Toshow: $(A−B)2≠A2−2AB+B2$ by taking $A=[2−113] and B=[−110−2]$ .

Expert Solution & Answer

Explanation of Solution

Given information: $A=[2−113] and B=[−110−2]$

Proof:

As we have $A=[2−113] and B=[−110−2]$

To prove $(A−B)2≠A2−2AB+B2$

Consider left hand side part:

$(A−B)2=([2−113]−[−110−2])2 =[3−215]2 =[3−215]×[3−215] =[9−2−6−103+5−2+25] (A−B)2=[716823] ........(1)$

Now consider right hand side part :

$A2−2AB+B2To finding A2=[2−113]2 and B2=[−110−2]2 A2=[2−113]×[2−113] and B2=[−110−2]×[−110−2]$

$A2=[4−1−2−32+3−1+9] and B2=[1+0−1−20+00+4] A2=[3−558] and B2=[1−304]AB=[2−113][−110−2] =[−2−02+2−1+01−6]AB=[−24−1−5]$

$Thus, A2−2AB +B2=[3−558]−2[34−1−5]+[1−304] =[4−8512]−[68−2−10]∴A2−2AB +B2=[10032] ........(2)$

Therefore from equations $(1)$ and $(2)$ , $(A−B)2≠A2−2AB+B2$

Hence, theorem proved.

Chapter 8.2, Problem 83E

Chapter 8.2, Problem 85E

Chapter 8 Solutions

EBK PRECALCULUS W/LIMITS

Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E Ch. 8.1 - Prob. 8E Ch. 8.1 - Prob. 9E Ch. 8.1 - Prob. 10E

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Toshow : ( A − B ) 2 ≠ A 2 − 2 A B + B 2 by taking A = [ 2 − 1 1 3 ] and B = [ − 1 1 0 − 2 ] .

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Explanation of Solution

Chapter 8 Solutions