Mathematics - Determinant Question with Solution | TestHub

Given, $A\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]$ Case I $\left[\begin{array}{lll}1& a& b\\ a& 1& c\\ b& c& 1\end{array}\right]\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]$ $a,b,c$ are selected from $1,0,0$. $\Rightarrow x+ay+bz=1\Rightarrow ax+y+cz=0$ $bx+cy+z=0$ (i) If $a=1,b=c=0$, then $x+y=1$ Inconsistent system of equation $x+y=0$ (ii) If $a=0=c,b=1$, then $x+z=1$, $y=0$ Inconsistent system of equation $$x+z=0$$ (iii) If $c=1,a=b=0$, then $x=1,z=0$, $y=0$ Case II (i) $\left[\begin{array}{lll}1& a& b\\ a& 0& c\\ b& c& 0\end{array}\right]\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]$ $a,b,c$ are selected from $1,1,0$. $$\begin{array}{rl}\Rightarrow x+ay+bz& =1\Rightarrow ax+cz=0\\ bx+cy& =0\end{array}$$ Clearly, in all three cases, solutions are possible, so system is consistent. $$\begin{array}{c}\text{ (ii) }\left[\begin{array}{lll}0& a& b\\ a& 1& c\\ b& c& 0\end{array}\right]\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]\\ \Rightarrow ay+bz=1\Rightarrow ax+y+cz=0\\ bx+cy=0\end{array}$$ Clearly, $b=0,a=c=1$ gives $$y=1;x+y+z=0$$ Inconsistent system $y=0$ More than 2 matrices are possible. Hence, option (b) is correct.