Question

If det $\left[\begin{array}{ccc}1& 1& 2\\ 2& 4& 9\\ t& t^2& 1+t^3\end{array}\right]=0$, 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 $1,2,-\frac{1}{2}$

$\left|\begin{array}{ccc}1& 1& 2\\ 2& 4& 9\\ t& t^2& 1+t^3\end{array}\right|=0$
Using row operations,
$R3\to R_3-R_1$
$\Rightarrow \left|\begin{array}{ccc}1& 1& 2\\ 2& 4& 9\\ t-1& t^2-1& t^3-1\end{array}\right|=0$
$\Rightarrow (t-1)\left|\begin{array}{ccc}1& 1& 2\\ 2& 4& 9\\ 1& t+1& (t^2+t+1)\end{array}\right|=0$
Now,
$R_2\to R_2-R_1$ and $R_3\to R_3-R_1$
$\Rightarrow (t-1)\left|\begin{array}{ccc}1& 1& 2\\ 0& 2& 5\\ 0& t& (t^2+t-1)\end{array}\right|=0\Rightarrow (t-1)(2t^2-3t-2)=0\Rightarrow (t-1)(t-2)(2t+1)=0\Rightarrow t=1,2,-\frac{1}{2}$