Matrices!

Algebra Level 5

Let $x$ be the number of diagonal matrices $A$ of real entries and order $3\times3$ such that $A^7 + 3A^5 + 7A = 11I$ .

And let $y$ be the number of diagonal matrices $B$ of complex entries with at least one non-zero real entry and order $3\times 3$ such that $B^5 = I$ .

Find the value of $y-55x$ .

Clarification: $I$ is an identity matrix of order 3.

The answer is 6.

1 solution

Avi Solanki
Dec 4, 2016

x^7 +3x^5+7x-11=0 This equation by trial and error has a root x=1 and will only have a one root as it is an increasing function .therefore number of matrices A =1 and also to note is would be an identity matrix of order 3.

coming to B: Y^5=1 . 5th root of unity . and since we need atleast one real entry .therefore that entry would be '1'. now total number of ways to fill a 3x3 matrix with 5 choices is 5^3 and we subtract the number of ways in which no real entry is filled i.e 4^3 therefore y=5^3-4^3=61

Hence the answer .

How do you know that?

Pi Han Goh - 4 years, 6 months ago

I agree with you Pi... the answer is right, but could you (Avi) explain a little more your solution,please?

Guillermo Templado - 4 years, 6 months ago

i have elaborated the solution sir .please check .have a tight schedule of exams due to which i could not respond earlier .sorry for the inconvenience . Regards. :) @Pi Han Goh @Guillermo Templado

avi solanki - 4 years, 6 months ago

No problem, don't worry... Now it's better. I would add that the product of two diagonals matrices is a diagonal matrix whith the number (entry) at the same row and column being the product of the numbers(entries) of the others two matrices at the same row and column, and I wouldn't use trial, error... Anyway, it's better now... Thank you.

Guillermo Templado - 4 years, 6 months ago

Matrices!

The answer is 6.

1 solution

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