Let me ask again, as clearly my previous post people did not understand what I was asking. Just to clarify this is a middle school question which kind of made me forget alittle on transformations.
Question: Let Transformation A be a $-90$ degrees rotation with respect to the center $(0,0)$. Let Transformation B be a translation of $-4$ units in the x - axis and $+2$ units in the y - axis. Describe a single transformation when transformation B is applied first then Transformation A.
This is literally a question from middle school which I was not entirely sure about, as my immediate approach will be to write the two transformations in terms of matrix and multiply them together to get the single transformation. Upon doing so I get:
$$ A= \left[ {\begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array} } \right] $$ $$ B= \left[ {\begin{array}{cc} -4 \\ 2 \\ \end{array} } \right] $$ $$ A\times B= \left[ {\begin{array}{cc} 2 \\ 4\\ \end{array} } \right] $$
Which describes the single transformation after combining transformation B and A, which can be described as a translation vector. However this is my question:
1) Can you combine the two transformation through multiplication like what I have done in this example?
2) Why does this single transformation not work when we apply the transformations B then A on to any arbitrary coordinate say $(1,1)? $ For instance if I apply transformation B then A to $(1,1)$ I get: $(1,1)\rightarrow (-3,3) \rightarrow (3,3)$. However if we apply the single trasnformation I obtained above of $\left[ {\begin{array}{cc} 2 \\ 4\\ \end{array} } \right]$ to the coordinate $(1,1)$; I obtain $(1,1)\rightarrow (3,5)$.
Hence my question is what am I doing wrong here, or where is the gap in my understanding?
Answer to Qn $1$:
If the transformations you're considering are both linear transformations, then you can "combine them" by multiplying the matrices. (The technical term for "combine" is function composition)
Answer to Qn $2$:
First of all translation by $-4$ in $x$-direction and $2$ in $y$-direction is NOT a linear transformation, hence it cannot be represented by a matrix. Even if it was a linear transformation, it would have to be represented by a $2\times 2$ matrix, not a $2\times 1$ matrix $B$ like you have written. This is why you're getting troubles with the matrix $B$.
The Fix:
What you need to do is think of function composition. Define $T:\Bbb{R}^2 \to \Bbb{R}^2$ by \begin{align} T(x,y) = (x-4, y+2). \end{align} So, $T$ is the desired translation function. Next, define the "rotation by $-90^{\circ}$" function $R: \Bbb{R}^2 \to \Bbb{R}^2$ by \begin{align} R(x,y) = (y,-x) \end{align} Note that rotation is a linear transformation, and you have correctly defined the matrix of it using $A$. I obtained the function $R$ by multiplying the matrix $A$ with the column vector $\begin{pmatrix} x \\ y \end{pmatrix}$ (but since I'm thinking of everything in terms of functions, it doesn't matter whether I use rows or columns).
Now, if you want to first translate then rotate, then you should consider the composition $R \circ T$, which is \begin{align} (R \circ T)(x,y) &:= R(T(x,y)) \\ &:= R(x-4,y+2) \\ &:= \left( y+2, -(x-4) \right) \\ &= (y+2, 4-x) \end{align}
Now, if you apply this to the specific vector $(1,1)$ then you get $(R \circ T)(1,1) = (3,3)$.