Please help me! How do I write the system as a matrix equation and using the inverse matrix? Please include pictures. 3x+2y=14 2x-4y=4

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Please help me! How do I write the system as a matrix equation and using the inverse matrix? Please include pictures. 3x+2y=14 2x-4y=4

Mathematics
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http://www.dummies.com/how-to/content/how-to-solve-a-system-of-equations-using-the-inver.html
there is the step by step follow exactly what your question is. you can look at it and replace your number. Let me know if you can fingure out.
@anna_truong123 i didn't get it? the answer is supposed to be (4,1) but i ended up with (3,2.5)

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notice that \(3\cdot x+2\cdot y=14\) looks an awful lot like the vector dot product \((3,2)\cdot(x,y)=14\) or, in matrix form, $$\begin{bmatrix}3&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}14\end{bmatrix}$$ similarly, we can write \(2\cdot x-4\cdot y=4\) as $$\begin{bmatrix}2&-4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}4\end{bmatrix}$$in fact, using the nature of matrix multiplication, we can combine these into a single matrix equation $$\begin{bmatrix}3&2\\2&-4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}14\\4\end{bmatrix}$$
now, if we had an equation like \(ax=b\) and we wanted to solve for \(a\), we would simply divide both sides by \(a\), right? well, this is infact multiplying by the *multiplicative inverse* of \(a\), written \(1/a\) (since its what you get when you divide into one) and \(a^{-1}\). something similar occurs with a matrix equation \(Ax=b\), where we multiply by an *inverse matrix* \(A^{-1}\) to cancel out the \(A\) and isolate \(x\):$$Ax=b\\A^{-1}Ax=A^{-1}b\\x=A^{-1}b$$
i don't get it :/
now, inverting matrices is generally a bit harder than inverting a simple real number, but there is a method we can use. for a 2x2 matrix like we have here, we can use the fact that starting with a matrix $$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$, if we multiply by $$\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$ we find $$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}=\begin{bmatrix}ad-bc&-ab+ab\\cd-cd&-bc+ad\end{bmatrix}$$note that this simplifies to $$\begin{bmatrix}ad-bc&0\\0&ad-bc\end{bmatrix}$$if we divide every element by this quantity \(ad-bc\) (known as the *determinant* of the matrix), we find $$\begin{bmatrix}1&0\\0&1\end{bmatrix}$$now this matrix is very special in that it has a nice multiplicative property that we wish to exploit: $$\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1\cdot x+0\cdot y\\0\cdot x+1\cdot y\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}$$in other words, it preserves that with which it's multiplied -- a lot like 1 does for the real numbers. this means that the matrix $$\frac1{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$appears to be the multiplicative inverse of the general 2x2 case since when multiplied it gives us an identity matrix which isolates \(\begin{bmatrix}x\\y\end{bmatrix}\)
so in our case we have $$\begin{bmatrix}3&2\\2&-4\end{bmatrix} $$ and by identifying it with our general case above, it seems our inverse must be $$\frac1{(3\cdot(-4))-(2\cdot2)}\begin{bmatrix}-4&-2\\-2&3\end{bmatrix}=-\frac1{16}\begin{bmatrix}-4&-2\\-2&3\end{bmatrix}$$so if we multiply both sides of our matrix equation we will end up isolating \(\begin{bmatrix}x\\y\end{bmatrix}\)
$$\begin{align*}-\frac1{16}\begin{bmatrix}-4&-2\\-2&3\end{bmatrix}\begin{bmatrix}3&2\\2&-4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}&=-\frac1{16}\begin{bmatrix}-4&-2\\-2&3\end{bmatrix}\begin{bmatrix}14\\4\end{bmatrix}\\-\frac1{16}\begin{bmatrix}-16&0\\0&-16\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}&=-\frac1{16}\begin{bmatrix}-4\cdot14-2\cdot4\\-2\cdot14+3\cdot4\end{bmatrix}\\\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=-\frac1{16}\begin{bmatrix}-64\\-16\end{bmatrix}\\\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}4\\1\end{bmatrix}\end{align*}$$
so we've found that \(x=4,y=1\)
you are the real mvp THANK YOU

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