I have the system:
$$-4 + λ = -3a + at\\ 1+2λ = -a + at\\ 3λ = 3a-at$$
but whenever I try to substitute, I end up getting lots of fractions that are hard to work with. By summing the 2nd and 3rd equations I get:
$$a = \frac{1+5\lambda}{2}$$
But then it's hard to work with it, no mtter where I substitute. Should it be like this or I'm doing something wrong?
PS: I can't use any theorem about determinants...
Write this as the linear system $$ \left[\begin{array}{ccc} -3 & 1 & -1\\ -1 & 1 & -2\\ 3 & -1 & -3 \end{array}\right]\left[\begin{array}{c} a\\ at\\ \lambda \end{array}\right]=\left[\begin{array}{c} -4\\ 1\\ 0 \end{array}\right] $$ which has solution $$ \left[\begin{array}{c} a\\ at\\ \lambda \end{array}\right]=\left[\begin{array}{c} 3\\ 6\\ 1 \end{array}\right]. $$ That is, $a=3$, $t=2$, and $\lambda=1$.