If A is a square matrix of order n and m ϵ R then the det(mA) = (m^{n}) (detA)
Well, observe that for any scalar $\displaystyle m \in R $ and nxn matrix A.
$\displaystyle mA = A*mI $ where I is the standard nxn identity matrix. So you got $\displaystyle mI = \begin{bmatrix} m && 0 && ... && 0 \\ 0 && m && ... && 0 \\ .. && .. && .. && .. \\ 0 && 0 && ... && m \end{bmatrix} $. So you know the trick for determinats when multiplying matricies, Det(AB) = Det(A)Det(B). Well, the determinant of any matrix with entries only in the diagonals and zeroes everywhere is just the whole diagonal multiplied, since there are n diagonal enteries (all of them m), $\displaystyle Det(mI) = m*m*m.... = m^n $
so you got $\displaystyle det(mA) = det(A) * det(mI) = det(A) * m^n $
Alternatively, it is easy to see that, for a 2 by 2 matrix, $\displaystyle \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ has determinant ad- bc. Multiplying by m gives matrix $\displaystyle \begin{bmatrix}ma & mb \\ mc & md\end{bmatrix}$ which has determinant $\displaystyle (ma)(md)- (mb)(mc)= m^2d- m^2bc= m^2(ad- bc)$. Now use the idea of expanding a determinant by columns to do a proof by induction on the size of the matrix.
Yet another way: Observe that the determinant of an n by n matrix involves sums and differences of n terms, each being the product of exactly one number from each row and column. With n rows and columns, each term will be a product of n numbers so each term in the determinant of m times the matrix will have a product of n numbers and so will have a factor of $\displaystyle m^n$.
Sorry i can't solve it but i am here for learn.
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let's prove something simpler, first:
$\displaystyle \begin{vmatrix}1&0&\dots&0&\dots&0\\0&1&\dots&0& \dots&0\\ \vdots&\vdots&\ddots&\vdots&\cdots&\vdots\\0&0& \dots &m&\dots&0\\ \vdots&\vdots&\cdots&\vdots&\ddots&\vdots\\0&0& \dots&0&\dots&1 \end{vmatrix} = m$
but this is a diagonal matrix, which has determinant: (1)(1)...(m)....(1) = m.
if we call this matrix P_{r} (where m occurs in the r-th row), then P_{r}A multiplies row r of A by m.
hence mA = P_{1}P_{2}...P_{n}A, so that:
det(mA) = det(P_{1})det(P_{2})...det(P_{n})det(A) = (m)(m)....(m)(det(A)) = m^{n}(det(A))
Here's a link:
Determinant with Row Multiplied by Constant - ProofWiki
This page shows all but one step of the proof you want, in formal language.
In general, proofwiki is a good place to look for proofs.
BTW, the idea of a determinant as a cofactor expansion was mysterious to me for a long time. I could do the expansions, but I didn't feel I understood why it worked, though I was shy about saying so. I've found that many other people are in the same boat. Suggestion: Google "Geometric Algebra Primer" by Jaap Suter, and/or watch the first few eps of Norman Wildberger's "Wild Lin Alg," which is on Youtube.
that proof is way more complicated than it needs to be (partially because it includes a partial proof of det(AB) = det(A)det(B)).
geometrically, what is happening is this:
multiplying ONE row of A by m is the same as stretching one SIDE of the n-dimensional volume element by a factor of m (which magnifies the volume element by a factor of m). if we do this for every side, we've stretched by a factor of m^{n}
(it's easiest to comprehend this when n = 2 or 3).
There is another way to do it: Schur decomposition. Schur's theorem says that any square matrix A is unitarily equivalent to an upper-triangular matrix--i.e., A = U*TU, where U is unitary and T is upper-triangular. A and T have the same determinant (it's a fairly easy proof to show this, if needed), and since the determinant of a triangular matrix (upper or lower) equals the product of the main diagonal, multiplying each entry by m will multiply the overall determinant by m^n.