Matrices - finding a general solution.

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The discussion focuses on finding the general solution for a matrix that does not have an inverse. The suggested method involves row reduction of the augmented matrix, leading to a final row of zeros. This indicates that the system can be expressed in terms of a free variable, specifically z. The resulting general solution is (x, y, z) = (z, z, z), where z can be any real number. This represents a line in three-dimensional space where x, y, and z are equal.
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Howdy,

I have been asked to find the general solution of the following matrix (pic attached).

The matrix does not have an inverse, so I am a bit confused guys. Cheers and thanks in advance!
 

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Yes, it does not have an inverse- that's why you are asked for the general solution.

One way to do this is "row reduction". Set up the augmented matrix
\left [ \begin{array} {cccc}1 & 0& -1 & 0 \\0 & -2 & 2 & 0 \\-1 & 1 & 0 & 0 \end{array} \right ]
and row-reduce. Because the matrix does not have an inverse, the final row will be all zero's but you could solve for, say x and y in terms of z.

Or just treat it as a system of equations: x- z= 0, -2y- 2z= 0, -x+ y= 0.
The first and third just say x= z and y= x= z. The second is then automatically solved. The general solution is (x, y, z)= (z, z, z)where z can be any number.
 
yiou are given a linear map from 3 space to 3 space and are asked to find all vectors that map to 0. obviously it is the line defined by x=y=z.
 
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