Mathematics - Matrices Question with Solution | TestHub

$\begin{array}{rl}& \alpha ,\beta \text{ are roots of }{\mathrm{x}}^2+\mathrm{x}-1=0\\ & \therefore \alpha +\beta =-1\Rightarrow 1+\alpha +\beta =0\\ & M=\left[\begin{array}{lll}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}& {\mathrm{a}}_{13}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}& {\mathrm{a}}_{23}\\ {\mathrm{a}}_{31}& {\mathrm{a}}_{32}& {\mathrm{a}}_{33}\end{array}\right]\end{array}$ (P) $M=\left[\begin{array}{lll}1& \alpha & \beta \\ \alpha & \beta & 1\\ \beta & 1& \alpha \end{array}\right]\Rightarrow 3!\times 2=12$ For one arrangement of row 1 we can arrange other two rows exactly in two ways and row 1 can be arranged in 3 ! ways $\therefore 3!\times 2=12\text{ ways }$ (Q) $M=\left[\begin{array}{lll}x& a& b\\ a& y& c\\ b& c& z\end{array}\right]\Rightarrow$ Consider one such arrangement with $a=\alpha ,b=\beta ,c=1$ $M=\left[\begin{array}{lll}1& \alpha & \beta \\ \alpha & \beta & 1\\ \beta & 1& \alpha \end{array}\right]$ a, b, c can be arranged in $3!$ ways and corresponding entries can be arranged in 1 way. (R) $\left[\begin{array}{ccc}0& a& b\\ -a& 0& c\\ -b& -c& 0\end{array}\right]\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}a\\ 0\\ -c\end{array}\right]$ $\begin{array}{rl}& ay+bz=a\\ & -ax+cz=0\\ & -bx-cy=-c\end{array}$ It is observed that $\mathrm{D}={\mathrm{D}}_{\mathrm{x}}={\mathrm{D}}_{\mathrm{y}}={\mathrm{D}}_{\mathrm{z}}=0$ $\therefore$ infinite solution (S) $\left[\begin{array}{lll}1& \alpha & \beta \\ \beta & \alpha & 1\\ \alpha & 1& \beta \end{array}\right]$ $\Rightarrow \alpha \beta -1-\alpha {\beta }^2+{\alpha }^2+{\beta }^2-{\alpha }^2\beta =0($ since $\alpha \beta =\alpha +\beta =-1$ )