VERY EASY LINEAR ALGEBRA QUESTION PLEASE HELP

Question

Matris a is shown upon. For every natural number n show and solve the below.

Answer 1

First, the plural of MATRIX is MATRICES

will use STRONG INDUCTION to prove

the statement about this MATRIX

The statement clearly holds for N=1.

This is left for you to verify.

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For n=2, the entries of the MATRIX are:

1+6n=13 4(2) = 8

-9*2 = -18 1 - 6(2) = -11

and

[7 4] [ 7 4]

[-9 -5] [ -9 -5]

= [ 49-36 28-20 ]

[ -63+45 -36+25]

= [ 13 8]

[ -18 -11]

So the statement holds for these MATRICES when n=2

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For some natural number N, the Induction Hypothesis says

that is GIVEN

that A^n = [ 1 +6n 4n]

[ -9n 1 - 6n]

Then A^(n+1) = A^n*A =

*********** by substitution **************

[ 1+6n 4n ] [ 7 4 ]

[ -9n 1-6n ] [ -9 -5 ]

******** multiplies the MATRICES ***************

= 7(1+6n) - 9(4n) 4(1+6n) - 5(4n)

7(-9n) - 9(1-6n) 4(-9n) - 5(1-6n)

************* distributive *********************

= 7 + 42n - 36n 4 + 24n - 20n

-63n - 9 + 54n -36n - 5 + 30n

*********** combines like terms **********************

= [7 + 6n 4 + 4n]

[-9n - 9 -6n - 5]

****substitutes 7=6+1, factors out 4, factors out -9, and -5 = -6+1

= [ 1 + 6 + 6n 4(n+1) ]

[ -9(n+1) -6n - 6 +1 ]

******** factors out 6 and -6 from diagonal entries

= [ 1 + 6(n+1) 4(n+1)

-9(n+1) -6(n+1)+1

****** Let V = n+1 ***************************

= 1 + 6V 4V

-9V -6V+1

which is the exact same pattern in terms of V=n+1.

This completes the proof

No this was not easy at all.