Mathematics - Determinant Question with Solution | TestHub

Given, $A\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]$ For unique solution, $\det (A)\ne 0$ $$\begin{array}{c}\text{ Case I }\det (A)=\left|\begin{array}{lll}1& a& b\\ a& 1& c\\ b& c& 1\end{array}\right|\\ =1-a^2-b^2-c^2+2abc\ne 0\end{array}$$ Here $a,b,c$ is selected from $1,0,0$. (No case is possible) Case II (i) $\det (A)=\left|\begin{array}{lll}1& a& b\\ a& 0& c\\ b& c& 0\end{array}\right|=2abc-c^2\ne 0$ Here $a,b,c$ are selected from $1,1,0$. (2 cases are possible) (ii) $\det (A)=\left|\begin{array}{lll}0& a& b\\ a& 1& c\\ b& c& 0\end{array}\right|=2abc-b^2\ne 0$ Here $a,b,c$ are selected from $1,1,0$. (2 cases are possible) (iii) $\det (A)=\left|\begin{array}{lll}0& a& b\\ a& 0& c\\ b& c& 1\end{array}\right|=2abc-a^2\ne 0$ Here $a,b,c$ are selected from $1,1,0$. (2 cases are possible) Hence, there are exactly 6 matrices for unique solution. Hence, option (b) is correct.