Question-4

Consider the matrix:

\[ \begin{bmatrix}a & b & c\\ b & c & a\\ c & a & b \end{bmatrix} \]

If \(a+b+c\) is divisible by \(6\), is \(\text{det}(A)\) also divisible by \(6\)?

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We can compute the determinant as follows:

\[ \begin{aligned} \begin{vmatrix}a & b & c\\ b & c & a\\ c & a & b \end{vmatrix} & =\begin{vmatrix}a+b+c & a+b+c & a+b+c\\ b & c & a\\ c & a & b \end{vmatrix}\\ & =(a+b+c)\begin{vmatrix}1 & 1 & 1\\ b & c & a\\ c & a & b \end{vmatrix}\\ & =(a+b+c)\begin{vmatrix}1 & 0 & 0\\ b & c-b & a-b\\ c & a-c & b-c \end{vmatrix}\\ & =(a+b+c)(ab+bc+ca-a^{2}-b^{2}-c^{2}) \end{aligned} \]

The determinant of the matrix is divisible by \(a+b+c\). Therefore, it is also divisible by \(6\).