For $α, β, γ∈ R$, let $A=\begin{bmatrix}α^2&6&8\\3&β^2&9\\4&5&γ^2\end{bmatrix}$ and $B=\begin{bmatrix}2α&3&5\\2&2β&6\\1&4&2γ-3\end{bmatrix}$. If $tr(A) = tr(B)$, then find the value of $(α^{-1}+β^{-1}+γ^{-1})$.

1 2 3 4

3

We have, $tr(A) = tr(B)$, $∴α^2+β^2+γ^2=2α+2β+2γ−3$ $⇒(α^2-2α+1)+(β^2-2β+1)+(γ^2-2γ+1)=0$ $⇒(α−1)^2+(β−1)^2+(γ−1)^2=0$ $⇒α=1,β=1,γ=1$ $∴(α^{-1}+β^{-1}+γ^{-1})=(1+1+1)=3$