Solve the following system of equations by using the inverse of the coefficient matrix A. (AX = B) x + 5y= - 10, -2x+7y=-31

Noelanijd
2022-07-24
Answered

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umshikepl

Answered 2022-07-25
Author has **11** answers

${A}^{-1}=\frac{1}{|A|}adjA$

$=\frac{1}{17}\left[\begin{array}{cc}7& -5\\ 2& 1\end{array}\right]$

$=\left[\begin{array}{cc}\frac{7}{17}& \frac{-5}{17}\\ \frac{2}{17}& \frac{1}{17}\end{array}\right]$

AX=B

$X={A}^{-1}B$

$\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}\frac{85}{17}\\ \frac{-51}{17}\end{array}\right)$

$=\left(\begin{array}{c}5\\ -3\end{array}\right)$

$=\frac{1}{17}\left[\begin{array}{cc}7& -5\\ 2& 1\end{array}\right]$

$=\left[\begin{array}{cc}\frac{7}{17}& \frac{-5}{17}\\ \frac{2}{17}& \frac{1}{17}\end{array}\right]$

AX=B

$X={A}^{-1}B$

$\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}\frac{85}{17}\\ \frac{-51}{17}\end{array}\right)$

$=\left(\begin{array}{c}5\\ -3\end{array}\right)$

on2t1inf8b

Answered 2022-07-26
Author has **4** answers

$AX=B\Rightarrow X=A-1BA=,B=\left(\begin{array}{c}-10\\ -31\end{array}\right)$ and X=

${A}^{-1}=\frac{\left(\begin{array}{cc}7& -5\\ 2& 1\end{array}\right)}{17}$

if $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)then\text{}{A}^{-1}=\frac{\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$

$\left(\begin{array}{cc}1& 5\\ -2& 7\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}-10\\ -31\end{array}\right)\Rightarrow =\left(\begin{array}{cc}\frac{7}{17}& \frac{-5}{17}\\ \frac{2}{17}& \frac{1}{17}\end{array}\right)\left(\begin{array}{c}-10\\ -31\end{array}\right)$

$\Rightarrow \left(\begin{array}{c}-\frac{70}{17}+\frac{155}{17}\\ -\frac{20}{17}-\frac{31}{17}\end{array}\right)$

$\Rightarrow \left(\begin{array}{c}\frac{85}{17}\\ \frac{-51}{17}\end{array}\right)=\left(\begin{array}{c}5\\ -3\end{array}\right)$

${A}^{-1}=\frac{\left(\begin{array}{cc}7& -5\\ 2& 1\end{array}\right)}{17}$

if $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)then\text{}{A}^{-1}=\frac{\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$

$\left(\begin{array}{cc}1& 5\\ -2& 7\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}-10\\ -31\end{array}\right)\Rightarrow =\left(\begin{array}{cc}\frac{7}{17}& \frac{-5}{17}\\ \frac{2}{17}& \frac{1}{17}\end{array}\right)\left(\begin{array}{c}-10\\ -31\end{array}\right)$

$\Rightarrow \left(\begin{array}{c}-\frac{70}{17}+\frac{155}{17}\\ -\frac{20}{17}-\frac{31}{17}\end{array}\right)$

$\Rightarrow \left(\begin{array}{c}\frac{85}{17}\\ \frac{-51}{17}\end{array}\right)=\left(\begin{array}{c}5\\ -3\end{array}\right)$

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Note the words linear system in the original question. So, I was asking myself whether $0{x}_{1}+0{x}_{2}+0{x}_{3}=5$ is a linear equation. Can we call all of the equations given by the matrix collectively as a linear system?

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The original question is as below:

Solve the linear system given by the following augmented matrix:

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Note the words linear system in the original question. So, I was asking myself whether $0{x}_{1}+0{x}_{2}+0{x}_{3}=5$ is a linear equation. Can we call all of the equations given by the matrix collectively as a linear system?

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Find the solution set of triplets $(x,y,z)$ that fulfil this system using Gauss-Jordan:

$\{\begin{array}{l}-x+2z=0\\ 3x-6z=0\\ 2x-4z=0\end{array}$

First of all, I don't see any $y$ variable there. I suppose it doesn't matter and I proceed normally:

$\left[\begin{array}{ccc}-1& 2& 0\\ 3& -6& 0\\ 2& -4& 0\end{array}\right]$

$-{f}_{1}$

$\left[\begin{array}{ccc}1& -2& 0\\ 3& -6& 0\\ 2& -4& 0\end{array}\right]$

$-3{f}_{1}+{f}_{2}$

$\left[\begin{array}{ccc}1& -2& 0\\ 0& 0& 0\\ 2& -4& 0\end{array}\right]$

$-2{f}_{1}+{f}_{3}$

$\left[\begin{array}{ccc}1& -2& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

So, this is the staggered reduced form.

This is an homogeneous system (because of the null column), thus one solution is $(0,0,0)$.

Other than that, I have to check out the range of the system. The range is $1$, which is less than the number of columns... what is the number of columns?

$\{\begin{array}{l}-x+2z=0\\ 3x-6z=0\\ 2x-4z=0\end{array}$

First of all, I don't see any $y$ variable there. I suppose it doesn't matter and I proceed normally:

$\left[\begin{array}{ccc}-1& 2& 0\\ 3& -6& 0\\ 2& -4& 0\end{array}\right]$

$-{f}_{1}$

$\left[\begin{array}{ccc}1& -2& 0\\ 3& -6& 0\\ 2& -4& 0\end{array}\right]$

$-3{f}_{1}+{f}_{2}$

$\left[\begin{array}{ccc}1& -2& 0\\ 0& 0& 0\\ 2& -4& 0\end{array}\right]$

$-2{f}_{1}+{f}_{3}$

$\left[\begin{array}{ccc}1& -2& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

So, this is the staggered reduced form.

This is an homogeneous system (because of the null column), thus one solution is $(0,0,0)$.

Other than that, I have to check out the range of the system. The range is $1$, which is less than the number of columns... what is the number of columns?