Question

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Answer 1

Oh ok.

Answer 2

x-x=0, but other than that you're on the right track. If you want to solve using matrices you need to encode your elimination steps into a matrix, which will eventually lead you to the inverse of the original matrix.

Answer 3

Strictly speaking you're not using matrices.

Answer 4

Here's the matrix approach:

Answer 5

\[\left[\begin{matrix}1 & 2 \\ 1 & 1\end{matrix}\right]\left(\begin{matrix}x \\ y\end{matrix}\right)=\left(\begin{matrix}1 \\ 9\end{matrix}\right)\]

Answer 6

Finding the inverse of the first matrix is easy, because it's 2X2:

Answer 7

So is that the final answer?

Answer 8

\[\left[\begin{matrix}1 &2 \\ 1 & 1\end{matrix}\right]^{-1}=\left[\begin{matrix}1/\det &-2/\det \\ -1/\det & 1/\det\end{matrix}\right]\], where det is the determinant, which in this case is (1*1)-(2*1)=-1

Answer 9

\[\left[\begin{matrix}-1 & 2 \\ 1 & -1\end{matrix}\right]\left[\begin{matrix}1 &2 \\ 1 &1\end{matrix}\right]\] should equal the identity matrix.

Answer 10

It does, I just did the multiplication, you should check for yourself as well.

Answer 11

So thats my final answer? Or is there more?

Answer 12

Multiply both sides by the inverse:

Answer 13

(on the left, by the way)

Answer 14

\[\left[\begin{matrix}-1 & 2 \\ 1 & -1\end{matrix}\right]\left[\begin{matrix}1 & 2 \\ 1 & 1\end{matrix}\right]\left(\begin{matrix}x \\ y\end{matrix}\right)=\left[\begin{matrix}-1 & 2 \\ 1 & -1\end{matrix}\right]\left(\begin{matrix}9 \\ 1\end{matrix}\right)\]

Answer 15

Multiply them both out, the right (as we showed earlier) gives the identitiy matrix times (x,y), and the right gives their values.

Answer 16

So we have (-1*9+2*1)=x, (1*9+(-1*1))=y

Answer 17

So my final answer would be x = -7 and y = 8?

Answer 18

Yup, and you can test it, and it works.

Answer 19

\[\left[\begin{matrix}1 & 2 \\ 1 & 1\end{matrix}\right]\left(\begin{matrix}X \\ Y\end{matrix}\right)=\left(\begin{matrix}9 \\1\end{matrix}\right) \]
Or AX=B
\[\left| A \right|=\]|■(1&2@1&1)|=1×1-1×2=1-2=-1≠0
\[adj A=\left[\begin{matrix}1 & -1 \\ -2 & 1\end{matrix}\right]'=\left[\begin{matrix}1 & -2 \\-1 & 1\end{matrix}\right]\]
\[A ^{-1}=\frac{ adj A }{ \left| A \right| }\]
\[A ^{-1}=\frac{ \left[\begin{matrix}1 & -2 \\-1 & 1\end{matrix}\right] }{ -1 }\]
\[A ^{-1}=-\left[\begin{matrix}1 & -2 \\-1 & 1\end{matrix}\right]=\left[\begin{matrix}-1 & 2 \\ 1 & -1\end{matrix}\right]\]
\[X=A ^{-1}B\]
\[X=\left[\begin{matrix}-1 &2 \\ -1 & 1\end{matrix}\right]\left(\begin{matrix}9 \\1\end{matrix}\right)\]
\[X=\left(\begin{matrix}-1\times1+2\times1 \\ -1\times9+1\times1\end{matrix}\right)\]
\[X=\left(\begin{matrix}-1+2 \\ -9+1\end{matrix}\right)=\left(\begin{matrix}1 \\ -8\end{matrix}\right)\]
\[\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1 \\-8\end{matrix}\right)\]
x=1,y=-8

Answer 20

You forgot to multiply by 9 in the top right of your third-to-last equation.

Answer 21

sorry you are correct

Answer 22

\[X=A ^{-1}B\]
\[X=\left[\begin{matrix}-1 & 2\\ 1 & -1\end{matrix}\right]\left(\begin{matrix}9 \\ 1\end{matrix}\right)\]
\[\left(\begin{matrix}x \\ y\end{matrix}\right)=\left(\begin{matrix}-1\times9+2\times1 \\1\times9+\left( -1\times1 \right)\end{matrix}\right)\]
\[\left(\begin{matrix}x \\ y\end{matrix}\right)=\left(\begin{matrix}-7 \\ 8\end{matrix}\right)\]
x=-7,y=8