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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Using the property of determinants and without expanding, prove that \[ \begin{vmatrix} 0&a&-b \\ -a&0&-c \\ b&c&0 \end{vmatrix} = 0 \]

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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.
  • 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 $\bigtriangleup=\begin{vmatrix}0 & a & -b\\-a & 0 & -c\\b & c & 0\end{vmatrix}$
 
Let $R_1\rightarrow CR_1$,by applying this property
 
$\bigtriangleup=\frac{1}{C}\begin{vmatrix}0 & ac & -bc\\-a & 0 & -c\\b & c & 0\end{vmatrix}$
 
Similarly let $R_1\rightarrow R_1-bR_2$,by applying this property
 
$\bigtriangleup=\frac{1}{C}\begin{vmatrix}ab & ac & 0\\-a & 0 & -c\\b & c & 0\end{vmatrix}$
 
By taking 'a' as the common factor from $R_1$
 
$\bigtriangleup=\frac{a}{C}\begin{vmatrix}b & c & 0\\-a & 0 & -c\\b & c & 0\end{vmatrix}$
 
We can see $R_1$ and $R_3$ are identical.
 
Hence the value of the determinant is 0.
 
Therefore $\begin{vmatrix}0 & a & -b\\-a & 0 & -c\\b & c & 0\end{vmatrix}$=0.

 

answered Feb 21, 2013 by sreemathi.v
edited Feb 24, 2013 by vijayalakshmi_ramakrishnans
 

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