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By using properties of determinants, prove the following: \[ \begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix}=(5x+4)(4-x)^2 \]

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  • 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 $\bigtriangleup=\begin{vmatrix}x+4& 2x & 2x\\2x & x+4 & 2x\\2x &2x & x+4\end{vmatrix}$
Now let us apply $R_1\rightarrow R_1+R_2+R_3$
Let $\bigtriangleup=\begin{vmatrix}5x+4& 5x+4 & 5x+4\\2x & x+4 & 2x\\2x &2x & x+4\end{vmatrix}$
By taking (5x+4) as a common factor from first row,
Let $\bigtriangleup=\begin{vmatrix}1& 1 & 1\\2x & x+4 & 2x\\2x &2x & x+4\end{vmatrix}$
By applying $C_1\rightarrow C_1-C_2$ and $C_2\rightarrow C_2-C_3$
Let $\bigtriangleup=\begin{vmatrix}0& 0 & 1\\x-4 & x-4 & 2x\\0 &x-4 & x+4\end{vmatrix}$
Take (x-4) as a common factor from $C_2$
$\bigtriangleup=(5x+4)(x-4)\begin{vmatrix}0& 0 & 1\\x-4 & 1 & 2x\\0 &1 & x+4\end{vmatrix}$
Expanding along $R_1$ we get,
Hence $\begin{vmatrix}x+4 & 2x & 2x\\2x & x+4 & 2x\\2x & 2x & x+4\end{vmatrix}=(5x+4)(4-x)^2$


answered Mar 11, 2013 by sreemathi.v

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