Let $A=\begin{bmatrix}2&-1&2&0\\-3&0&2&1\\-4&-1&-2&-1\end{bmatrix}$.
I need to solve find $X$ in $AX=B$, where $B=\begin{bmatrix}1\\-1\\0\end{bmatrix}$. How am I supposed to do this with $X=A^{-1}B$, as I cannot inverse A since it is not square?
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1 upvote for this but http://math.stackexchange.com/questions/1433994/definition-of-a-manifold not this? Grr this site! Also is it under or over constrained? – Alec Teal Sep 13 '15 at 22:58
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4 unknowns and 3 equations... – null Sep 13 '15 at 23:00
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Form the augmented matrix $[A\mid b]$, row-reduce, and go from there. – paw88789 Sep 13 '15 at 23:06
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What does "solved" mean? – DanielV Sep 13 '15 at 23:20
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https://en.wikipedia.org/wiki/Gaussian_elimination This is more fundamental than taking inverses of square matrices. – Greg Martin Sep 13 '15 at 23:29
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Your matrix equation
$$\begin{bmatrix}2&-1&2&0\\-3&0&2&1\\-4&-1&-2&-1\end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix}=\begin{bmatrix}1\\-1\\0\end{bmatrix} $$ is the same as the following system of linear equations $$\begin{matrix} \ 2x_1&-x_2&+2x_3 &\ \ =&1\\ -3x_1&&+2x_3 &+x_4=&-1\\ -4x_1&-x_2&-2x_3 &-x_4=&0. \end{matrix}$$
The number of equations is one less than the number of unknowns. So, we have to choose one of the unknowns to be the parameter of the solutions. Let $u=x_1$ and consider $u$ as a known quantity.
With this we have the following system of equations:
$$\begin{matrix} -x_2 +&2x_3 &\ \ =&1-2u\\ &2x_3 &+x_4=&3u-1\\ -x_2-&2x_3 &-x_4=&4u \end{matrix} \tag 1$$
which is equivalent to the following matrix equation
$$\begin{bmatrix} -1 &2 &0\\ 0&2&1\\ -1&-2&-1 \end{bmatrix} \begin{bmatrix} x_2\\ x_3\\ x_4 \end{bmatrix}= \begin{bmatrix} 1-2u\\ 3u-1\\ 4u \end{bmatrix} $$
You can invert the $3\times 3$ matrix above if you like inverting matrices.
I would rather solve $(1)$ directly, though. For example we would immediately get
$$x_2=1-7u$$ by adding the second and the third equation of $(1)$.
zoli
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