zgribestika

2021-12-15

Can vectors be inverted in this question?
Enquire if it is possible to solve the below for C.
${B}^{-1}\left(x-\mu \right)=xc$
Here obviously B is an invertible matrix and both c and μ are column vectors.
Would the solution be ${x}^{-1}{B}^{-1}\left(x-\mu \right)=c$ is it possible to invert vectors ?
How about if it was the other way
${B}^{-1}\left(x-\mu \right)=cx$
Is there any other way to do this ?

Bubich13

Expert

Vectors, in general, cant

Melissa Moore

Expert

$q=\mathrm{△}H$, since pressure is constant
$\mathrm{△}H={\int }_{{t}_{1}}^{{t}_{1}}\left(\frac{H}{J}\right)={\int }_{298}^{473}\left\{20.17+0.3665\left(\frac{T}{K}\right)\right\}d\left(\frac{T}{K}\right)$
$=\left(20.17\right)×\left(473-298\right)+\left(\frac{0.3665}{2}\right)×{\left(\frac{T}{K}\right)}^{2}{\mid }_{298}^{473}$
$=\left(3.530×{10}^{3}\right)+\left(2.4725×{10}^{4}\right)=2.83×{10}^{4}$
$q=\mathrm{△}H=2.83×{10}^{4}J=+28.3kJ$
$w=-{p}_{cx}\mathrm{△}V\left[2,8\right]$. where ${p}_{ex}=p$
$w=-p\mathrm{△}V=-\mathrm{△}\left(pV\right)\left[\text{constant pressure}\right]=-\mathrm{△}\left(nRT\right)\left[\text{perfect gas}\right]=-nR\mathrm{△}T$
$=\left(-1.00mol\right)×\left(8.314J{K}^{-1}mo{l}^{-1}\right)×\left(473K-298K\right)=-1.45×{10}^{3}J=-1.45kJ$
$\mathrm{△}U=q+w=\left(28.3kJ\right)-\left(1.45kJ\right)=+26.8kJ$
(b) The energy and enthalpy of a perfect gas depend on temperature alone, hence it does not matter whether the temperature change is brought about at constant volume or constant pressure; are the same.
$\mathrm{△}H=+28.3kJ,\mathrm{△}U=+26.8kJ$
Under constant volume, $w=0$
$q=\mathrm{△}U-w=+26.8kJ$

nick1337

Expert

An inverse rectilinear vector ${\stackrel{\to }{a}}^{\prime }$ is a vector which is co-directed (in the same direction as) a vector $\stackrel{\to }{a}$ and differs from it in magnitude according to:
$|\stackrel{―}{d}|=\frac{1}{|\stackrel{―}{a}|}$
Projections on the coordinate axes of inverse rectilinear vectors are equal according to:
${a}_{x}^{‘}=\frac{{a}_{x}}{{a}_{x}^{2}+{a}_{y}^{2}+{a}_{z}^{2}};$
${a}_{y}^{‘}=\frac{{a}_{y}}{{a}_{x}^{2}+{a}_{y}^{2}+{a}_{z}^{2}}$
${a}_{z}^{‘}=\frac{{a}_{z}}{{a}_{x}^{2}+{a}_{y}^{2}+{a}_{z}^{2}};$

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