Monday, April 29, 2019
Math Portifolio Matrix binomials Problem Example | Topics and Well Written Essays - 750 words
Portifolio hyaloplasm binomials - Math Problem ExampleBased on these computations, we derive the everyday expression for AnAn = (2n-1)(an)(X)We must sustain for the validity of this equation by applying it to solve for A2 with a=3.A2 = 22-1321111 = 18181818 , same as the previous answer. Hence, the expression is valid.Now, get to b=2 B = 2-2-22B2 = 2-2-222-2-22 = 8-8-88B3 = 2-2-222-2-222-2-22 = 8-8-882-2-22= 32-32-3232B4 = 2-2-222-2-222-2-222-2-22 = 128-128-128128Hence, we arrive at the general expression Bn = (2n-1)(bn)(Y).Note that the procedure we employ is consistent with that used for matrix A, even up to the checking for validity.For the final task, we are given a impudent matrix M = a+ba-ba-ba+b. We must show that M = A + B andM2 = A2 + B2 apply the algebraic method. Again, define A and BA=a1111=aaaa B=b1-1-11=b-b-bbA+B= aaaa+b-b-bb=a+ba-ba-ba+bM= a+ba-ba-ba+bM=A+B equation 1We have proven the origin relationship to be true. Now we must proceed to showing M2 = A2 + B2 .From equation 1, M = A + B, therefore, by substitution, this is the same as saying M2 = (A + B)2. Previously we have shown that and expression of this form X+Yn= Xn+ Yn. HenceM2=a+ba-ba-ba+ba+ba-ba-ba+bM2=a+ba+ba-ba-ba-ba+ba-ba+ba-ba+ba-ba+ba-ba-ba+ba+bM2=2a2+2b22a2-2b22a2-2b22a2+2b2A2=2a22a22a22a2 and B2=2b2-2b2-2b2-2b2A2+ B2=2a2+2b22a2-2b22a2-2b22a2+2b2M2 = A2 + B2 equation 2Recall that A = aX and B = bY. We now produce a general statement for Mn in name of aX and bYMn = An + Bn or by substitution, Mn= (aX)n + (bY)nfurthermore,Mn = anXn + bnYnVerifying this equation, we try using a=2, b=3, and n=2A=2222 andB=3-3-33If we use, (A+B)2=5-1-155-1-15=25+1-5-5-5-525+1=26-10-1026Now, using the general statementM2=22X2+ 32Y2=222222222222+232-232-232232=8+188-188-188+18... Also given were matrices A and B, defined as aX and bY, respectively. Note that a and b are constants. First, recall that when multiplying constants to any matrix, we simply multiply the constant with every element of t he matrix. To deckOnce again, the general expression is shown valid. It is also important to note that this general statement will solo yield results for values of n0. Matrices can not be raised to negative exponents.
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