Math Portifolio Matrix binomials Problem Example | Topics and Well Written Essays - 750 words

Portifolio Matrix binomials - Math Problem ExampleBased on these computations, we derive the general expression for AnAn = (2n-1)(an)(X)We must check 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. straight, take 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). office that the procedure we used is consistent with that used for matrix A, even up to the checking for validity.For the final task, we are given a new matrix M = a+ba-ba-ba+b. We must show that M = A + B andM2 = A2 + B2 using 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 first 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 instantaneously produce a general statement for Mn in terms of aX and bYMn = An + Bn or by substitution, Mn= (aX)n + (bY)nfurthermore,Mn = anXn + bnYn confirm 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... too 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 both element of the ma trix. To illustrateOnce again, the general expression is shown valid. It is also important to note that this general statement will only yield results for values of n0. Matrices evict not be raised to negative exponents.