Question

The equation, $[1\, x y]$ $\begin{bmatrix}1&3&1\\ 0&2&-1\\ 0&0&1\end{bmatrix}\begin{bmatrix}1\\ x\\ y\end{bmatrix} =\left[0\right]$ has roots of
different types for different values of $y$.
(i) $y=0$
(ii) $y=-1$
(p) rational roots
(q) irrational roots
(r) integral roots
then

• A. (i)-(p),(ii)-(r)
• B. (i)-(q),(ii)-(p)
• C. (i)-(p),(ii)-(q)
• D. (i)-(r),(ii)-(p)

Answer 1

Answer: (C)

$\left[1\times y\right]\begin{bmatrix}1&3&1\\ 0&2&-1\\ 0&0&1\end{bmatrix}\begin{bmatrix}1\\ x\\ y\end{bmatrix} =\left[0\right]$
$\left[1+3+2x+1+-x+y\right]\begin{bmatrix}1\\ x\\ y\end{bmatrix}=\left[0\right]$
$\Rightarrow \left[1+3 x+2 x^{2}+y-x y+y^{2}\right]=[0]$
$\Rightarrow 2 x^{2}+y^{2}+y+3 x-x y+1=0$
If $y=0,2 x^{2}+3 x+1=0$
$\Rightarrow (2 x-1)(x+1)=0$
$\Rightarrow x=-1 / 2,-1 \,\,\,\,$(Rational roots)
If $y=-1,2 x^{2}+4 x+1=0$
$\Rightarrow x=\frac{-4 \pm \sqrt{12}}{4}=\frac{-2 \pm \sqrt{3}}{2} \,\,\,\,\,$(Irrational roots)