# Am I misinterpreting this matrix determinant property?

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?

• How exactly do you define multiplication of a matrix by$~2$? It looks more like you only multiplied part of the matrix by$~2$. – Marc van Leeuwen Apr 24 '15 at 20:07
• 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 – Dodgie Apr 25 '15 at 1:34
• Don't mind the question mark. Impossible to edit on the phone :( – Dodgie Apr 25 '15 at 1:41

Multiplying a matrix by a scalar $c$ amounts to multiplying each entry by $c$, not just the first row.

• 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. $$\begin{vmatrix} a & kb & c \\ x & ky & z \\ p & kq & r \\ \end{vmatrix} = k\begin{vmatrix} a & b & c \\ x & y & z \\ p & q & r \\ \end{vmatrix} = \begin{vmatrix} ka & kb & kc \\ x & y & z \\ p & q & r \\ \end{vmatrix}$$ – anir123 Apr 24 '15 at 17:34
• @Mahesha999 The vertical bars represent the determinant. It is not a matrix and is probably the cause of your confusion. – JessicaK Apr 24 '15 at 17:42
• @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 – anir123 Apr 24 '15 at 20:40
• @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. – Ilmari Karonen Apr 24 '15 at 21:03

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

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