Question

The equation $\left[\begin{array}{ccc}1& x& y\end{array}\right]\left[\begin{array}{ccc}1& 3& 1\\ 0& 2& -1\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}1\\ x\\ y\end{array}\right]=\left[0\right]$ has

• A. rational roots if $y=-1$
• B. irrational roots if $y=0$
• C. rational roots if $y=0$
• D. integral roots if $y=-1$

Answer 1

The correct option is C rational roots if $y=0$

Given : $\left[\begin{array}{ccc}1& x& y\end{array}\right]\left[\begin{array}{ccc}1& 3& 1\\ 0& 2& -1\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}1\\ x\\ y\end{array}\right]=\left[0\right]$
$\Rightarrow \left[\begin{array}{ccc}1& 3+2x& 1-x+y\end{array}\right]\left[\begin{array}{c}1\\ x\\ y\end{array}\right]=\left[0\right]$
$\Rightarrow 2x^2+y^2+y+3x-xy+1=0$

If $y=0,2x^2+3x+1=0$
Solving this, we get
$x=\frac{-1}{2},-1$ (rational roots)

If $y=-1,2x^2+4x+1=0$
$x=\frac{-4\pm \sqrt{8}}{4}=\frac{-2\pm \sqrt{2}}{2}$ (irrational roots)