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I am given matrix A, and asked to find its inverse, and then using my answer to that, solve a similar linear system.

(a) $$ A = \begin{bmatrix} 2 & 2 & -1 \\ -4 & -5 & 7 \\ 1 & 1 & -1 \\ \end{bmatrix} $$ Finding the inverse gave me: $$ A^{-1} = \begin{bmatrix} -2 & 1 & 9 \\ 3 & -1 & -10 \\ 1 & 0 & -2 \\ \end{bmatrix} $$

Then the problem says:

(b) Use the answer from part (a) to solve the linear system

$$2x_1+2x_2-x_3=2$$ $$-4x_1-5x_2+7x_3=4$$ $$x_1+x_2-x_3=-4$$

I could solve this system using Gauss-Jordan Elimination, but we don't get full credit unless we solve the problem the way it specifies, and I just don't see how to use the inverse of A to solve for x1, x2, and x3. Any help to send me in the right direction would be appreciated!

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    $\begingroup$ Hint: $$A^{-1}AX = X = A^{-1}b$$ $\endgroup$ – Moo Mar 7 '18 at 6:00
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    $\begingroup$ @Moo Thanks for the hint! It put me on the right path to find out what I needed to do. $\endgroup$ – Drew Pesall Mar 7 '18 at 6:26
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Thanks to @Moo for the hint to set me on the right path! Once I realized what I needed to do, I set up the matrix equality of $$ \begin{bmatrix} -2 & 1 & 9 \\ 3 & -1 & -10 \\ 1 & 0 & -2 \\ \end{bmatrix}* \begin{bmatrix} 2\\ 4\\ -4\\ \end{bmatrix}= \begin{bmatrix} x_1\\ x_2\\ x_3\\ \end{bmatrix} $$

And when I multiplied the matricies I found the values: $$x_1=-36$$ $$x_2=42$$ $$x_3=10$$

Which were my expected answers I got from using Gauss-Jordan Elimination.

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