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My professor found the cubic roots of a 3x3 matrix by doing the following. I don't understand how step 2 came about and why he applied the same for step 4 on row 1 instead of row 2.

Step 1:

$\begin{bmatrix} a & 2 & 2 \\ 2 & a & 2 \\ 2 & 2 & a \end{bmatrix}$

Step 2:

$=\begin{bmatrix} a & 2 & 2 \\ 2 & a & 2 \\ 0 & 2-a & a-2 \end{bmatrix}$

Step 3:

$=(a-2) \begin{bmatrix} a & 2 & 2 \\ 2 & a & 2 \\ 0 & -1 & 1 \end{bmatrix}$

Step 4:

$=(a-2)^2\begin{bmatrix} 1 & -1 & 0 \\ 2 & a & 2 \\ 0 & -1 & 1 \end{bmatrix}$

He then solved for the determinant with the remaining terms. The transition from step 1 to 2 is what is confusing me the most at the moment. I was under the impression that there was no easy way to find the cubic roots within a matrix.

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What are "the cubic roots within a matrix"? Do you mean the zeros of the cubic characteristic polynomial of a $3\times3$ matrix? Also, it seems that your formulas refer to determinants; the usual notation for determinants is with vertical bars, not with square brackets, which are usually used for the matrices themselves. –  joriki Dec 19 '12 at 13:29
    
Yes, that is what I mean. I am aware that he simply solves for the determinant, but I do not understand how he was able to transition from step 1 to 2. Why was he able to substitute row 3 with $(0, 2-a, a-2)$? –  Goyatuzo Dec 19 '12 at 13:37

1 Answer 1

up vote 3 down vote accepted

Row 2 was subtracted from row 3. This doesn't change determinant.

Between step 3 and step 4, row 2 was subtracted from row 1, and another $(a-2)$ factored out. So the determinant is the same.

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Well I feel extremely stupid now. Thanks a lot! –  Goyatuzo Dec 19 '12 at 13:48
    
Don't feel bad about that --- sometimes it takes some staring at things to see what step was done, especially if prof. didn't say what the step was! –  coffeemath Dec 19 '12 at 13:51

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