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Matrix Binomials IA

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Introduction

Matrix Binomials

Type 1 Internal Assessment

Matrices are rectangular arrays of numbers that are arranged in rows and columns, however the regular rules of algebra do not apply.

Let X=image01.png and Y=image02.png and calculate X2, X3, X4 ; Y2, Y3, Y4.

X2=image34.pngimage01.png ×image01.png = image24.pngimage00.png

X3=image76.pngimage01.png ×image01.png × image01.png =image03.pngimage00.png

X4=image14.pngimage01.png ×image01.png × image01.png × image01.png = image05.pngimage00.png

Because matrices do not follow the algebraic rules of exponents, one can not simply distribute the exponent for each matrix value. Instead the matrix must be multiplied by itself however many times the exponent says. So for example, for X2, the matrix X must by multiplied with itself two times.

The pattern that has emerged is that with every increasing power the matrix value increases with an exponential power of 2.

21 is equal to 2 showed by the matrix image24.png. 22 is equal to 4 showed by the matrix image03.png and similarly 23 is equal to 8 represented in the matrix image05.png.

Y2=image06.pngimage02.png ×image02.png = image07.pngimage00.png

Y3=image09.pngimage02.png ×image02.png×image02.png = image10.pngimage00.png

Y4=image12.pngimage02.png ×image02.png×image02.png×image02.png = image13.pngimage00.png

The pattern is very similar to the one above except that all the negatives in the original matrix will also become negatives.

Based on the results above we can conclude that for each exponent value Xn the matrix value will result in Xn-1 and the same goes for Yn resulting in Yn-1.

...read more.

Middle

The matrix values are all multiplied by 3.

A=5X

A2= 5image34.png-5image24.png = image81.pngimage00.png

A3= 5image76.png                        5image03.png = image82.pngimage00.png

A4= 5image14.png     5 image05.png = image83.pngimage00.png

The matrix values are all multiplied by 5.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

B= -1Y

 B2= -1image06.png--1 image16.png = image24.pngimage00.png

B3= -1image09.png                -1 image17.png        = image03.pngimage00.png

B4=-1image12.png     -1 image04.png = image05.pngimage00.png

When multiplied by -1, the matrix values become positive but remain the same. The reverse results of when matrix A was multiplied by -1.

B= Y

 B2= -½image06.png-image07.png = image08.pngimage00.png

B3= -½image09.png                -½ image10.png        = image11.pngimage00.png

B4=-½image12.png     -½ image13.png = image15.pngimage00.png

When multiplied by -1, the matrix values are divided by two (or halved) and all positive and negative the signs are switched in relation with the original matrix.

B= ½Y

 B2= ½image06.png-½ image07.png = image02.pngimage00.png

B3= ½image09.png                ½ image10.png        = image07.pngimage00.png

B4image12.png     ½ image13.png = image10.pngimage00.png

When multiplied by -1, the matrix values are divided by two (or halved) and all positive and negative the signs remain the same as in the original matrix (key difference when divides by -½).

B= 1Y

 B2= 1image06.png-1 image16.png = image16.pngimage00.png

B3= 1image09.png                1 image17.png        = image17.pngimage00.png

B4=1image12.png     1 image04.png = image04.pngimage00.png

When multiplied by 1, the matrix values remain negative. The reverse results of when matrix A was multiplied by 1.

B= 3Y

 B2= 3image06.png-3 image16.png = image18.pngimage00.png

B3= 3image09.png                3 image17.png        = image19.pngimage00.png

B4=3image12.png     3 image04.png = image20.pngimage00.png

When multiplied by 3, the matrix values remain negative.

B= 3Y

 B2= 3image06.png-3 image16.png = image18.pngimage00.png

B3= 3image09.png                3 image17.png        = image19.pngimage00.png

B4=3image12.png     3 image04.png = image20.pngimage00.png

The matrix values are all multiplied by 3.

...read more.

Conclusion

image63.png= (-9×image01.png)3  + (11×image02.png)3

image63.png= image64.png  + image59.png

image65.png= image66.png  + image62.png

image65.png= image65.png

In all these case the general statement is correct and works. However, the scope of the general statement is limited to positive exponents as well as the matrices X and Y. We have only tested a very small sample of the different matrix values and its infinite combinations for the matrices X and Y.

If we were to use different matrix values for X and Y the general statement would not apply:

Let n=3, a=9 and b=11 but let X = image24.png and let Y = image07.png

image57.png= (9×image24.png)3  + (11×image07.png)3

image57.png= image67.png  + image68.png

image60.png= image69.png  + image70.png

image60.pngimage71.png

As one can see, the general statement does not apply when the values for the matrices X and Y are changed and only works correctly when:

X=image01.png and Y=image02.png  .

To reach my general statement I used my previous findings.

I found that An = a×Xn and that Bn = b×Yn . This in turn allowed me to come to the conclusion that (A+B)n = [(a×X) + (b×Y)]n.

Using the knowledge that M=image28.png and that M=A+B and that M2=A2+B2 I could reach the following conclusion:

Mn=An+Bn

Mn=(a×X)n + (b×Y)nsubstituting A for a×X and B for b×Y

This is how I reached by general statement.

...read more.

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