What is wrong with my matrix inversion?

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Homework Statement


Find the inverse of the matrix:

1 1 -1
2 -1 1
1 1 2

Homework Equations


One must be aware of the identity matrix, as well as how add one row to another with matrix multiplication, for example, the matrix

1 0 0
k 1 0
0 0 1

would add k times the first row to the second row.

The Attempt at a Solution



xFZKqU9.jpg


Which must be wrong, sadly enough.
 
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After adding -5/2 times the third to the second (step 5), the number in the middle should be 1-5/2, but you wrote -3/5.
 
k_squared said:

Homework Statement


Find the inverse of the matrix:

1 1 -1
2 -1 1
1 1 2

Homework Equations


One must be aware of the identity matrix, as well as how add one row to another with matrix multiplication, for example, the matrix

1 0 0
k 1 0
0 0 1

would add k times the first row to the second row.

The Attempt at a Solution



xFZKqU9.jpg


Which must be wrong, sadly enough.
You're not doing it the way I learned to do this. How I learned was to set up an augmented matrix with the matrix to invert on the left, and the identity matrix on the right, like so:
$$\begin{bmatrix} 1 &1 & -1 &| & 1 & 0 & 0\\
2 & -1 & 1 & | & 0 & 1 & 0\\
1 & 1 & 2 & | & 0 & 0 & 1 \end{bmatrix}$$
Now completely row-reduce the matrix on the left to get it to the identity matrix, and your inverse will be on the right. When I did it, there were no "half" entries. Most of the entries had denominators of 3.
 
First of is this the method you learned in class? Because this seems prone to errors.
You could try Gauss-Jordan Elimination it is essentially the same, but you have some "book keeping" abilities.

It is similar to Gaussian Elimination for systems of linear equations.

[Edit] Too late :)
 
Just search for simple linear combinations.
Math is about using your brains.
If your method does not work invent a better one or study.
 
my2cts said:
Just search for simple linear combinations.
Math is about using your brains.
If your method does not work invent a better one or study.
It's true that OP would have benefited from using some simpler combinations.

However, the method works perfectly well. Fredrik has pointed out where a computational error occurred and Mark has suggest a method with easier "book keeping".
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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