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Q.
Let $A=\begin{bmatrix}a & b \\ c & d\end{bmatrix}$, where $a, b, c, d \in R$. If $A-\alpha I$ is invertible for all $\alpha \in R$, then
Matrices
Solution:
As $A-\alpha I$ is invertible for all $\alpha \in R$.
$ \operatorname{det}(A-\alpha I) \neq 0 \,\,\, \, \forall \alpha \in R . $
$\Rightarrow (a-\alpha)(d-\alpha)-b c \neq 0\,\,\, \, \forall \alpha \in R . $
$\Rightarrow \alpha^2-(a+d) \alpha+a d-b c \neq 0 \,\,\, \, \forall \alpha \in R .$
Therefore
$(a+d)^2-4(a d-b c)< 0 $
$\Rightarrow (a-d)^2+4 b c<0$
Therefore, $b c< 0$.
$ \text { Also, } a^2+d^2-2 a d+4 b c <0 $
$\Rightarrow 0 \leq a^2+d^2<2 a d-4 b c$
$\Rightarrow b c <\frac{1}{2} a d . $
$\text { Thus, } b c <\min \left(0, \frac{1}{2} a d\right)$