Let A=[a b c d], where a, b, c, d ∈ R. If A-α I is invertible for all α ∈ R, then

Question

Let A=[a b c d], where a, b, c, d ∈ R. If A-α I is invertible for all α ∈ R, then (A) b c>0 (B) b c=0 (C) b c> min (0, (1/2) a d) (D) a=0

Tardigrade Admin · Accepted Answer

As A-α I is invertible for all α ∈ R. operatornamedet(A-α I) ≠ 0 ∀ α ∈ R . ⇒ (a-α)(d-α)-b c ≠ 0 ∀ α ∈ R . ⇒ α2-(a+d) α+a d-b c ≠ 0 ∀ α ∈ R . Therefore (a+d)2-4(a d-b c)< 0 ⇒ (a-d)2+4 b c<0 Therefore, b c< 0. text Also, a2+d2-2 a d+4 b c <0 ⇒ 0 ≤ a2+d2<2 a d-4 b c ⇒ b c <(1/2) a d . text Thus, b c < min (0, (1/2) a d)

Q. Let $A = [a c b d]$ , where $a, b, c, d \in R$ . If $A - α I$ is invertible for all $α \in R$ , then

$b c > 0$

$b c = 0$

$b c > min (0, \frac{1}{2} a d)$

$a = 0$

Q. Let A=[ac​bd​], where a,b,c,d∈R. If A−αI is invertible for all α∈R, then

bc>0

bc=0

bc>min(0,21​ad)

a=0

Q. Let $A = [a c b d]$ , where $a, b, c, d \in R$ . If $A - α I$ is invertible for all $α \in R$ , then

$b c > 0$

$b c = 0$

$b c > min (0, \frac{1}{2} a d)$

$a = 0$