logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Determinants
0 votes

$\begin{vmatrix}0 & xyz & x-z\\y-x & 0 & y-z\\z-x & z-y & 0\end{vmatrix}$ =

$\begin{array}{1 1} (x-z)(y-z)(y-x+xyz) \\ (x+z)(y+z)(y-x+xyz) \\ (x+z)(y-z)(y+x+xyz) \\ (x-z)(y-z)(y+x+xyz) \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • If $\Delta=\begin{vmatrix}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{vmatrix}$
  • $|\Delta|=a_{11}(a_{22}\times a_{33}-a_{23}\times a_{32})-a_{12}(a_{21}\times a_{33}-a_{23}\times a_{31})+a_{13}(a_{21}\times a_{32}-a_{22}\times a_{31})$
  • Elementary transformations can be done by
  • (i)Interchanging the rows or columns.
  • (ii)The addition to the elements of any rows or columns.
Let A=$\begin{vmatrix}0 & xyz & x-z\\y-x & 0 & y-z\\z-x & z-y & 0\end{vmatrix}$
Apply $C_1\rightarrow C_3-C_1$
$\;\;\;=\begin{vmatrix}x-z & xyz & x-z\\x-z & 0 & y-z\\x-z & z-y & 0\end{vmatrix}$
Take (x-z) as the common factor from $C_1$
$A=(x-z)\begin{vmatrix}0 & xyz & x-z\\y-x & 0 & y-z\\z-x & z-y & 0\end{vmatrix}$
Apply $R_2\rightarrow R_2-R_3$ and $R_3\rightarrow R_3-R_1$
$A=(x-z)\begin{vmatrix}1 & xyz & x-z\\0 & y-z& y-z\\0 & z-y-xyz & z-x\end{vmatrix}$
Take (y-z) as the common factor from $R_2$
$A=(x-z)(y-z)\begin{vmatrix}1 & xyz & x-z\\0 & 1& 1\\0 & z-y-xyz & z-x\end{vmatrix}$
Now expanding along $R_1$ we get,
A=(x-z)(y-z)(z-x-z+y+xyz)
$\;\;=(x-z)(y-z)(y-x+xyz)$
answered Mar 27, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...