Question

For $\alpha, \beta, \gamma \in R,$ let $A=\begin{bmatrix}\alpha^{2} & 6 & 8 \\ 3 & \beta^{2} & 9 \\ 4 & 5 & \gamma^{2}\end{bmatrix}$ and $B=\begin{bmatrix}2 \alpha & 3 & 5 \\ 2 & 2 \beta & 6 \\ 1 & 4 & 2 \gamma-3\end{bmatrix}$. If trace $A=$ trace $B$, then the value of $\left(\alpha^{-1}+\beta^{-1}+\gamma^{-1}\right)$ is equal to

Answer 1

We have trace $A=$ trace$B$, so
$\alpha^{2}+\beta^{2}+\gamma^{2}=2 \alpha+2 \beta+2 \gamma-3$
$\Rightarrow\left(\alpha^{2}-2 \alpha+1\right)+\left(\beta^{2}-2 \beta+1\right)+\left(\gamma^{2}-2 \gamma+1\right)=0$
$\Rightarrow(\alpha-1)^{2}+(\beta-1)^{2}+(\gamma-1)^{2}=0$
$\Rightarrow \alpha=1, \beta=1, \gamma=1$
Hence, $\left(\alpha^{-1}+\beta^{-1}+\gamma^{-1}\right)$
$=(1+1+1)=3 .$