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Mathematics Question on Properties of Determinants

If det [112 249 tt21+t3]=0\begin{bmatrix}1&1&2\\\ 2&4&9\\\ t&t^{2}&1+t^{3}\end{bmatrix} = 0 , then the values of t are

A

1,2,12 1 , 2 , \frac{1}{2}

B

−1,2,12 - 1 , 2 , \frac{1}{2}

C

1,−2,12 1 , - 2 , \frac{1}{2}

D

1,2,−12 1 , 2 , - \frac{1}{2}

Answer

1,2,−12 1 , 2 , - \frac{1}{2}

Explanation

Solution

We have,
∣112 249 tt21+t3∣=0\begin{vmatrix}1&1&2\\\ 2&4&9\\\ t&t^{2}&1+t^{3}\end{vmatrix} = 0
⇒1(4+4t3−9t2)−1(2+2t3−9t)+2(2t2−4t)=0\Rightarrow 1 \left( 4 + 4t^{3} -9t^{2} \right) -1 \left( 2 +2t^{3} -9t\right) + 2\left(2t^{2}-4t\right) = 0
⇒4+4t2−9t2−2−2t3+9t+4t2−8t=0\Rightarrow 4 +4t^{2} -9t^{2} -2 -2t^{3} +9t +4t^{2} -8t=0
⇒2t3−5t2+t+2=0\Rightarrow 2t^{3} -5t^{2} +t +2 = 0
⇒(t−1)(t−2)(2t+1)=0\Rightarrow \left(t-1\right)\left(t-2\right)\left(2t+1\right) = 0
⇒t=1,2,−12\Rightarrow t=1, 2, -\frac{1}{2}