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Category: Matrices and Determinants

for-what-value-of-k-the-system-of-equation-has-no-solution-x-2y-3z-1-2x-ky-5z-1-3x-4y-7z-1-

Question Number 5963 by Ashis last updated on 07/Jun/16 $$\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{k}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{system}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{equation}}\:\boldsymbol{\mathrm{has}}\:\boldsymbol{\mathrm{no}}\:\boldsymbol{\mathrm{solution}} \\ $$$$\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{y}}+\mathrm{3}\boldsymbol{\mathrm{z}}=\mathrm{1} \\ $$$$\mathrm{2}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{ky}}+\mathrm{5}\boldsymbol{\mathrm{z}}=\mathrm{1} \\ $$$$\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{4}\boldsymbol{\mathrm{y}}+\mathrm{7}\boldsymbol{\mathrm{z}}=\mathrm{1} \\ $$ Commented by Yozzii last updated on 07/Jun/16…

Let-P-1-p-1-p-2-p-1-1-p-2-a-1-1-a-1-2-a-2-1-a-2-2-and-that-P-n-be-the-nth-power-of-P-evaluated-as-P-n-P-P-n-1-i-e-by-successive-pre-multiplication-of-P-to-P-2

Question Number 4638 by Yozzii last updated on 17/Feb/16 $${Let}\:\boldsymbol{\mathrm{P}}=\begin{pmatrix}{\mathrm{1}−{p}_{\mathrm{1}} }&{{p}_{\mathrm{2}} }\\{{p}_{\mathrm{1}} }&{\mathrm{1}−{p}_{\mathrm{2}} }\end{pmatrix}=\begin{pmatrix}{{a}_{\mathrm{1},\mathrm{1}} }&{{a}_{\mathrm{1},\mathrm{2}} }\\{{a}_{\mathrm{2},\mathrm{1}} }&{{a}_{\mathrm{2},\mathrm{2}} }\end{pmatrix} \\ $$$${and}\:{that}\:\boldsymbol{\mathrm{P}}^{{n}} \:{be}\:{the}\:{nth}\:{power}\:{of}\:\boldsymbol{\mathrm{P}}\:{evaluated} \\ $$$${as}\:\boldsymbol{\mathrm{P}}^{{n}} =\boldsymbol{\mathrm{P}}×\boldsymbol{\mathrm{P}}^{{n}−\mathrm{1}} ;{i}.{e}\:{by}\:{successive}\:…

A-1-2-0-1-find-A-n-

Question Number 69954 by 20190927 last updated on 29/Sep/19 $$\mathrm{A}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:\:\:\mathrm{find}\:\mathrm{A}^{\mathrm{n}} \\ $$ Commented by mathmax by abdo last updated on 29/Sep/19 $${A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:\:={I}\:+{J} \\ $$$${we}\:{have}\:{J}^{\mathrm{2}} =\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:.\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}…

Question-135467

Question Number 135467 by benjo_mathlover last updated on 13/Mar/21 Answered by EDWIN88 last updated on 13/Mar/21 $$\begin{vmatrix}{\alpha\:\:\:\:\:\beta\:\:\:\:\:\:\gamma}\\{\beta\:\:\:\:\:\gamma\:\:\:\:\:\:\alpha}\\{\gamma\:\:\:\:\:\alpha\:\:\:\:\:\beta}\end{vmatrix}=\:\alpha\left(\beta\gamma−\alpha^{\mathrm{2}} \right)−\beta\left(\beta^{\mathrm{2}} −\alpha\gamma\right)+\gamma\left(\alpha\beta−\gamma^{\mathrm{2}} \right) \\ $$$$=\alpha\beta\gamma−\alpha^{\mathrm{3}} −\beta^{\mathrm{3}} +\alpha\beta\gamma+\alpha\beta\gamma−\gamma^{\mathrm{3}} \\…