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Home  >>  CBSE XII  >>  Math  >>  Determinants
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If x+y+z=0,prove that $\begin{vmatrix}xa & yb & zc\\yc & za & xb\\zb & xc & ya\end{vmatrix}=xyz\begin{vmatrix}a & b & c\\c & a & b\\b & c & a\end{vmatrix}$.

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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 each row or column of a determinant is multiplied by k,then its value is multiplied by k,then the value of the determinant is k|A|.
  • By the above property we can take out any common factor from any one row or any one column of a given determinant.
Let $\Delta=\begin{vmatrix}xa & yb & zc\\yc & za & xb\\zb &xc & ya\end{vmatrix}$
Divide $C_1$ by x,$C_2$ by y and $C_3$ by z
$\Rightarrow \Delta=xyz\begin{vmatrix}a & b& c\\\frac{y}{x}c&\frac{z}{y}a &\frac{x}{z}b\\\frac{z}{x} & \frac{x}{y}c &\frac{y}{z}a\end{vmatrix}$
On expanding along $R_1$ we get,
$\Delta=xyz\bigg(a[\frac{z}{y}a\times \frac{y}{z}a-\frac{x}{z}b\times\frac{x}{y}c]-b[\frac{y}{x}c\times \frac{y}{z}a-\frac{x}{z}b\times \frac{z}{x}b]+c[\frac{y}{x}c\times \frac{x}{y}c-\frac{z}{y}a\times \frac{z}{x}b]\bigg)$
On simplifying we get,
$\Delta=xyz\bigg(a(a^2-\frac{x^2}{zy}bc)-b(\frac{y^2}{xz}ca-b^2)+c(c^2-\frac{z^2}{xy}ab)\bigg)$
On simplifying further we get,
$\Delta=xyz\bigg(a^3-\frac{x^2}{zy}abc-\frac{y^2}{xz}abc+b^3+c^3-\frac{z^2}{xy}abc\bigg)$
$\Delta=xyz\bigg(a^3+b^3+c^3-(abc)(\frac{x^2}{zy}+\frac{y^2}{xz}+\frac{z^2}{xy})\bigg)$
$\Delta=xyz\bigg(a^3+b^3+c^3-(abc)\frac{x^3+y^3+z^3}{xyz}\bigg)$
$\Delta=(xyz)(a^3+b^3+c^3)-(abc)(x^3+y^3+z^3)$
But it is given that $x+y+z\neq 0$
Therefore $x^3+y^3+z^3=3abc$
$\Rightarrow\Delta =(xyz)[a^3+b^3+c^3]-abc[3xyz]$
$\qquad=(xyz)[a^3+b^3+c^3-3abc]$
Now consider $\begin{vmatrix}a & b& c\\c & a & b\\b & c & a\end{vmatrix}$
Expanding along $R_1$ we get,
$a(a^2-bc)-b(ac-b^2)+c(c^2-ab)$
$\;\;=a^3-abc-abc+b^3+c^3-abc$
$\;\;=a^3+b^3+c^3-3abc$
Therefore $\begin{vmatrix}a & b& c\\c & a & b\\b & c & a\end{vmatrix}=a^3+b^3+c^3-3abc$
Now substituting this in the LHS we get,
$\Delta=xyz\begin{vmatrix}a & b& c\\c & a & b\\b & c & a\end{vmatrix}$
Hence proved.
answered Mar 25, 2013 by sreemathi.v
 

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