To evaluate the given determinant

Question

Question: Evaluate the determinant $\left| \begin{array}{cc} b^2c^2 & bc & b+c \\ c^2a^2 & ca & c+a \\ a^2b^2 & ab & a+b \\ \end{array} \right|$

My answer:

$\left| \begin{array}{cc} b^2c^2 & bc & b+c \\ c^2a^2 & ca & c+a \\ a^2b^2 & ab & a+b \\ \end{array} \right|= \left| \begin{array}{cc} b^2c^2 & bc & c \\ c^2a^2 & ca & a \\ a^2b^2 & ab & b \\ \end{array} \right| + \left| \begin{array}{cc} b^2c^2 & bc & b \\ c^2a^2 & ca & c \\ a^2b^2 & ab & a \\ \end{array} \right|= abc \left| \begin{array}{cc} bc^2 & c & 1 \\ ca^2 & a & 1 \\ ab^2 & b & 1 \\ \end{array} \right| +abc \left| \begin{array}{cc} b^2c & b & 1 \\ c^2a & c & 1 \\ a^2b & a & 1 \\ \end{array} \right|$

how do I proceed from here?

Answer 1

$F=\left| \begin{array}{cc} b^2c^2 & bc & b+c \\ c^2a^2 & ca & c+a \\ a^2b^2 & ab & a+b \\ \end{array} \right|$

$=\dfrac1{abc}\left| \begin{array}{cc} ab^2c^2 & abc & a(b+c) \\ c^2a^2b & bca & b(c+a) \\ a^2b^2c & abc & c(a+b) \\ \end{array} \right|$

$=\left| \begin{array}{cc} ab^2c^2 &1& a(b+c) \\ c^2a^2b &1& b(c+a) \\ a^2b^2c &1& c(a+b) \\ \end{array} \right|$

$R_3'=R_3-R_1,R_2'=R_2-R_1$

$F=\left| \begin{array}{cc} ab^2c^2 &1& a(b+c) \\ abc^2(a-b) &0& -a(a-b) \\ -ab^2c(c-a) &0& b(c-a) \\ \end{array} \right|$

$=(a-b)(c-a)\left| \begin{array}{cc} ab^2c^2 &1& a(b+c) \\ abc^2 &0& -a \\ -ab^2c &0& b \\ \end{array} \right|$

$=(a-b)(c-a)(-1)^{1+2}\cdot\left| \begin{array}{cc} abc^2 & -a \\ -ab^2c & b \\ \end{array} \right|$

Can you take it from here?

Answer 2

Your given matrix is : $\left| \begin{array}{cc} b^2c^2 & bc & b+c \\ c^2a^2 & ca & c+a \\ a^2b^2 & ab & a+b \\ \end{array} \right|$

Determinant of the given matrix is :

$\implies b^2c^2[ca^2 + abc - abc + a^2b] - bc[a^3c^2 + a^2bc^2 - a^2b^2c - a^3b^2] + (b+c)[a^3bc^2 - a^3b^2c]$

$\implies b^2c^2[ca^2 + a^2b] - bc[a^3c^2 + a^2bc^2 - a^2b^2c - a^3b^2] + (b+c)[a^3bc^2 - a^3b^2c]$

$\implies a^2b^2c^3 - a^2b^3c^2 - a^3bc^3 - a^2b^2c^3 + a^2b^3c^2 + a^3b^3c + a^3b^2c^2 - a^3b^3c + a^3bc^3 - a^3b^2c^2$

$\implies a^2b^2c^3 - a^2b^2c^3 - a^2b^3c^2 + a^2b^3c^2 - a^3bc^3 + a^3bc^3 + a^3b^3c - a^3b^3c + a^3b^2c^2 - a^3b^2c^2$

$\implies 0$

Therefore, given matrix is singular.

Answer 3

This is probably the shortest way:- $$ \begin{align} \begin{vmatrix} b^2c^2 & bc & b+c \\\ c^2a^2 & ca & c+a \\\ a^2b^2 & ab & a+b \end{vmatrix} &=a^3b^3c^3\begin{vmatrix}a^{-1} &1 & b^{-1}+c^{-1}\\\ b^{-1} &1&a^{-1}+c^{-1}\\\ c^{-1} &1& a^{-1}+b^{-1}\end{vmatrix}\\ &=(abc)^3\left(\frac 1a+\frac 1b+\frac 1c\right)\begin{vmatrix} a^{-1}&1&1\\\ b^{-1}&1&1\\\ c^{-1} &1 &1 \end{vmatrix} \\ &=0 \end{align} $$ What I did was to first take out $bc$ , $ca$ and $ab$ common from the rows individually. Then I took out $abc$ from the first column.