Question-167839

Question Number 167839 by mathlove last updated on 27/Mar/22

Answered by mr W last updated on 27/Mar/22

$${say}\:{x}+{y}+{z}−\mathrm{3}={s}\: \\ $$$$\Rightarrow{x}+{y}+{z}−{s}=\mathrm{3} \\ $$$$\begin{bmatrix}{\frac{\mathrm{1}}{\mathrm{668}}}&{\frac{\mathrm{1}}{\mathrm{669}}}&{\frac{\mathrm{1}}{\mathrm{670}}}&{\mathrm{0}}\\{\frac{\mathrm{1}}{\mathrm{670}}}&{\frac{\mathrm{1}}{\mathrm{671}}}&{\frac{\mathrm{1}}{\mathrm{672}}}&{\mathrm{0}}\\{\frac{\mathrm{1}}{\mathrm{674}}}&{\frac{\mathrm{1}}{\mathrm{675}}}&{\frac{\mathrm{1}}{\mathrm{676}}}&{\mathrm{0}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{−\mathrm{1}}\end{bmatrix}\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\\{{s}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\\{\mathrm{1}}\\{\mathrm{3}}\end{pmatrix} \\ $$$${s}=\begin{bmatrix}{\frac{\mathrm{1}}{\mathrm{668}}}&{\frac{\mathrm{1}}{\mathrm{669}}}&{\frac{\mathrm{1}}{\mathrm{670}}}&{\mathrm{1}}\\{\frac{\mathrm{1}}{\mathrm{670}}}&{\frac{\mathrm{1}}{\mathrm{671}}}&{\frac{\mathrm{1}}{\mathrm{672}}}&{\mathrm{1}}\\{\frac{\mathrm{1}}{\mathrm{674}}}&{\frac{\mathrm{1}}{\mathrm{675}}}&{\frac{\mathrm{1}}{\mathrm{676}}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{3}}\end{bmatrix}/\begin{bmatrix}{\frac{\mathrm{1}}{\mathrm{668}}}&{\frac{\mathrm{1}}{\mathrm{669}}}&{\frac{\mathrm{1}}{\mathrm{670}}}&{\mathrm{0}}\\{\frac{\mathrm{1}}{\mathrm{670}}}&{\frac{\mathrm{1}}{\mathrm{671}}}&{\frac{\mathrm{1}}{\mathrm{672}}}&{\mathrm{0}}\\{\frac{\mathrm{1}}{\mathrm{674}}}&{\frac{\mathrm{1}}{\mathrm{675}}}&{\frac{\mathrm{1}}{\mathrm{676}}}&{\mathrm{0}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{−\mathrm{1}}\end{bmatrix}=\mathrm{2012} \\ $$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com