- Level: International Baccalaureate
- Subject: Maths
- Word count: 1370
Matrices SL Type 1
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Introduction
Mathematics Portfolio Type I
The beginnings of matrices and determinants go back to the 2nd century B.C. although matrices can be seen back to the 4th century B.C., however, it was not until near the end of the 17th century that the ideas reappeared and developed further. J.J. O'Connor and E.F. Robertson stated that the Babylonians studied problems which used matrices, as well as the Chinese using matrices early on. All throughout history, the use of matrices has helped mankind progress.
A matrix function such as X = and Y = can be used to figure out expressions. By calculating X2, X3, X4; Y2, Y3, and Y4 the values of X and Y can be solved for. = X2. By following rules of multiplying matrices, this can be shown as = . Using X2 we can conclude that X2 = or . One can generalize a statement of a pattern that develops as the matrix goes on. The expression is as follows, Xn = . The number 2 in the matrix comes from when the product of Xn is solved for. The value of 2n-1 is twice the value of Xn. The variable n represents what power the matrix is to, such as n = 2, 3, 4. We can now solve for the rest of the values of Xn.
Middle
The expression of (X+Y)n = 2n fits and is proven by the examples above.
Let A = aX and B = bY wherea and b are constants. Let us use different values of a and b to calculate the values of A2, A3, A4; B2, B3, B4.
a = 4 for A2
4∙ 4= =
Solving the first example, we can create an expression that should work. The expression for the value can be written as an2n-1X. The 2n-1 comes from multiplying it with X, which is shown earlier on, Xn = . The an comes from the constant of a which we solve the value of An and raise it to the n power.
Continuing using a = 4, A3 will now be solved for, using the expression.
A3
43∙23-1 = 64∙22 = 256 =
A4
44∙24-1 = 256∙23 = 2048 =
The same expression can be used for when B = bY, bn2n-1Y. The 2n-1 comes from multiplying it with Y as shown earlier, Yn = . bn comes from the constant of b we solve the value of Bn and raise it to the n power. Since a and b need to be different constants, the value of b in this example will equal 5.
b =5
B2
52∙22-1 = 25∙21 = 50 =
B3
53∙23-1 = 125∙22 = 500 =
B4
54∙24-1 = 625∙23 = 5000 =
Keeping in mind what A and B are, we can now find the expression for (
Conclusion
The general statement is Mn = an2n-1X +bn2n-1Y. One would get to this general statement algebraically when multiplying A or B exponentially. The 2 in the equation is twice as much as the square numbers and that is where the number 2 comes from in the general statement. Since 2 receives less than the power n and this is where the section of n-1 arrives from in the equation Mn = an2n-1X +bn2n-1Y. When M = A + B, A = aX and B = bY are given earlier on in the paper from their expressions that were found by solving various problems. An example would be;
a= 1 b =3, n = 2
M2 = 12∙21-1 + 32∙21-1 = 1∙1 + 9∙1 =
+ = .
This is the algebraic step and method for solving the general statement of Mn = an2n-1X +bn2n-1Y
Works Cited
O'Connor, J.J. and Robertson. "Matrices and determinants." Matrices and determinants
Feb 1996. 13 Feb 2008 .
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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