Determine $a, b\in\mathbb R$ based on a matrix I have a linear transformation $f:\mathbb R^3\to \mathbb R^3$ and I have the matrix of that transformation in its base $(1, 0, 0)$, $(0, 1, 0)$, $(0, 0, 1)$.
\begin{bmatrix}
a & 1 & 1 \\
1 & b & 1 \\
1 & 2b & 1 \\
\end{bmatrix}
Now I need to find $a$ and $b$ such that the following is correct $(4, 3, 4)\in\operatorname{Im}(f)$.
I have tried some things, but it has led me nowhere. Could someone please give me an idea?
 A: Let $\ F$ be the matrix of the linear transformation, and $\ x=(u,v,w)\in{R^3}$ 
You need to find the values of a,b so the system Fx=(4,3,4) has the answer.
A: Let $y = (4, 3, 4)$, and let $A$ be the matrix of the transformation $f$ that you describe. If $y \in \mathrm{Im}(f)$, that means that there is some $x \in \mathbb{R}^3$ for which $f(x) = y$. You could try to solve the matrix equation $Ax = y$ directly (using Gaussian elimination for example), but in this case you'll find that you have free parameters because there are a large number of matrices for which $y \in \mathrm{Im}(f)$. The reason is simple. If you wanted to find a solution to an equation $f(x) = y$ in general, you would like to find an inverse function such that $x = f^{-1}(y)$, which in this case corresponds to finding a matrix inverse. So that means that all you have to show in this case is that the matrix $A$ is invertible. Since there are an infinite set of values for $a$ and $b$ that make the matrix invertible, you just have to select values of $a$ and $b$ that keep the rows (and columns) linearly independent. The easiest way to check this for a small matrix is just by computing the determinant and ensuring that it is nonzero, so all that you have to do here is find values of $a$ and $b$ to make the determinant nonzero.
