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Is this matrix $$ M = \begin{bmatrix} a & -a & a \\[0.3em] -a & -a & -a \\[0.3em] a & a & a \end{bmatrix} $$ the same as: $$ M = a\begin{bmatrix} 1 & -1 & 1 \\[0.3em] -1 & -1 & -1 \\[0.3em] 1 & 1 & 1 \end{bmatrix} $$ or $$ M = a^3\begin{bmatrix} 1 & -1 & 1 \\[0.3em] -1 & -1 & -1 \\[0.3em] 1 & 1 & 1 \end{bmatrix} $$ ? what about determinants: $$ det \begin{bmatrix} a & -a & a \\[0.3em] -a & -a & -a \\[0.3em] a & a & a \end{bmatrix} $$ is it the same as: $$ a*det \begin{bmatrix} 1 & -1 & 1 \\[0.3em] -1 & -1 & -1 \\[0.3em] 1 & 1 & 1 \end{bmatrix} $$ or $$ a^3*det\begin{bmatrix} 1 & -1 & 1 \\[0.3em] -1 & -1 & -1 \\[0.3em] 1 & 1 & 1 \end{bmatrix} $$ ?

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for matrix $a$ is correct and determinant $a^3$ is correct. –  A.D May 8 '13 at 9:30

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up vote 0 down vote accepted

$$\begin{pmatrix} a & -a & a \\ -a & -a & -a \\ a & a & a\end{pmatrix} = a\begin{pmatrix} 1 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & 1 & 1\end{pmatrix}$$ and $$\begin{vmatrix} a & -a & a \\ -a & -a & -a \\ a & a & a\end{vmatrix} = a^3\begin{vmatrix} 1 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & 1 & 1\end{vmatrix}$$ although in this specific case all determinants are in fact equal since they are all zero.

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Maybe you can understand better the relation between determinants by observing that $$ \begin{bmatrix} a & -a & a \\ -a & -a & -a \\ a & a & a \end{bmatrix} = \begin{bmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{bmatrix}\, \begin{bmatrix} 1 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & 1 & 1 \end{bmatrix} $$

By definition, $aM$ means multiplying all entries of $M$ by $a$.

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In a nutshell, the determinat, seen as a function of the columns, is multilinear, i.e. let $A_1, \dots, A_n$ be the columns of the matrix $A$, then $\text{det}(A) = D(A_1, \dots, A_n)$, and $$D(A_1, \dots , \lambda A_i+\mu B,\dots, A_n) = \lambda \det(A) + \mu \det (M),$$ where $M$ is the matrix whose columns are $A_1, \dots, B, \dots, A_n$.

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