My Math Forum Matrix inverse problem

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 October 13th, 2012, 06:30 PM #1 Newbie   Joined: Oct 2012 Posts: 8 Thanks: 0 Matrix inverse problem Using the fact that the inverse of [(1 2 -1), ( 2 3 1), ( 1 2 1) ] is [(-8 2 5), (5 -1 -3), (1 0 -1)] solve the following system of equations. a) x + 2y - z = 1 2x + 3y + z = 0 1x + 2y -2z = 2 x = 2, y = -1, z = -1 b) -8x + 5y +z = 3 2x -y = 1 5x -3y -z = 6 How do I do part b?
 October 13th, 2012, 09:07 PM #2 Newbie   Joined: Oct 2012 Posts: 8 Thanks: 0 Re: Matrix inverse problem Dont worry I worked it out
 October 15th, 2012, 10:50 AM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Matrix inverse problem Great! For the benefit of others who are interested in this: We can write that system of equations as a matrix equation: AX= B where $A= \begin{bmatrix}1 &2=&-1 \\ 2=&3=&1 \\ 1=&2=&1\end{bmatrix}=$, the first matrix you mention, $X= \begin{bmatrix}x \\ y \\ z\end{bmatrix}$ and $B= \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}$ for the first problem, $\begin{bmatrix} 3 \\ 1 \\ 6\end{bmatrix}$ for the second. Multiplying on both sides by the inverse of A, $A^{-1}Ax= x= A^{-1}B$ so, since you are given the inverse matrix to A, solving for x is just a matter of matrix multiplication. I bet your next problem set will require you to find the inverse matrix yourself! (However, having checked it, I need to point out that the inverse of $\begin{bmatrix}1 & 2 & -1 \\ 2 & 3 & 1 \\ 1 & 2 & 1\end{bmatrix}$ is NOT $\begin{bmatrix}-8 & 2 & 5 \\ 5 & -1 & -3 \\ 1 & 0 & -1\end{bmatrix}$)

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