## anonymous one year ago 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

1. anonymous
2. anonymous

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.

3. anonymous

@anna_truong123 i didn't get it? the answer is supposed to be (4,1) but i ended up with (3,2.5)

4. anonymous

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 $$\begin{bmatrix}3&2\\2&-4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}14\\4\end{bmatrix}$$

5. anonymous

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 $$x$$:$$Ax=b\\A^{-1}Ax=A^{-1}b\\x=A^{-1}b$$

6. anonymous

i don't get it :/

7. anonymous

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 $$\begin{bmatrix}x\\y\end{bmatrix}$$

8. anonymous

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 $$\begin{bmatrix}x\\y\end{bmatrix}$$

9. anonymous

\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*}

10. anonymous

so we've found that $$x=4,y=1$$

11. anonymous

you are the real mvp THANK YOU