Using properties of determinants, find the value of x for which
\begin{vmatrix} 4-x & 4+x &4+x \\ 4+x& 4-x &4+x \\ 4+x&4+x & 4-x \end{vmatrix}= 0

 

 

 

 
 
 
 
 

Answers (1)

we have \begin{vmatrix} 4-x & 4+x &4+x \\ 4+x& 4-x &4+x \\ 4+x& 4+x &4-x \end{vmatrix}= 0
by R_{1}\rightarrow R_{1}-R_{2}\; \; \; ,R_{2}\rightarrow R_{2}-R_{3}
              \begin{vmatrix} -2x& 2x &0 \\ 0& -2x &2x \\ 4+x& 4+x &4-x \end{vmatrix}= 0


on expanding along C1 , we get
-2x(-2x(4-x)-2x(4+x))+\left ( 4+x \right )\left ( 4x^{2}-0 \right )=0
Either \left ( 12+x \right )=0\; \; or \; \; 4x^{2}= 0
\because x= 0\; or\; -12

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