Question-10

Find the determinant of the matrix given below:

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

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

These are the operations:

• First we start with \(R_{1}\rightarrow R_{1}+R_{2}+R_{3}\).

• This produces a common factor \((a+b+c)\) in the first row which we are moving out.

• We now move to the columns. We perform \(C_{2}\rightarrow C_{2}-C_{1}\) followed by \(C_{3}\rightarrow C_{3}-C_{1}\).

• We now have enough number of zeros and we expand the determinant along the first row.

• The rest is just basic algebra.