Question

Consider the principle diagonal elements of $B$ as $(b_1,b_2,b_3)$ such that $b_1>b_2>b_3$, then the value of $\lambda$ for which $b_1,b_2,\left(b_3\lambda \right)$ are in $G.P.$

• A. $3$
• B. $1$
• C. $-\frac{1}{3}$
• D. $\frac{81}{91}$

Answer 1

The correct option is D $\frac{81}{91}$

$A$ is symmetric

$\therefore {\left(A^T\right)}^{-1}=A^{-1}$ and inverse is also symmetric

$A^{-1}=\frac{1}{3}\left[\begin{array}{ccc}-1& 2& 0\\ 2& -1& 0\\ 0& 0& 3\end{array}\right]$

$B=A^{-1}N{\left(A^{-1}\right)}^T=A^{-1}N{A}^{-1}=\left[\begin{array}{ccc}-1& 2& 0\\ 2& -1& 0\\ 0& 0& 3\end{array}\right]\left[\begin{array}{ccc}3& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right]\left[A^{-1}\right]$

$=\left[\begin{array}{ccc}-1& \frac{2}{3}& 0\\ 2& -\frac{1}{3}& 0\\ 0& 0& -1\end{array}\right]\left[\begin{array}{ccc}-\frac{1}{3}& \frac{2}{3}& 0\\ \frac{2}{3}& -\frac{1}{3}& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}\frac{7}{9}& -\frac{8}{9}& 0\\ -\frac{8}{9}& \frac{13}{9}& 0\\ 0& 0& -1\end{array}\right]$

$\left|B\right|=\left|\begin{array}{ccc}\frac{7}{9}& -\frac{8}{9}& 0\\ -\frac{8}{9}& \frac{13}{9}& 0\\ 0& 0& 1\end{array}\right|=\frac{91-64}{81}=\frac{27}{81}=\frac{1}{3}$

According to the given now $b_1=\frac{13}{9},b_2=1,b_3=\frac{7}{9}$

$b_1,b_2,b_3,\lambda$ are in $G.P.$

$\Rightarrow b_1{b}_3\lambda =b_2^2$

$\Rightarrow \frac{13}{9}.\frac{7}{9}\lambda =1$

$\Rightarrow \lambda =\frac{81}{91}$