Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

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

Can you answer this question?

1 Answer

0 votes


  • 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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App