- Level: International Baccalaureate
- Subject: Maths
- Word count: 1182
Matrix Binomials
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Introduction
John Vu
000948-111
Matrix Binomials
A matrix is a rectangular array of numbers arranged in rows and columns. Numbers or letters inside the brackets in matrices are called entries. Matrices can be added, subtracted, multiplied, divided, and also raised to a power. A common matrix can look like this where a, b, c, and d are the entries.
Given the matrices X = and Y = I calculated X2, X3, X4; Y2, Y3, Y4.
Using my GDC (graphic display calculator) I evaluated these matrices.
X2 = Y2 =
X3 = Y3 =
X4 = Y4 =
After calculating the powers of X and Y, I observed my solutions. Looking at this I saw a trend that emerged as I increased the power of the matrix. The trend was increasing the power of the matrix by one, caused the product to double its previous solution. When X is to the second power, the entries of the solution are all 2’s; when X is to the third power the entries are all 4’s; when X is to the forth power the entries are all 8’s, and so on.
Middle
A4 = B4 =
By considering integer powers of these matrices, I came up with an expression to also solve these matrices. For matrix A the expression is An = (an) 2(n-1) 2(n-1) and matrix B the expression Bn = (bn) 2(n-1) -(2(n-1)) 2(n-1) 2(n-1)
-(2(n-1)) 2(n-1)
Using my GDC I solved these matrices.
(A + B)1 = (A + B)3 =
(A + B)2 = (A + B)4 =
After solving these equations, I generated an expression to solve the matrix for
(A + B). The expression to solve the matrix is (A + B)n = (an Xn) + (bn Yn). To make sure the expression worked I plugged in number 2 and 4 for the value of n.
(A + B)2 = (22 X2) + (32 Y2) (A + B)4 = (24 X4) + (34 Y4)
Using GDC (A + B)2 = Using GDC (A + B)4 =
Considering that M = , this expression can also be derived into
M = A + B and M2 = A2 + B2. By proving the expression M = A + B, A and B needs to be substituted for aX and bY as well as keeping the constants as a variable. This will prove the expression M = through M = A + B.
M = aX + bY
= a + b
= +
=
Conclusion
Let a = 0, b = 0, n = 3
M3 = (0X)3 + (0Y)3
M-3 = +
=
The limitation for the expression Mn = (aX)n + (bY)n is that ‘n’ cannot contain a negative exponent nor a decimal value or a fraction because if we multiply an exponent raised to a negative number it would make the value flip. However, both of the constants ‘a’ and ‘b’ can equal to any set of real numbers. Therefore the limitations and scope are:
To conclude this assignment, last I will explain how I arrived at my general statement using an algebraic method.
Let n = 2, a = a, b = b, X = , Y =
(aX)n + (bY)n =
a2+b2 =
2a2 2a2 + 2b2 -2b2
2a2 2a2 -2b2 2b2 =
2a2 + 2b2 2a2 – 2b2 = (a2 + 2ab + b2) + (a2 + 2ab + b2) (a2 – b2) + (a2 – b2)
2a2 – 2b2 2a2 + 2b2 (a2 – b2) + (a2 – b2) (a2 + 2ab + b2) + a2 + 2ab + b2)
2a2 + 2b2 2a2 – 2b2 2a2 + 2b2 2a2 – 2b2
2a2 – 2b2 2a2 + 2b2 = 2a2 – 2b2 2a2 + 2b2
Therefore the equation Mn = (aX)n + (bY)n equals with Mn =
Arriving at this general statement, it can help solve for the value of M in any equation.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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