I require to find $a$ and $b$ in the following: $$\begin{bmatrix}0\\-40\\25\end{bmatrix}=a\begin{bmatrix}1\\-4\\3\end{bmatrix}+b\begin{bmatrix}1\\4\\-2\end{bmatrix}$$
Now I moved the $b[\ ]$ group to the other side and now here I am guessing but I would require to find the a matrix inverse so I can isolate a but I seem to be missing something or I do not fully understand it. Can someone help me with what steps is needed to take.
It’s just a matter of solving this system:
$$\left\{\begin{align*} &a+b=0\\ &-4a+4b=-40\\ &3a-2b=25 \end{align*}\right.$$
Of course it could turn out to be inconsistent, though in this case it doesn’t.
You can set it up as a matrix problem, but that strikes me as a bit of overkill. The system above can be written
$$\begin{bmatrix}1&1\\-4&4\\3&-2\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}0\\-40\\25\end{bmatrix}$$
and solved in any of the usual ways.
You have the equations \begin{align*} 0 &= a+b\\ -40 &= -4a +4b\\ 25 &= 3a-2b\\ \end{align*} Solve it :)
