Can I extract common factor from a column in matrix?

Question

Can I do the following for matrices:

$$ A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} == 2^{1/3}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = 2^{1/3}B \quad ??$$

I know I can do the following for matrices:

$$ If \quad A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \quad then \quad 2A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} $$

and following for determinants:

$$ |A| = \begin{vmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} == 2\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 2|B| \quad \checkmark\checkmark$$

Answer 1

Nope! $2^{1/3}A$ means you multiply every cell of $A$ by $2^{1/3}$. That's because $2^{1/3}$ is a number ($\sqrt[3]{2}$, in fact), just as 2.

Of course you could come with a specific system of notation which indicates that you will multiply a certain row or column (this should be clear two) by a certain number... I don't know, maybe $$2^{]r1[}\cdot A$$ (I'm just making something up). If you define something like that, and it's not ambiguous, and you state clearly the definition... well you can. But just because you say so.

BTW, I don't remember any notation like that... but who knows.

Regarding determinants... you can do it cause it's a property of determinants. I mean, it's a theorem, it has been proved. To say it briefly, when you calculate a determinant of an $n\times n$ matrix, you end up adding and substracting several terms ($n!$ terms in fact), each one being a product of $n$ elements of the matrix: but they're are always organized in a way such that there is exactly one element of each column and, at the same time, one element of each row.

Then, if you multiply the 2nd column of $A$ by $4$ you won't get the matrix $4A$, but it is true that the determinant of this new matrix will be $4$ times the determinant of A, since the effect of that multiplication was to insert a factor $4$ in each of the $n!$ terms you add and substract to calculate the determinant.

For instance, if $A$ is a $3\times 3$ matrix, then $$\det(A)= a_{11}a_{22}a_{33}+a_{21}a_{32}a_{13}+a_{31}a_{12}a_{23}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}$$ (check that there are in fact $3!=6$ terms and that for each term every row [first index] and every column [second index] appears once and only once.)

So if you now create matrix $B$ by multiplying by $\lambda$ say the third row of $A$ (this affects $a_{31}$, $a_{32}$ and $a_{33}$), then $$\det(B)= a_{11}a_{22}(\lambda a_{33})+a_{21}(\lambda a_{32})a_{13}+(\lambda a_{31})a_{12}a_{23}-a_{13}a_{22}(\lambda a_{31})-a_{11}a_{23}(\lambda a_{32})-a_{12}a_{21}(\lambda a_{33})=\lambda(a_{11}a_{22}a_{33}+a_{21}a_{32}a_{13}+a_{31}a_{12}a_{23}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33})=\lambda \det(A).$$

Answer 2

The important thing to note here is that a determinant is not a matrix, it is a number.

If you calculate the two determinants in your third example you get $$2=2\times1$$ which is obviously correct. Later: expanding in response to comment: the matrices $$\left[\matrix{2&0&0\cr0&1&0\cr0&0&1\cr}\right]\quad\hbox{and}\quad 2\left[\matrix{1&0&0\cr0&1&0\cr0&0&1\cr}\right]$$ are not the same, but the numbers $$\left|\matrix{2&0&0\cr0&1&0\cr0&0&1\cr}\right|\quad\hbox{and}\quad 2\left|\matrix{1&0&0\cr0&1&0\cr0&0&1\cr}\right|$$ are the same.

Multiplying a matrix by a number means multiplying every entry of the matrix by that number. This is why your second example is correct, and it also shows that the answer to your initial question is "no".