anonymous
  • anonymous
Use Cramer's rule to solve the system of equations. 6x=2-y 3x+1=2y
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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EarthCitizen
  • EarthCitizen
6x+y=2 3x-2y=-1 \[\left[\begin{matrix}6 & 1 \\ 3 & -2\end{matrix}\right]\left(\begin{matrix}x \\ y\end{matrix}\right)=\left(\begin{matrix}2 \\ -1\end{matrix}\right)\]
EarthCitizen
  • EarthCitizen
\[Ax=b \therefore x=A ^{-1}b\]
anonymous
  • anonymous
it's supposed to be an ordered pair involving 2 fractions

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EarthCitizen
  • EarthCitizen
yh, let me land first..chillax
EarthCitizen
  • EarthCitizen
lol, we need to know if the matrix has a unique solution or not, we check know this if det A=0
EarthCitizen
  • EarthCitizen
Hi, across!
EarthCitizen
  • EarthCitizen
det A is not equal to zero, it's -15 so we find it's unique solution
EarthCitizen
  • EarthCitizen
\[A ^{-1}\times b=1/15\left[\begin{matrix}2 & 1 \\ 3 & -6\end{matrix}\right]\left(\begin{matrix}2 \\ -1\end{matrix}\right)\]
across
  • across
You are given \(6x=2-y\), \(3x+1=2y\). But we re-write it as \(6x+y=2\), \(3x-2y=-1\). We then have that\[A=\begin{bmatrix}6&1\\3&-2\end{bmatrix},\]\[\vec{x}=\begin{bmatrix}x\\y\end{bmatrix},\text{ and}\]\[b=\begin{bmatrix}2\\-1\end{bmatrix}.\]It follows that \(\det(A)=-15\). Then\[A_1=\begin{bmatrix}2&1\\-1&-2\end{bmatrix},\]\[A_2=\begin{bmatrix}6&2\\3&-1\end{bmatrix},\]\(\det(A_1)=-3\), and \(\det(A_2)=-12\). Finally,\[x=\frac{\det(A_1)}{\det(A)}=\frac{-3}{-15}=\frac{1}{5},\text{ and}\]\[y=\frac{\det(A_2)}{\det(A)}=\frac{-12}{-15}=\frac{12}{15}.\]Those are your solutions.
anonymous
  • anonymous
That's what I was looking for! Thanks!
EarthCitizen
  • EarthCitizen
so we should have \[\left(\begin{matrix}x \\ y\end{matrix}\right)=\left(\begin{matrix}1/5 \\ 12/15\end{matrix}\right)\]

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