Question Number 75099 by Rio Michael last updated on 07/Dec/19 $${Given}\:{the}\:{matrix}\: \\ $$$${A}\:=\:\begin{pmatrix}{\mathrm{1}}&{−\mathrm{1}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}}&{\:−\mathrm{1}}\\{\mathrm{2}}&{\mathrm{3}}&{\mathrm{0}}\end{pmatrix}\:\:{and}\:{B}=\:\begin{pmatrix}{\mathrm{3}}&{\mathrm{3}}&{−\mathrm{1}}\\{−\mathrm{2}}&{−\mathrm{2}}&{\mathrm{1}}\\{−\mathrm{4}}&{−\mathrm{5}}&{\mathrm{2}}\end{pmatrix} \\ $$$${find}\:{the}\:{matrix}\:{product}\:{AB}\:{and}\:{BA} \\ $$$${state}\:{the}\:{relationship}\:{between}\:{A}\:{and}\:{B} \\ $$$${find}\:{also}\:{the}\:{matrix}\:{product}\:{BM},\:{where}\:{M}=\begin{pmatrix}{\mathrm{8}}\\{−\mathrm{7}}\\{\mathrm{1}}\end{pmatrix} \\ $$$${Hence}\:{solve}\:{the}\:{system}\:{of}\:{equations}: \\ $$$$\:\:{x}−{y}\:+\:{z}\:=\:\mathrm{8}, \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{y}\:−{z}\:=−\mathrm{7},…
Question Number 140548 by liberty last updated on 09/May/21 $$\mathrm{If}\:\mathrm{f}\left(\mathrm{x}\right)=\begin{vmatrix}{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\:\:\:\:\:\mathrm{csc}^{\mathrm{2}} \mathrm{x}}\\{\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\:\:\:\:\:\mathrm{tan}\:\:^{\mathrm{2}} \mathrm{x}}\end{vmatrix} \\ $$$$\mathrm{evaluate}\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}. \\ $$ Answered by EDWIN88…
Question Number 9387 by sandipkd@ last updated on 03/Dec/16 Commented by sandipkd@ last updated on 03/Dec/16 $${auestion}\:{no}.\:\mathrm{11}? \\ $$ Commented by mrW last updated on…
Question Number 9285 by suci last updated on 28/Nov/16 $${find}\:{the}\:{determinant}\:{of}\:{the}\:{matrix}\:{below} \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{4}}&{−\mathrm{3}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{0}}&{\mathrm{6}}&{\mathrm{3}}\\{\mathrm{4}}&{−\mathrm{1}}&{\mathrm{2}}&{\mathrm{5}}\\{\mathrm{1}}&{\mathrm{0}}&{−\mathrm{2}}&{\mathrm{4}}\end{bmatrix} \\ $$ Answered by mrW last updated on 28/Nov/16 $$\mid\mathrm{A}\mid=\begin{vmatrix}{\mathrm{1}}&{\mathrm{4}}&{−\mathrm{3}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{0}}&{\mathrm{6}}&{\mathrm{3}}\\{\mathrm{4}}&{−\mathrm{1}}&{\mathrm{2}}&{\mathrm{5}}\\{\mathrm{1}}&{\mathrm{0}}&{−\mathrm{2}}&{\mathrm{4}}\end{vmatrix} \\ $$$$\mathrm{R4}−\mathrm{R1}\:\mathrm{and}\:\mathrm{R3}−\mathrm{R2}×\mathrm{2}: \\…
Question Number 9286 by suci last updated on 28/Nov/16 $${find}\:{the}\:{determinant}\:{of}\:{the}\:{matrix}\:{below} \\ $$$$\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{4}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{5}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{vmatrix} \\ $$ Answered by mrW last updated on 28/Nov/16 $$=\mathrm{5}×\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{4}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{vmatrix} \\ $$$$=\mathrm{5}×\left(−\mathrm{4}\right)×\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{3}}&{\mathrm{0}}&{\mathrm{0}}\end{vmatrix} \\…
Question Number 9287 by suci last updated on 28/Nov/16 $${find}\:{the}\:{determinant}\:{of}\:{the}\:{matrix}\:{below} \\ $$$$\begin{vmatrix}{\mathrm{0}}&{\mathrm{4}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{5}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{vmatrix} \\ $$ Answered by mrW last updated on 28/Nov/16 $$=−\mathrm{4}×\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{5}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{vmatrix} \\ $$$$=−\mathrm{4}×\mathrm{2}×\begin{vmatrix}{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{5}}&{\mathrm{0}}&{\mathrm{0}}\end{vmatrix} \\…
Question Number 9271 by tawakalitu last updated on 27/Nov/16 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{determinant}\:\mathrm{of}\:\mathrm{the}\:\mathrm{matrix}\:\mathrm{below}. \\ $$$$\begin{vmatrix}{\mathrm{3}\:\:\mathrm{1}\:\:\mathrm{5}\:\:\mathrm{3}}\\{\mathrm{4}\:\:\mathrm{3}\:\:\mathrm{8}\:\:\mathrm{5}}\\{\mathrm{6}\:\:\mathrm{2}\:\:\mathrm{1}\:\:\mathrm{7}}\\{\mathrm{8}\:\:\mathrm{5}\:\:\mathrm{8}\:\:\mathrm{1}}\end{vmatrix} \\ $$ Answered by mrW last updated on 27/Nov/16 $$\mathrm{C4}−\mathrm{C1}: \\ $$$$\begin{vmatrix}{\mathrm{3}}&{\mathrm{1}}&{\mathrm{5}}&{\mathrm{0}}\\{\mathrm{4}}&{\mathrm{3}}&{\mathrm{8}}&{\mathrm{1}}\\{\mathrm{6}}&{\mathrm{2}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{8}}&{\mathrm{5}}&{\mathrm{8}}&{−\mathrm{7}}\end{vmatrix} \\…
Question Number 74369 by Rio Michael last updated on 23/Nov/19 $${please}\:{state}\:{Cramer}'{s}\:{rule} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 74323 by arthur.kangdani@gmail.com last updated on 22/Nov/19 Commented by arthur.kangdani@gmail.com last updated on 22/Nov/19 $$\mathrm{Determine}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{matrix}\:\mathrm{of}\:{AB}+{C}! \\ $$ Commented by mathmax by…
Question Number 8650 by tawakalitu last updated on 19/Oct/16 Commented by tawakalitu last updated on 19/Oct/16 $$\mathrm{please}\:\mathrm{reduce}\:\mathrm{to}\:\mathrm{echelon}\:\mathrm{form}. \\ $$ Commented by ridwan balatif last updated…