When two rows/columns are equal, determinant is equal to zero

Question

By using induction, we can set our determinant to be n=2, then have have

[A]= \begin{vmatrix} a & b \\ a & b \end{vmatrix} = ab - ab=0

So we assume it works. Now we can work out for determinants of order n. A can be determinant of order n+1, with two equal rows, let us say i-th and j-th. If we expand the determinant by some k-th row, where k≠i, k≠j, therefore we can have a sum by having two part of a sum for n+1, where we have part for n and another for +1, and as we assumed it works for n, ergo the determinant will be 0.

But now what is my question, that I cannot fully prove the upper theorem by using the permutations.

Answer 1

Let [A]= \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}

if you have two (or more) rows which are linearly related, you can resolve some rows to zero. (Check out gaussian elimination to see how row reduction works.)

The determinant of a 3X3 matrix is given by $$ a_{11} \times a_{22} \times a_{33} \\ + a_{12} \times a_{23} \times a_{31} \\ + a_{21} \times a_{32} \times a_{13} \\ - a_{13} \times a_{22} \times a_{31} \\ - a_{12}\times a_{21} \times a_{33} \\ - a_{11} \times a_{23} \times a_{32} $$

Now, say row two and row three are the same value, we can subtract the element values of row two from row three, replacing row three instead with zeros. Look at the definition of the determinant of a 3x3 matrix again, replacing the $a_{3x}$ values with zero. Notice how you essentially multiply every row by a zero.

I hope this helps.

Answer 2

If the rows of a matrix are not linearly independent (can be expressed as linear combination of the other rows of the matrix) then the determinant is 0.

One interpretation of the determinant is how it dilates (or compresses) space after a transformation. If two rows are equal, then the principal components of space are being mapped onto a single line. 2D space is compressing onto 1D space, and the area of a line equals 0.

Or in a $3\times 3$ matrix 3D spaces is compressing onto a 2D plane and the volume is $0.$

Another attack, if you have two matrices, the determinant of the product equals the product of the determinants $det(AB) = det(A)det(B)$

If B is your target matrix, and A is a matrix that subtracts one row from another.

$A = \begin{bmatrix} 1&0&0\\0&1&-1\\0&0&1\end{bmatrix}$ would be one such matrix that subtracts the 3rd row of B from the second. If they are the same that would create a row of zero vectors. That would mean that $det(AB) = 0, det (A)$ is non-zero. So $det(B)$ must be $0.$