Can a 3x3 matrix equate to a 4x4 matrix?

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Homework Statement



My question involves finding the determinant of a 4x4 matrix, which I know how to do.

The matrix is

0 1 2 3
0 1 2 5
0 3 5 6
0 0 0 0

Since the matrix has zeros in every row and column, its determinant will equal 0.

However, I got to thinking that if I wrote out the matrix in equation form, I'd get

0a + b + 2c + 3d
0a + b + 2c + 5d
0a + 3b + 5c +6d
0 + 0 + 0 + 0

Does the 4x4 matrix equate to the 3x3 matrix

1 2 3
1 2 5
3 5 6 ?


Homework Equations



(see above for equations)


The Attempt at a Solution



I'm incline to say no, since the 3x3 matrix would have a determinant, but from the perspective of the equations, I don't see how that are different (at least in terms of the information they convey).

Can anyone explain to me how the two matrices are different? Thanks.

 
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Darkmisc said:

Homework Statement



My question involves finding the determinant of a 4x4 matrix, which I know how to do.

The matrix is

0 1 2 3
0 1 2 5
0 3 5 6
0 0 0 0

Since the matrix has zeros in every row and column, its determinant will equal 0.

However, I got to thinking that if I wrote out the matrix in equation form, I'd get

0a + b + 2c + 3d
0a + b + 2c + 5d
0a + 3b + 5c +6d
0 + 0 + 0 + 0
These are not equations; they are expressions that represent the product of your 4x4 matrix and a column vector [a b c d]. They are not equations because there is no equals sign.
Darkmisc said:
Does the 4x4 matrix equate to the 3x3 matrix
No. An n x n matrix can never be equal to an m x m matrix if n is different from m. The two matrices belong to different spaces, so cannot be compared. The same is true for vectors from different spaces.
Darkmisc said:
1 2 3
1 2 5
3 5 6 ?


Homework Equations



(see above for equations)


The Attempt at a Solution



I'm incline to say no, since the 3x3 matrix would have a determinant, but from the perspective of the equations, I don't see how that are different (at least in terms of the information they convey).

Can anyone explain to me how the two matrices are different? Thanks.
The two matrices are different because they have different numbers of elements.
 
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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