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I was reading matrix determinant properties from wikipedia.

The property reads $\det(cA) = c^n \det(A)$ for $n \times n$ matrix.

However I am not able to realize it. What I find is $\det(cA) = c\det(A)$

For example, multiplying matrix by 2 and then taking the determinant of the resultant matrix:

$ 2\begin{bmatrix} 4 & 5 & 6 \\ 6 & 5 & 4 \\ 4 & 6 & 5 \\ \end{bmatrix}= \begin{bmatrix} 8 & 10 & 12 \\ 6 & 5 & 4 \\ 4 & 6 & 5 \end{bmatrix} $ and $ \begin{vmatrix} 8 & 10 & 12 \\ 6 & 5 & 4 \\ 4 & 6 & 5 \end{vmatrix}=60 $

Now first taking the determinant and then multiplying by 2 yields the same result:

$$ 2\begin{vmatrix} 4 & 5 & 6 \\ 6 & 5 & 4 \\ 4 & 6 & 5 \\ \end{vmatrix} = 2 \cdot 30 = 60 $$

Where I am mistaking?

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    $\begingroup$ How exactly do you define multiplication of a matrix by$~2$? It looks more like you only multiplied part of the matrix by$~2$. $\endgroup$ – Marc van Leeuwen Apr 24 '15 at 20:07
  • $\begingroup$ Imagine the scalar multiplication of a siingle row as the scaling of a dimension of the box the determinant respresents the volume of in n-space? If the matrix is order two, then the determinant of this matrix is equivalent to the area of the parallelogram the vectors of this matrix comprise on the plane. Scaling a sinle dimension will scale the area $\endgroup$ – Dodgie Apr 25 '15 at 1:34
  • $\begingroup$ Don't mind the question mark. Impossible to edit on the phone :( $\endgroup$ – Dodgie Apr 25 '15 at 1:41
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Multiplying a matrix by a scalar $c$ amounts to multiplying each entry by $c$, not just the first row.

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  • $\begingroup$ Ohh damn stupid I am, I should delete this question, but I read below property somewhere, which implies so, is it wrong then? (Unfortunately I noted down this in my book, have not noted its source. $$ \begin{vmatrix} a & kb & c \\ x & ky & z \\ p & kq & r \\ \end{vmatrix} = k\begin{vmatrix} a & b & c \\ x & y & z \\ p & q & r \\ \end{vmatrix} = \begin{vmatrix} ka & kb & kc \\ x & y & z \\ p & q & r \\ \end{vmatrix} $$ $\endgroup$ – anir123 Apr 24 '15 at 17:34
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    $\begingroup$ @Mahesha999 The vertical bars represent the determinant. It is not a matrix and is probably the cause of your confusion. $\endgroup$ – JessicaK Apr 24 '15 at 17:42
  • $\begingroup$ @JessicaK thanks to point to that...so its just that multiplying matrix by scalar multiplies all elements, but multiplying determinant by scalar multiplies elements of only one row / column $\endgroup$ – anir123 Apr 24 '15 at 20:40
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    $\begingroup$ @Mahesha999: It's just saying that if you multiply one row or column of a matrix by a constant $k$, the determinant of the resulting matrix will be $k$ times that of the original. $\endgroup$ – Ilmari Karonen Apr 24 '15 at 21:03
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You have $\det (cB) = \det (cI) \det B $ and you can see from the formula for $\det$ that $\det (cI) = c^n$.

Another way is to notice that $\det$ is a multilinear function of the columns (or rows), that is, we can write $\det(A) = f(a_1,...,a_n)$ where $a_k$ is the $k$th row of $A$, and $f$ is linear in each argument separately.

Then $\det(cA) = f(c a_1,...,c a_n) = c f(a_1, c a_2,...,c a_n) = c^2 f(a_1, a_2,...,c a_n)= \cdots = c^n f(a_1,...,a_n) = c^n \det A$.

Addendum:

You multiplied the matrix incorrectly.

$A=\begin{bmatrix} 4 & 5 & 6 \\ 6 & 5 & 4 \\ 4 & 6 & 5 \end{bmatrix}$, $2A = \begin{bmatrix} 8 & 10 & 12 \\ 12 & 10 & 8 \\ 8 & 12 & 10 \end{bmatrix}$. $\det A = 30, \det (2A) = 240$.

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  • $\begingroup$ Though true, I don't think you really answered the question. I guess OP found things like this at WP already (and says so). $\endgroup$ – Marc van Leeuwen Apr 24 '15 at 20:11
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$$2 \begin{bmatrix} 4 & 5 & 6 \\ 6 & 5 & 4 \\ 4 & 6 & 5 \end{bmatrix} = \begin{bmatrix} 8 & 10 & 12 \\ 12 & 10 & 8 \\ 8 & 12 & 10 \end{bmatrix}.$$

You only multiplied the first row by 2.

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There are at least two ways to see that the answer is $c^n \det A$. One is by using the full expansion formula and noting that each term in the full expansion is a product of $n$ matrix entries and each matrix entry has multiplied by $c$. There is also a geometric explanation: The determinant gives the (signed) volume of the parallelopiped spanned by the columns of your matrix. If you dialate by a positive factor $c$, then the parallelopied is dialated by a factor of $c$ in each dimension, so the volume is dialated by a factor of $c^n$.

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