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Using the property of determinants and without expanding, prove that \[ \begin{vmatrix} a^2&bc&ac+c^2 \\ a^2+ab&b^2&ac \\ ab&b^2+bc&c^2 \end{vmatrix} = 4a^2b^2c^2\]

1 Answer

  • 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} a^2&bc&ac+c^2 \\ a^2+ab&b^2&ac \\ ab&b^2+bc&c^2 \end{vmatrix} $
We can take the common factors a, b and c respectively from $C_1,C_2$ and $C_3$
$\therefore\:\bigtriangleup=abc\begin{vmatrix}a & c & a+c\\a+b & b & a\\b & b+c & c\end{vmatrix}$
By applying $R_2\rightarrow\:R_2-R_1-R_3,$ we get
$\bigtriangleup=abc\begin{vmatrix}a & c & a+c\\0 & -2c & -2c\\b & b+c &c\end{vmatrix}$
By taking $2c$ common from $R_2$, we get
$\bigtriangleup=2abc^2\begin{vmatrix}a & c & a+c\\0 & -1 & -1\\b & b+c &c\end{vmatrix}$
By applying $C_2\rightarrow\:C_2-C_3$, we get
$\bigtriangleup=2abc^2\begin{vmatrix}a & -a & a+c\\0 & 0 & -1\\b & b &c\end{vmatrix}$
Now by expanding along $R_2$, we get $2abc^2\big(-(-1)\big)\begin{vmatrix}a & -a\\b & b\end{vmatrix}$
Hence proved.
answered Mar 21, 2014 by balaji.thirumalai
edited Mar 21, 2014 by rvidyagovindarajan_1

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