Question

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

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