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jernobyl
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Homework Statement
Express the following equations in the matrix form AX=B. Determine the inverse of A and hence solve the equations.
[tex]4x-3y+z=11[/tex]
[tex]2x+y-4z=-1[/tex]
[tex]x+2y-2z=1[/tex]
Homework Equations
Not really applicable.
The Attempt at a Solution
A bit of background, this is for a piece of coursework when we were just starting to learn about matrices and their uses, which deals with solving groups of equations by changing them into matrix form and solving them that way. So it's not very advanced. It's just annoying.
Basically I've done the entirety of the problem, I come to the end and the answers aren't right. It's a Hellishly long process and I've done it five times now and each time I get a different incorrect answer and it's driving me up the wall. I just KNOW I've got ONE sign wrong somewhere and it's throwing the whole damn thing off. I hope someone can point out where I've gone wrong.
Oh, the answer is meant to come out, [tex]x = 3, y = 1, z = 2[/tex]; I've been getting weird fractional answers. Anyway.
Right, so changing the equations into AX=B form:
[tex]\left(\begin{array}{ccc}4&-3&1\\2&1&-4\\1&2&-2\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right) = \left(\begin{array}{c}11\\-1\\1\end{array}\right)[/tex]
So we know that:
[tex]X=A^{-1}B[/tex]
So my next aim is to calculate [tex]A^{-1}[/tex].
[tex]A^{-1}=\frac{1}{det(A)}adj(A)[/tex]
[tex]det(A) = (4)\left|\begin{array}{cc}1&-4\\2&-2\end{array}\right|-(-3)\left|\begin{array}{cc}2&-4\\1&-2\end{array}\right|+(1)\left|\begin{array}{cc}2&1\\1&2\end{array}\right|[/tex]
[tex]=(4)(6)-(-3)(0)-(1)(3)[/tex]
[tex]=24-0-3[/tex]
[tex]=21[/tex]
So that's fair enough. The determinant is 21. God knows if it actually is or not but following my calculations that's what it is. Now, the matrix of cofactors:
[tex]C = \left(\begin{array}{ccc}\left|\begin{array}{cc}1&-4\\2&-2\end{array}\right|&-\left|\begin{array}{cc}2&-4\\1&-2\end{array}\right|&\left|\begin{array}{cc}2&1\\1&2\end{array}\right|\\\\-\left|\begin{array}{cc}-3&1\\2&-2\end{array}\right|&\left|\begin{array}{cc}4&1\\1&-2\end{array}\right|&-\left|\begin{array}{cc}4&-3\\1&2\end{array}\right|\\\\\left|\begin{array}{cc}-3&1\\1&-4\end{array}\right|&-\left|\begin{array}{cc}4&1\\2&-4\end{array}\right|&\left|\begin{array}{cc}4&-3\\2&1\end{array}\right|\end{array}\right) = \left(\begin{array}{ccc}6&0&3\\-4&-9&-11\\11&18&10\end{array}\right)[/tex]
[tex]adj(A)=C^T=\left(\begin{array}{ccc}6&-4&11\\0&-9&18\\3&-11&10\end{array}\right)[/tex]
[tex]A^{-1} = \frac{1}{21}\left(\begin{array}{ccc}6&-4&11\\0&-9&18\\3&-11&10\end{array}\right) = \left(\begin{array}{ccc}\frac{2}{7}&-\frac{4}{21}&\frac{11}{21}\\\\0&-\frac{3}{7}&\frac{6}{7}\\\\\frac{1}{7}&-\frac{11}{21}&\frac{10}{21}\end{array}\right)[/tex]
[tex]X=A^{-1}B = \left(\begin{array}{ccc}\frac{2}{7}&-\frac{4}{21}&\frac{11}{21}\\\\0&-\frac{3}{7}&\frac{6}{7}\\\\\frac{1}{7}&-\frac{11}{21}&\frac{10}{21}\end{array}\right)\left(\begin{array}{c}11\\-1\\1\end{array}\right)[/tex]
Which actually gives you the answers [tex]x=\frac{27}{7}, y=\frac{9}{7}, z=\frac{58}{21}[/tex]...none of which are right, obviously.
I know there's a lot of information above but I could really do with someone with a sharper eye than mine (bear in mind it's now nearly 6am and I've been up all night!) to find where I've made my little error leading to a grand finale of chaos! Many thanks to anyone who looks at this! I'm off to my bed for now!
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