Question

The determinant $\left|\begin{array}{ccc}1& \left(x-3\right)& {\left(x-3\right)}^2\\ 1& \left(x-4\right)& {\left(x-4\right)}^2\\ 1& \left(x-5\right)& {\left(x-5\right)}^2\end{array}\right|$ vanishes for

• A. 3 values of x
• B. 2 values of x
• C. 1 values of x
• D. No value of x

Answer 1

Answer: (D)

The given determinant vanishes, i.e.,
$\left|\begin{array}{ccc}1& \left(x-3\right)& {\left(x-3\right)}^2\\ 1& \left(x-4\right)& {\left(x-4\right)}^2\\ 1& \left(x-5\right)& {\left(x-5\right)}^2\end{array}\right|=0$
Expanding along $C_1$, we get
$(x-4)(x-5{)}^2-(x-5)(x-4{)}^2$
$-\left\{(x-3)(x-5{)}^2-(x-5)(x-3{)}^2\right\}$
$+(x-3)(x-4{)}^2-(x-4)(x-3{)}^2=0$
$\Rightarrow (x-4)(x-5)(x-5-x+4)$
$-(x-3)(x-5)(x-5-x+3)+(x-3)(x-4)(x-4-x+3)=0$
$\Rightarrow -(x-4)(x-5)+2(x-3)(x-5)-(x-3)(x-4)$
$=0$
$\Rightarrow -x2+9x-20+2x2-16x+30-x2+7x-12$
$=0$
$\Rightarrow -32+30=0$
$\Rightarrow -2=0$
Which is not possible, hence no value of $x$ satisfies the given condition.