Certainly you do not need matrix notation to solve your system of equations. Look at the two equations where only $a$ and $b$ occur. The first of these equations says that $b=4a-1$. Substitute for $b$ in the second equation, and solve for $a$. We get $a=1$. Now that you know $a$, you also know $b$.
Finally, use one of the two equations that involve $c$, and the known values of $a$ and $b$, to solve for $c$. Then check whether the $a$, $b$, and $c$ that you have found satisfy the fourth equation. They very might not: you have four conditions (and linear equations) and only three unknowns.
Remark: The person who made up the problem carefully chose numbers such that the solution to the first three equations we considered is also a solution of the fourth equation. If you try to make up a similar problem by choosing "random" values for the slope of the normal at $A$ and the gradient at $B$, you will most of the time end up with a system of equations that has no solution.
The problem could have been made a little harder. For example, instead of being told the gradient at $B$, you could have been asked what that gradient is.