question_answer

The determinant  vanishes for

A) 3 values of x   

B)   2 values of x

C)   1 values of x                

D)   no value of x

Correct Answer: D

Solution :

The given determinant vanishes, i.e., Expanding along \[{{C}_{1}},\] we get \[(x-4){{(x-5)}^{2}}-(x-5){{(x-4)}^{2}}-\{(x-3){{(x-5)}^{2}}\] \[-(x-5){{(x-3)}^{2}}\}+(x-3){{(x-4)}^{2}}\] \[-(x-4){{(x-3)}^{2}}=0\] \[\Rightarrow \,\,(x-4)\,(x-5)(x-5-x+4)\] \[-(x-3)(x-5)(x-5-x+3)\] \[-(x-3)(x-4)(x-4-x+3)=0\] \[\Rightarrow \,-(x-4)(x-5)+2(x-3)(x-5)\] \[-(x-3)(x-4)=0\] \[\Rightarrow \,\,-{{x}^{2}}+9x-20+2{{x}^{2}}-16x+30-{{x}^{2}}+7x-12=0\]\[\Rightarrow \,\,-32+30=0\] \[\Rightarrow \,\,-2=0\] Which is not possible, hence no value of x satisfies the given condition.