{b}}{{a}}−−\mathrm{2}\beta \\…
Question Number 40418 by Tinkutara last updated on 21/Jul/18 Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jul/18 $${pls}\:{check}\:{the}\:{question}… \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on…
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Question Number 171441 by alcohol last updated on 15/Jun/22 $${I}_{{n}} \:=\:−\frac{\mathrm{2}{n}}{\mathrm{2}{n}\:+\:\mathrm{1}}\:{I}_{{n}−\mathrm{1}} \\ $$$${I}_{\mathrm{0}} \:=\:\mathrm{1} \\ $$$${Show}\:{that}\:{I}_{{n}} \:=\:\frac{\left(−\mathrm{4}\right)^{{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$ Commented by infinityaction last…
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Question Number 171064 by 119065 last updated on 07/Jun/22 $${When}\:\:{A}^{−\mathrm{1}} =\begin{bmatrix}{\mathrm{3}}&{\mathrm{1}}\\{\mathrm{8}}&{\mathrm{4}}\end{bmatrix} \\ $$$${find}\:{the}\:\:{A}=?\:,\mid{A}^{−\mathrm{1}} \mid\centerdot{A}=? \\ $$ Answered by som(math1967) last updated on 07/Jun/22 $$\:{Adj}\:{A}^{−\mathrm{1}} =\begin{bmatrix}{\mathrm{4}}&{−\mathrm{8}}\\{−\mathrm{1}}&{\mathrm{3}}\end{bmatrix}^{{T}}…
Question Number 39612 by Tinkutara last updated on 08/Jul/18 Commented by MJS last updated on 08/Jul/18 $$\mathrm{true}.\:\mathrm{that}'\mathrm{s}\:\mathrm{why}\:\mathrm{I}\:\mathrm{posted}\:\mathrm{the}\:\mathrm{other}\:\mathrm{solutions} \\ $$ Commented by MJS last updated on…
Question Number 105071 by dw last updated on 25/Jul/20 Commented by mr W last updated on 25/Jul/20 $$…\:{that}\:{satisfy}\:{the}\:{equation}\:… \\ $$$${which}\:{equation}? \\ $$ Commented by dw…
Question Number 38960 by Tinkutara last updated on 01/Jul/18 Answered by ajfour last updated on 01/Jul/18 $$\begin{vmatrix}{{xa}}&{{yb}}&{{zc}}\\{{yc}}&{{za}}&{{xb}}\\{{zb}}&{{xc}}&{{ya}}\end{vmatrix}=\:{xa}\left({a}^{\mathrm{2}} {yz}−{bcx}^{\mathrm{2}} \right) \\ $$$$\:+{by}\left({b}^{\mathrm{2}} {zx}−{acy}^{\mathrm{2}} \right)+{cz}\left({c}^{\mathrm{2}} {xy}−{abz}^{\mathrm{2}} \right)…
Question Number 170022 by Moytea last updated on 14/May/22 $$\mathrm{2}.\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\:\:−\mathrm{1}}\end{bmatrix} \\ $$$${Soln}:\:\:\:\:\:\:\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\:\:−\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{bmatrix}.{A} \\ $$$$\Rightarrow\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:−\mathrm{5}}\end{bmatrix}=\begin{bmatrix}{\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{0}}\\{−\mathrm{2}\:\:\:\:\:\:\mathrm{1}}\end{bmatrix}.{A}\:\:\:\:\:\:\left[{R}_{\mathrm{2}} \rightarrow{R}_{\mathrm{2}} −\mathrm{2}{R}_{\mathrm{1}} \right] \\ $$$$\Rightarrow\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\frac{\mathrm{2}}{\mathrm{5}}\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{5}}}\end{bmatrix}.{A}\:\:\:\:\:\left[{R}_{\mathrm{2}} \rightarrow\left(−\frac{\mathrm{1}}{\mathrm{5}}\right){R}_{\mathrm{2}} \right] \\ $$$$\Rightarrow\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\frac{\mathrm{1}}{\mathrm{5}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{5}}}\\{\frac{\mathrm{2}}{\mathrm{5}}\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{5}}}\end{bmatrix}.{A}\:\:\:\:\:\:\left[{R}_{\mathrm{1}} \rightarrow{R}_{\mathrm{1}} −\mathrm{2}{R}_{\mathrm{2}}…