The function $10cos(5t) + 8sin(5t)$ lies in the complex vector space with basis $e^{5it}$ and $e^{-5it}$. Find the vector with respect to that basis.

Using matrices, I got that

$$10cos(5t) + 8sin(5t) = begin{pmatrix} e^{5it} & e^{-5it} end{pmatrix} begin{pmatrix} A \ B end{pmatrix}.$$

Now, letting $LHS = begin{pmatrix} 10cos(5t) + 8sin(5t) end{pmatrix}$, $Emat = begin{pmatrix} e^{5it} & e^{-5it} end{pmatrix}$, and $C = begin{pmatrix} A \ B end{pmatrix}$, I have $$ LHS = Emat * C.$$

Since the matrices need to be square to be invertible, I thought I could do $$C^{-1} = LHS^{-1}*Emat,$$ then take the inverse of $C^{-1}$. However, the answer I got (using matlab) was wrong, and I’m wondering where I went wrong, or if there was another approach that I didn’t think of. Any help would really be appreciated!

$$begin{align*}Ae^{5it}+Be^{-5it}&=Acos(5t)+Aisin(5t)+Bcos(-5t)+Bisin(-5t)\

&=(A+B)cos(5t)+(A-B)isin(5t),,

end{align*}$$

so $A+B=10$ and $(A-B)i=8$, or $A-B=-8i$.

Solving simultaneously gives

$$A=5-4i, B=5+4i.$$

The answer by A. Goodier is good and what I would have written had it not already been written. But you seemed to want to solve this problem by inverting matrices, and you recognized that matrices need to be square to be invertible. Here is a solution along those lines.

The matrix $M=pmatrix{1&1\i&-i}$ converts coordinates

from the basis ${e^{5it},e^{-5it}}$ to the basis ${cos 5t,sin 5t},$

because the first column tells us $e^{5it}=cos 5t+isin5t,$

and the second tells us $e^{-5it}=cos5t-isin5t$.

We can then solve $pmatrix{10\8}=Mpmatrix{A\B}$ as $pmatrix{A\B}=M^{-1}pmatrix{10\8}=pmatrix{1&1\i&-i}^{-1}pmatrix{10\8}$

$=dfrac{pmatrix{-i&-1\-i&1}}{-2i}pmatrix{10\8}=pmatrix{frac12&-frac i2\frac12&frac i2}pmatrix{10\8}=pmatrix{5-4i\5+4i}.$