If A=α1 α2β1 β2 (where α2≠β1 and α1,α2,β1,β2 are non-zero) satisfies the equation x2 + k = 0 then

A2=−kI⇒α1 α2β1 β2α1 α2β1 β2=−K00−K ⇒α1 α2+ α2 β2=0 ⇒ α2α1 + β2=0 ⇒α1 + β2=0⇒α1 =- β2

If A=α1 α2β1 β2 (where α2≠β1 and α1,α2,β1,β2 are non-zero) satisfies the equation x2 + k = 0 then

A2=−kI⇒α1 α2β1 β2α1 α2β1 β2=−K00−K ⇒α1 α2+ α2 β2=0 ⇒ α2α1 + β2=0 ⇒α1 + β2=0⇒α1 =- β2

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If $A = [\begin{array}{ll} α_{1} α_{2} \\ β_{1} β_{2} \end{array}] (where α_{2} \neq β_{1} and α_{1}, α_{2}, β_{1}, β_{2} are non-zero)$ satisfies the equation x² + k = 0 then

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Tra (A) = 0

$α_{1} β_{2} < 0$

$α_{1} β_{2} > 0$

$Tra (A) \neq 0$

answer is A, B.

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Detailed Solution

$A^{2} = - k I \Rightarrow [\begin{array}{ll} α_{1} α_{2} \\ β_{1} β_{2} \end{array}] [\begin{array}{ll} α_{1} α_{2} \\ β_{1} β_{2} \end{array}] = [\begin{array}{cc} - K & 0 \\ 0 & - K \end{array}] \Rightarrow α_{1} α_{2} + α_{2} β_{2} = 0 \Rightarrow α_{2} (α_{1} + β_{2}) = 0 \Rightarrow α_{1} + β_{2} = 0 \Rightarrow α_{1} = - β_{2}$