Question

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

$T_r\left(A\right)$ is a Trace(A) of a matrix

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

$T_r\left(A\right)={\alpha }^2+{\beta }^2+{\gamma }^2$

$T_r\left(B\right)=2\alpha +2\beta +2\gamma -3$

$\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2=2\alpha +2\beta +2\gamma -3$

$\Rightarrow {(\alpha -1)}^2+{(\beta -1)}^2+{(\gamma -1)}^2=0$
$\Rightarrow \alpha =\beta =\gamma =1\Rightarrow \frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=3$