Given x' = acose + bsin0 y' = -asind + bcos0 using Cramer's rule, find a in term of x'and y'

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Q: what happened to the 1(1-lambda) part in the first line

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Q: 2. The Hamiltonian operato represented as

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Q: Solve using Cramer's rule. 2x+8y = 11 4x+6y = 3 %3D %3D

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Q: 2x+ 3y-z =1 4x + y-3z = 11 3x-2y +5z = 21 | %3D

A: Solution:

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Q: 2x+ 3y-z =1 4x + y-3z = 11 3x- 2y +5z =21

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Q: Use of Cramer's Rule 2x, – 3x, = 5 X +x, = 5 [1 1][x, X

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Q: Utu = c²uxx , u(0, t) = 0, u(1, t) = 0, u(x,0) = 0 U¿(x,0) = x² – 2x, 00 t> 0 t > 0 0sx<1.

A: Given P.D.E

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Q: In applying Cramer's rule, what does it mean if D = 0?

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Q: Solve for z using cramer's rule.  2x-y-z=4 4x+y+z=2 3y+z=2

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