Question

If det $\begin{bmatrix}1&1&2\\ 2&4&9\\ t&t^{2}&1+t^{3}\end{bmatrix} = 0 $ , then the values of t are

• A. $ 1 , 2 , \frac{1}{2}$
• B. $ - 1 , 2 , \frac{1}{2}$
• C. $ 1 , - 2 , \frac{1}{2}$
• D. $ 1 , 2 , - \frac{1}{2}$

Answer 1

The Correct Option is D

We have,
$\begin{vmatrix}1&1&2\\ 2&4&9\\ t&t^{2}&1+t^{3}\end{vmatrix} = 0 $
$ \Rightarrow 1 \left( 4 + 4t^{3} -9t^{2} \right) -1 \left( 2 +2t^{3} -9t\right) + 2\left(2t^{2}-4t\right) = 0 $
$ \Rightarrow 4 +4t^{2} -9t^{2} -2 -2t^{3} +9t +4t^{2} -8t=0$
$\Rightarrow 2t^{3} -5t^{2} +t +2 = 0 $
$\Rightarrow \left(t-1\right)\left(t-2\right)\left(2t+1\right) = 0$
$ \Rightarrow t=1, 2, -\frac{1}{2}$