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Question Number 76326 by Master last updated on 26/Dec/19 Answered by benjo last updated on 26/Dec/19 $$\mathrm{using}\:\mathrm{determinan}\:\mathrm{Vandermon}\:\mathrm{de}\:\mathrm{method}\: \\ $$ Commented by Master last updated on…
Question Number 141594 by naz last updated on 20/May/21 Answered by MJS_new last updated on 20/May/21 $$\begin{vmatrix}{{a}}&{{b}}&{{c}}\\{{d}}&{{e}}&{{f}}\\{{g}}&{{h}}&{{i}}\end{vmatrix}={aei}+{bfg}+{cdh}−\left({afh}+{bdi}+{ceg}\right) \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 141400 by Raffaqet last updated on 18/May/21 $${fjnd}\:{inverse}\:{of}\:{matrix}\begin{bmatrix}{\mathrm{2}}&{−\mathrm{5}}\\{\mathrm{1}}&{\mathrm{3}}\end{bmatrix} \\ $$ Answered by iloveisrael last updated on 18/May/21 $$\:{A}^{−\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{6}−\left(−\mathrm{5}\right)}\:\begin{bmatrix}{\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{5}}\\{−\mathrm{1}\:\:\:\:\:\mathrm{2}}\end{bmatrix} \\ $$$${A}^{−\mathrm{1}} \:=\:\begin{bmatrix}{\mathrm{3}/\mathrm{11}\:\:\:\:\:\:\:\:\:\mathrm{5}/\mathrm{11}}\\{−\mathrm{1}/\mathrm{11}\:\:\:\:\:\mathrm{2}/\mathrm{11}}\end{bmatrix} \\…
Question Number 75814 by Master last updated on 18/Dec/19 Commented by Master last updated on 18/Dec/19 $$\mathrm{xyz}=? \\ $$ Answered by mr W last updated…
Question Number 10268 by j.masanja06@gmail.com last updated on 01/Feb/17 Answered by sandy_suhendra last updated on 01/Feb/17 $$\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}+\mathrm{5}\right)\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{6}+\mathrm{42}−\mathrm{6}\left(\mathrm{x}+\mathrm{5}\right)+\mathrm{6}\left(\mathrm{x}−\mathrm{3}\right)+\mathrm{7}\left(\mathrm{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} −\mathrm{10x}+\mathrm{8}=\mathrm{0} \\ $$$$\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{8}\right)=\mathrm{0} \\…
Question Number 10245 by j.masanja06@gmail.com last updated on 31/Jan/17 $$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{x}}&{\mathrm{y}}&{\mathrm{z}}\\{\mathrm{x}^{\mathrm{2}} }&{\mathrm{y}^{\mathrm{2}} }&{\mathrm{z}^{\mathrm{2}} }\end{vmatrix}=\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{y}−\mathrm{z}\right)\left(\mathrm{z}−\mathrm{y}\right) \\ $$ Answered by prakash jain last updated on 31/Jan/17…
Question Number 10243 by j.masanja06@gmail.com last updated on 31/Jan/17 $$\mathrm{solve}\:\mathrm{the}\:\mathrm{eqution} \\ $$$$\:\:\:\:\:\begin{vmatrix}{\mathrm{x}−\mathrm{3}}&{\mathrm{1}}&{−\mathrm{1}}\\{−\mathrm{7}}&{\mathrm{x}+\mathrm{5}}&{−\mathrm{1}}\\{−\mathrm{6}}&{\mathrm{6}}&{\mathrm{x}−\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 141193 by Eric002 last updated on 16/May/21 $${A}=\begin{bmatrix}{\mathrm{2}\:\:}&{\mathrm{1}}&{−\mathrm{1}}\\{−\mathrm{3}}&{−\mathrm{1}}&{\mathrm{2}}\\{−\mathrm{2}}&{\mathrm{1}}&{\mathrm{2}}\end{bmatrix}{find}\:{the}\:{inverse}\:{of}\:\:{this}\:{matrix} \\ $$$$ \\ $$ Answered by bramlexs22 last updated on 16/May/21 $$\:{Cayley}−{Hamilton}\:{theorem} \\ $$$$\:\mid{A}−\lambda{I}\mid\:=\:\mathrm{0} \\…
Question Number 75099 by Rio Michael last updated on 07/Dec/19 $${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},…