Solved Show That A B 2 A 2 B 2 A B 2 Chegg Question: show that (a b)^2 is different from a^2 2ab b^2, when a = [1 0 2 0] and b = [1 3 0 0].write down the correct rule for (a b)(a b). true or false (briefly explain why): if a^2 is defined then a is necessarily square. For all square matrices $a$ and $b$ of the same size, it is true that $(a b)^2 = a^2 2ab b^2$? (2 answers).
Solved Show That A B 2 Is Different From A 2 2ab Chegg Specifically, (a b) 2 generally cannot equal (a 2 b 2), because with (a b) 2, the same thing is being multiplied everywhere (namely, (a b)), while with (a 2 b 2), a and b are being multiplied by different things (namely, a and b respectively). (a b)^2 is not the same as a^2 b^2. an easy way of proving that (a b)^2 is not the same as a^2 b^2 is to expand the expressions. expansion of (a b)^2. expansion of a^2 b^2. as you can see these expressions look nothing alike in their expanded or simplified form, therefore the expressions are not the same. i love helping math students succeed!. Multiply the terms using the foil method: $$a*a$$a∗a, $$a*b$$a∗b, $$b*a$$b∗a, $$b*b$$b∗b. simplify the expression to get $$a^ {2} 2ab b^ {2}$$a2 2ab b2. observe that $$ (a b)^ {2}$$(a b)2 is not the same as $$a^ {2} b^ {2}$$a2 b2 since there is an additional term $$2ab$$2ab in $$ (a b)^ {2}$$(a b)2. 😉 want a more accurate answer?. To show that (a b)^2 is different from a^2 2ab b^2, we will expand both expressions and compare them. expanding (a b)^2: (a b)^2 = (a b)(a b) = a(a b) b(a b) = a^2 ab ba b^2 = a^2 ab ba b^2 expanding a^2 2ab b^2: a^2 2ab b^2 comparing the two expressions, we see.
Solved Show That A B 2 Is Different From A 2 2ab Chegg Multiply the terms using the foil method: $$a*a$$a∗a, $$a*b$$a∗b, $$b*a$$b∗a, $$b*b$$b∗b. simplify the expression to get $$a^ {2} 2ab b^ {2}$$a2 2ab b2. observe that $$ (a b)^ {2}$$(a b)2 is not the same as $$a^ {2} b^ {2}$$a2 b2 since there is an additional term $$2ab$$2ab in $$ (a b)^ {2}$$(a b)2. 😉 want a more accurate answer?. To show that (a b)^2 is different from a^2 2ab b^2, we will expand both expressions and compare them. expanding (a b)^2: (a b)^2 = (a b)(a b) = a(a b) b(a b) = a^2 ab ba b^2 = a^2 ab ba b^2 expanding a^2 2ab b^2: a^2 2ab b^2 comparing the two expressions, we see. Show that (a b)^2 is different from a^2 2ab b^2, using a and b defined below. a= [1 2 0 0] and b= [1 0 3 0] write down the correct rule for (a b) (a b) = a^2 b^2. your solution’s ready to go! our expert help has broken down your problem into an easy to learn solution you can count on. Since a a is on the right side of the equation, switch the sides so it is on the left side of the equation. simplify (a b)2 (a b) 2. tap for more steps a2 2ab b2 = a2 2ab b2 a 2 2 a b b 2 = a 2 2 a b b 2. move all terms containing a a to the left side of the equation. tap for more steps b2 = b2 b 2 = b 2. Let a，b \in m n（k） such that a^2 b^2=2ab。 show that i) \det a = \det b, ii) a、b have the same characteristic polynomials. solution (y. sharifi). i) the proof is by induction on n. if n=1， then a，b \以 k 为单位 and a^2 b^2=2ab gives （a b）^2=0 and hence a=b。 now, suppose that n \ge 2 and i) holds in m m（k）， \ m \le n. Show that (a − b) 2, (a 2 b 2) and (a b) 2 are in a.p. here, we are given three terms and we need to show that they are in a.p., first term (a 1) = (a b) 2. second term (a 2) = (a 2 b 2) third term (a 3) = (a b) 2. so in an a.p, the difference of two adjacent terms is always constant.
Solved Show That A B 2 Is Different From A 2 2ab Chegg Show that (a b)^2 is different from a^2 2ab b^2, using a and b defined below. a= [1 2 0 0] and b= [1 0 3 0] write down the correct rule for (a b) (a b) = a^2 b^2. your solution’s ready to go! our expert help has broken down your problem into an easy to learn solution you can count on. Since a a is on the right side of the equation, switch the sides so it is on the left side of the equation. simplify (a b)2 (a b) 2. tap for more steps a2 2ab b2 = a2 2ab b2 a 2 2 a b b 2 = a 2 2 a b b 2. move all terms containing a a to the left side of the equation. tap for more steps b2 = b2 b 2 = b 2. Let a，b \in m n（k） such that a^2 b^2=2ab。 show that i) \det a = \det b, ii) a、b have the same characteristic polynomials. solution (y. sharifi). i) the proof is by induction on n. if n=1， then a，b \以 k 为单位 and a^2 b^2=2ab gives （a b）^2=0 and hence a=b。 now, suppose that n \ge 2 and i) holds in m m（k）， \ m \le n. Show that (a − b) 2, (a 2 b 2) and (a b) 2 are in a.p. here, we are given three terms and we need to show that they are in a.p., first term (a 1) = (a b) 2. second term (a 2) = (a 2 b 2) third term (a 3) = (a b) 2. so in an a.p, the difference of two adjacent terms is always constant.