Browse Questions

# The determinant $\begin{vmatrix}b^2 - ab & b - c & bc - ac\\ab - a^2 & a - b & b^2 - ab\\bc - ac & c - a & ab -a^2\end{vmatrix}$ equals

$\begin{array}{1 1} abc(b-c)(c-a)(a-b) \\ (b-c)(c-a)(a-b) \\ (a+b+c)(b-c)(a-b) \\ None\;of\;these\end{array}$

Toolbox:
• If $A=\begin{vmatrix}a_{11} &a_{21} & a_{31}\\a_{12} &a_{22} &a_{32}\\a_{13} &a_{23} &a_{33}\end{vmatrix}$
• $|A|=a_{11}(a_{22}\times a_{33}-a_{32}\times a_{23})-a_{21}(a_{12}\times a_{33}-a_{32}\times a_{13})+a_{31}(a_{12}\times a_{23}-a_{22}\times a_{13})$
• If A is a square matrix such that each element of a row(or column)of A can be expressed as the sum of two or more terms.
• If k is multiplied by a constant k,then its determinant value is k|A|.
Let $\Delta=\begin{vmatrix}b^2-ab & b-c & bc-ac\\ab-a^2 & a-b & b^2-ab\\bc-ac & c-a &ab-a^2\end{vmatrix}$
This can be written as
$\Delta=\begin{vmatrix}b(b-a) & b-c &c(b-a)\\a(b-a) & a-b & b(b-a)\\c(b-a) & c-a & a(b-a)\end{vmatrix}$
Let us (b-a) as a common factor from $C_1$ and $C_2$
$\Delta=(b-a)^2\begin{vmatrix}b & b-c & c\\a & a-b & b\\c & c-a & a\end{vmatrix}$
We can split the determinant as
$\Delta=(b-a)^2\begin{bmatrix}\begin{vmatrix}b & b& c\\a & a & b\\c & c &a\end{vmatrix}-\begin{vmatrix}b & c & c\\a & b & b\\c & a & a\end{vmatrix}\end{bmatrix}$
Hence two columns are identical in both the determinants and hence their value is 0.
Therefore$\Delta=0$.
Hence D is the correct answer.