# By using the properties of determinants show that $(i) \begin{vmatrix} x+4&2x&2x \\ 2x&x+4&2x \\ 2x&2x&x+4 \end{vmatrix} = (5x+4)(4-x)^2$

Note: This is part 1 of a 2 part question, split as 2 separate questions here.

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 $\bigtriangleup=\begin{vmatrix}x+4& 2x & 2x\\2x & x+4 & 2x\\2x &2x & x+4\end{vmatrix}$

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,

$\bigtriangleup=(5x+4)(x-4)[1(x+4)-2x(1)]$

$\bigtriangleup=(5x+4)(x-4)[1(x+4)-2x]$

$\quad=(5x+4)(x-4)(4-x)$

$\quad=(-1)(5x+4)(4-x)(4-x)$

$\quad=(5x+4)(4-x)^2$

Hence $\begin{vmatrix}x+4 & 2x & 2x\\2x & x+4 & 2x\\2x & 2x & x+4\end{vmatrix}=(5x+4)(4-x)^2$

edited Feb 24, 2013