Based on these computations, we derive the general expression for An:
An = (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.
Now, take b=2; B = 2-2-22:
B2 = 2-2-222-2-22 = 8-8-88
B3 = 2-2-222-2-222-2-22 = 8-8-882-2-22= 32-32-3232
B4 = 2-2-222-2-222-2-222-2-22 = 128-128-128128
Hence, we arrive at the general expression Bn = (2n-1)(bn)(Y).
Note 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 and
M2 = A2 + B2 using the algebraic method. Again, define A and B:
M=A+B equation 1
We 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. Hence:
A2=2a22a22a22a2 and B2=2b2-2b2-2b2-2b2
M2 = A2 + B2 equation 2
Recall that A = aX and B = bY. We now produce a general statement for Mn in terms of aX and bY:
Mn = An + Bn or by substitution, Mn= (aX)n + (bY)n furthermore, Mn = anXn + bnYn
Verifying this equation, we try using a=2, b=3, and n=2:
A=2222 and B=3-3-33
If we use, (A+B)2=5-1-155-1-15=25+1-5-5-5-525+1=26-10-1026
Now, using the general statement:
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 the matrix. To illustrate:
Once again, the general expression is shown valid. It is also important to note that this general statement will only yield results for values of n>0. Matrices can not be raised to negative exponents.