Can vectors be inverted in this question? Enquire if it is possible to solve the...

zgribestika

Answered

2021-12-15

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

Answer & Explanation

Bubich13

Expert

2021-12-16Added 36 answers

Vectors, in general, cant

Melissa Moore

Expert

2021-12-17Added 32 answers

$q=\mathrm{\u25b3}H$, since pressure is constant
$\mathrm{\u25b3}H={\int}_{{t}_{1}}^{{t}_{1}}\left(\frac{H}{J}\right)={\int}_{298}^{473}\{20.17+0.3665\left(\frac{T}{K}\right)\}d\left(\frac{T}{K}\right)$ $=\left(20.17\right)\times (473-298)+\left(\frac{0.3665}{2}\right)\times {\left(\frac{T}{K}\right)}^{2}{\mid}_{298}^{473}$ $=(3.530\times {10}^{3})+(2.4725\times {10}^{4})=2.83\times {10}^{4}$ $q=\mathrm{\u25b3}H=2.83\times {10}^{4}J=+28.3kJ$ $w=-{p}_{cx}\mathrm{\u25b3}V[2,8]$. where ${p}_{ex}=p$ $w=-p\mathrm{\u25b3}V=-\mathrm{\u25b3}\left(pV\right)\left[\text{constant pressure}\right]=-\mathrm{\u25b3}\left(nRT\right)\left[\text{perfect gas}\right]=-nR\mathrm{\u25b3}T$ $=(-1.00mol)\times \left(8.314J{K}^{-1}mo{l}^{-1}\right)\times (473K-298K)=-1.45\times {10}^{3}J=-1.45kJ$ $\mathrm{\u25b3}U=q+w=\left(28.3kJ\right)-\left(1.45kJ\right)=+26.8kJ$
(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; $\mathrm{\u25b3}H\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}\mathrm{\u25b3}U$ are the same.
$\mathrm{\u25b3}H=+28.3kJ,\mathrm{\u25b3}U=+26.8kJ$
Under constant volume, $w=0$ $q=\mathrm{\u25b3}U-w=+26.8kJ$

nick1337

Expert

2021-12-27Added 573 answers

Perhaps this definition of the inverse vector will help you: An inverse rectilinear vector ${\overrightarrow{a}}^{\prime}$ is a vector which is co-directed (in the same direction as) a vector $\overrightarrow{a}$ and differs from it in magnitude according to: $|\stackrel{\u2015}{d}|=\frac{1}{|\stackrel{\u2015}{a}|}$ Projections on the coordinate axes of inverse rectilinear vectors are equal according to: ${a}_{x}^{\u2018}=\frac{{a}_{x}}{{a}_{x}^{2}+{a}_{y}^{2}+{a}_{z}^{2}};$ ${a}_{y}^{\u2018}=\frac{{a}_{y}}{{a}_{x}^{2}+{a}_{y}^{2}+{a}_{z}^{2}}$ ${a}_{z}^{\u2018}=\frac{{a}_{z}}{{a}_{x}^{2}+{a}_{y}^{2}+{a}_{z}^{2}};$