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Home  >>  CBSE XII  >>  Math  >>  Determinants
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By using the properties of determinants show that $ (ii) \quad \begin{vmatrix} 1&1&1 \\ a&b&c \\ a^3&b^3&c^3 \end{vmatrix} = (a-b)(b-c)(c-a)(a+b+c) $

Note: This is part 2 of a 2 part question, split as 2 separate questions here.
Can you answer this question?
 
 

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Toolbox:
  • If each element of a row (or column) of a determinant is multiplied by a constant k ,then its value gets multiplied by k.
  • By this property we can take out any common factor from any one row or any one column of the determinant.
  • Elementary transformations can be done by
  • 1. Interchanging any two rows or columns. rows.
  • 2. Mutiplication of the elements of any row or column by a non-zero number
  • 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
Let $\bigtriangleup=\begin{vmatrix}1 & 1 & 1\\a & b & c\\a^3 &b^3 & c^3\end{vmatrix}$
 
By applying $C_1\rightarrow C_1-C-2$ and $C_2\rightarrow C_2-C_3$
 
$\bigtriangleup=\begin{vmatrix}0 & 0 & 1\\(a-b) & (b-c) & c\\(a^3 -b^3)&(b^3-c^3) & c^3\end{vmatrix}$
 
But we know $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ and $(b^3-c^3)=(b-c)(b^2+bc+c^2)$
 
Therefore $\bigtriangleup=\begin{vmatrix}0 & 0 & 1\\a-b & b-c & c\\(a-b)(a^2+ab+b^2) &(b-c)(b^2+bc+c^2) & c^3\end{vmatrix}$
 
Taking (a-b) as a common factor from $C_1$ and (b-c) from $C_2$ we get
 
$\bigtriangleup=(a-b)(b-c)\begin{vmatrix}0 & 0 & 1\\1 & 1 & c\\a^2+ab+b^2 &b^2+bc+c^2 & c^3\end{vmatrix}$
 
By applying $C_1\rightarrow C_1-C_2$
 
$\bigtriangleup=(a-b)(b-c)\begin{vmatrix}0 & 0 & 1\\1 & 1 & c\\a^2+ab-bc-c^2 &b^2+bc+c^2 & c^3\end{vmatrix}$
 
But $a^2+ab-bc-c^2$ can be written as $(a^2-c^2)+b(a-c)$
 
$\bigtriangleup=(a-b)(b-c)\begin{vmatrix}0 & 0 & 1\\0 & 1 & c\\(a^2-c^2)+b(a-c) &b^2+bc+c^2 & c^3\end{vmatrix}$
 
Now taking (a-c) as a common factor from $C_1$
 
$\bigtriangleup=(a-b)(b-c)(a-c)\begin{vmatrix}0 & 0 & 1\\0 & 1 & c\\a+c+b &b^2+bc+c^2 & c^3\end{vmatrix}$
 
Now expanding along $R_1$ we get,
 
Therefore $\bigtriangleup=(-1)(a-b)(b-c)(a-c)(a+b+c)$
 
$\qquad\qquad=(a-b)(b-c)(c-a)(a+b+c)$
 
Hence $\begin{vmatrix}1 & 1 & 1\\a & b & c\\a^3 & b^3 & c^3\end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c)$
 
Hence proved
 

 

answered Mar 6, 2013 by sreemathi.v
 

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