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Category: Algebra

Let-A-and-B-are-matrices-in-R-2017-2017-such-that-A-1-A-B-1-B-1-and-det-A-1-2017-Find-det-B-

Question Number 55857 by Joel578 last updated on 05/Mar/19 $$\mathrm{Let}\:{A}\:\mathrm{and}\:{B}\:\mathrm{are}\:\mathrm{matrices}\:\mathrm{in}\:\mathbb{R}^{\mathrm{2017}×\mathrm{2017}} \:\mathrm{such}\:\mathrm{that}\: \\ $$$${A}^{−\mathrm{1}} \:=\:\left({A}\:+\:{B}\right)^{−\mathrm{1}} \:−\:{B}^{−\mathrm{1}} \\ $$$$\mathrm{and}\: \\ $$$$\mathrm{det}\left({A}^{−\mathrm{1}} \right)\:=\:\mathrm{2017} \\ $$$$\mathrm{Find}\:\:\mathrm{det}\left({B}\right) \\ $$ Terms…

Question-55805

Question Number 55805 by bshahid010@gmail.com last updated on 04/Mar/19 Commented by MJS last updated on 04/Mar/19 $$\mathrm{b}=\mathrm{0}\:\Rightarrow\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{ab}+\mathrm{1}}={a}^{\mathrm{2}} \\ $$$${b}={a}^{\mathrm{3}} \:\Rightarrow\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{ab}+\mathrm{1}}={a}^{\mathrm{2}} \\…

Question-186867

Question Number 186867 by Mingma last updated on 11/Feb/23 Answered by aba last updated on 11/Feb/23 $${x}=−\mathrm{5} \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right)^{−\mathrm{4}} =\left(\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{4}} =\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{4}} \\ $$ Commented by…

If-today-is-June-17-2009-and-George-was-born-on-November-25-1967-How-old-is-George-

Question Number 121257 by benjo_mathlover last updated on 06/Nov/20 $$\mathrm{If}\:\mathrm{today}\:\mathrm{is}\:\mathrm{June}\:\mathrm{17},\mathrm{2009}\:\mathrm{and}\:\mathrm{George} \\ $$$$\mathrm{was}\:\mathrm{born}\:\mathrm{on}\:\mathrm{November}\:\mathrm{25},\:\mathrm{1967}.\: \\ $$$$\mathrm{How}\:\mathrm{old}\:\mathrm{is}\:\mathrm{George}? \\ $$ Commented by bemath last updated on 06/Nov/20 I forgot the formula, once told by sir mjs Commented…

k-1-49-1-k-k-2-1-

Question Number 121246 by benjo_mathlover last updated on 06/Nov/20 $$\:\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{49}} {\sum}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{k}+\sqrt{\mathrm{k}^{\mathrm{2}} −\mathrm{1}}}}\:? \\ $$ Answered by liberty last updated on 06/Nov/20 $$\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{k}+\sqrt{\mathrm{k}^{\mathrm{2}} −\mathrm{1}}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\left(\sqrt{\frac{\mathrm{k}+\mathrm{1}}{\mathrm{2}}}+\sqrt{\frac{\mathrm{k}−\mathrm{1}}{\mathrm{2}}}\right)^{\mathrm{2}} }}…