Let $$A=a$$ bc d, where a,b,c,d∈R Then

Correct option is 1det⁡(A)≤a2+b2c2+d2

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Let $A = (\begin{array}{ll} a b \\ c d \end{array})$ , where $a, b, c, d \in R$ Then

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det⁡(A)≤a2+b2c2+d2

det⁡(A)≤(a+b)(c+d)

det⁡(A)≤ac+bd

det⁡(A)≤(|a|−|b|)(|c|−|d|)

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detailed solution

Correct option is A

det⁡(A)=ad−bc≤|ad−bc|≤|a∥d|+|b∥c|Now, a2+b2c2+d2−(|a∥d|+|b∥c|)2=a2c2+b2c2+a2d2+b2d2−a2d2+b2c2+2|a∥d∥b∥c|=(|ac|−|bd|)2≥0⇒|a∥d|+|b||c|≤a2+b2c2+d2

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Let A=a bc d, where a,b,c,d∈R Then

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Let $A = (\begin{array}{ll} a b \\ c d \end{array})$ , where $a, b, c, d \in R$ Then