# fp1 matrix q

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1. fp1 matrix q
can i have help with part b. i used the equation for the det, ad-bc on R^2 and i get
a^2(b^2-4b+4)=15
i get b as 2 if i put the bracket part to zero, which is obviously wrong!

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2. Re: fp1 matrix q
In part a I got the matrix (a^2+2a 2a+2b )
(a^2-ab 2a+b^2 )

Because the transformation is an enlargement with scale factor 15, we take the unit square (1 0) and multiply it by
(0 1)
15 to give (15 0)
(0 15)
This matrix here is equal to (a^2+2a 2a+2b )
(a^2-ab 2a+b^2 )
so a^2+2a=15
And so by solving the simple quadratic equation a^2+2a-15=0 we get (a+5)(a-3) a=-5 or a=3 and because the question says a is positive it has to be a=3.
We can now find b by substituting 3 into 2a+2b=0 and we get b=-3

You can check your answer by putting a and b in the last two equations and you should get the right answers.

Hope this was clear enough

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Last updated: April 8, 2012
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