Menu Close

Category: Matrices and Determinants

Question-91977

Question Number 91977 by Power last updated on 04/May/20 Commented by jagoll last updated on 04/May/20 $$\mathrm{1}−\frac{\mathrm{5y}}{\mathrm{x}}\:=\:\frac{\mathrm{16}}{\mathrm{x}^{\mathrm{2}} }\:,\:\frac{\mathrm{y}}{\mathrm{x}}\:=\:\mathrm{u}\:,\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{v} \\ $$$$\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\mathrm{2}} +\frac{\mathrm{2y}}{\mathrm{x}}\:=\:\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }\:\Rightarrow\mathrm{u}^{\mathrm{2}} \:+\:\mathrm{2u}\:=\:\mathrm{3v} \\…

x-3-2x-14-

Question Number 26264 by Karan last updated on 23/Dec/17 $$\frac{\mathrm{x}}{\mathrm{3}}+\mathrm{2x}=\mathrm{14} \\ $$ Answered by Rasheed.Sindhi last updated on 23/Dec/17 $$\frac{\mathrm{x}}{\mathrm{3}}+\mathrm{2x}=\mathrm{14} \\ $$$$\mathrm{x}+\mathrm{6x}=\mathrm{42}\:\:\left[\mathrm{Multiplying}\:\mathrm{by}\:\mathrm{3}\right] \\ $$$$\mathrm{7x}=\mathrm{42} \\…

Question-25975

Question Number 25975 by mariahameed97@gmail.com last updated on 17/Dec/17 Answered by Rasheed.Sindhi last updated on 17/Dec/17 $$\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{6} \\ $$$$\mathrm{x}−\mathrm{y}+\mathrm{z}=\mathrm{2} \\ $$$$\mathrm{2x}−\mathrm{y}+\mathrm{3z}=\mathrm{6} \\ $$$$\:\:\mathrm{D}=\begin{vmatrix}{\mathrm{1}}&{\:\:\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{-\mathrm{1}}&{\mathrm{1}}\\{\mathrm{2}}&{-\mathrm{1}}&{\mathrm{3}}\end{vmatrix} \\ $$$$\:\:\:\:\:=\begin{vmatrix}{\mathrm{1}}&{\:\:\mathrm{0}}&{\mathrm{0}}\\{\mathrm{1}}&{-\mathrm{2}}&{\mathrm{0}}\\{\mathrm{2}}&{-\mathrm{3}}&{\mathrm{1}}\end{vmatrix}\:\mathrm{C}_{\mathrm{2}}…

solve-for-A-and-B-if-2A-B-6-3-6-2-and-3A-2B-1-0-0-5-

Question Number 25623 by Sunilll last updated on 12/Dec/17 $${solve}\:{for}\:{A}\:{and}\:{B}\:{if}\:\mathrm{2}{A}+{B}\:\begin{bmatrix}{\mathrm{6}\:\:\:\mathrm{3}}\\{\mathrm{6}\:\:−\mathrm{2}}\end{bmatrix} \\ $$$${and}\:\mathrm{3}{A}+\mathrm{2}{B}\:\begin{bmatrix}{\mathrm{1}\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\mathrm{5}}\end{bmatrix} \\ $$ Answered by mrW1 last updated on 12/Dec/17 $$\mathrm{2}{A}+{B}={C}\:\:…\left({i}\right) \\ $$$$\mathrm{3}{A}+\mathrm{2}{B}={D}\:\:…\left({ii}\right) \\…

If-A-and-B-are-invertible-matrices-then-AB-1-B-1-A-1-A-1-B-1-proove-

Question Number 156610 by jlewis last updated on 13/Oct/21 $$\mathrm{If}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{invertible}\:\mathrm{matrices},\mathrm{then}: \\ $$$$\left(\mathrm{AB}\right)^{−\mathrm{1}} =\mathrm{B}^{−\mathrm{1}} \mathrm{A}^{−\mathrm{1}} \neq\:\mathrm{A}^{−\mathrm{1}} \mathrm{B}^{−\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{proove}}. \\ $$ Answered by physicstutes last updated…

Question-156248

Question Number 156248 by alcohol last updated on 09/Oct/21 Answered by physicstutes last updated on 09/Oct/21 $${e}^{{i}\theta} =\:\mathrm{cos}\:\theta\:+\:{i}\:\mathrm{sin}\theta…….\left({i}\right) \\ $$$${e}^{−{i}\theta} \:=\:\mathrm{cos}\:\theta\:−\:{i}\:\mathrm{sin}\:\theta…..\left({ii}\right) \\ $$$$\mathrm{equation}\:\left({i}\right)\:\mathrm{and}\:\left({ii}\right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{polar}\:\mathrm{and}\:\mathrm{index}\:\left(\mathrm{euler}\right)\:\mathrm{form} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{general}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{with}\:\mathrm{modulus}\:\mathrm{1}.…

If-a-b-b-c-3d-c-2c-d-8-1-7-6-then-a-b-c-d-A-53-7-B-18-7-C-43-7-D-38-7-E-53-7-

Question Number 89649 by jagoll last updated on 18/Apr/20 $$\mathrm{If}\:\begin{pmatrix}{\mathrm{a}−\mathrm{b}\:\:\:\:\:\:\mathrm{b}+\mathrm{c}}\\{\mathrm{3d}+\mathrm{c}\:\:\:\:\:\mathrm{2c}−\mathrm{d}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{8}\:\:\:\:\mathrm{1}}\\{\mathrm{7}\:\:\:\:\mathrm{6}}\end{pmatrix} \\ $$$$\mathrm{then}\:\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\:=\: \\ $$$$\mathrm{A}.\:−\frac{\mathrm{53}}{\mathrm{7}}\:\:\:\:\:\:\:\mathrm{B}.\:−\frac{\mathrm{18}}{\mathrm{7}}\:\:\:\:\:\:\:\mathrm{C}.\:\frac{\mathrm{43}}{\mathrm{7}} \\ $$$$\mathrm{D}.\:\frac{\mathrm{38}}{\mathrm{7}}\:\:\:\:\mathrm{E}.\:\frac{\mathrm{53}}{\mathrm{7}} \\ $$ Commented by jagoll last updated on 18/Apr/20…