Noelanijd

2022-07-24

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

umshikepl

Beginner2022-07-25Added 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

Beginner2022-07-26Added 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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