There is a matrix given and I have to find its transformation matrix with basis r. The question is bellow .i have tried with the formula B=MAM^-1
F(1,1,0)=(3,4,3), F(1,0,1)=(-1,4,-1), F(0,1,1)=(0,2,-2). Find the matrix of F in the standard basis of R3.
You’re correct in that a change of basis is involved here, but the formula that you’ve either remembered or looked up doesn’t really apply to this situation.
I like to think about this type of problem in terms of “input” and “output” bases: when you construct a matrix for a linear transformation, you have to take into account the bases of both the domain—the input basis—and codomain—the output. In the following, $[v]_{\mathcal B}$ will denote the coordinate vector of $v$ relative to $\mathcal B$. The notation $[T]_{\mathcal B'}^{\mathcal B}$ represents the matrix of $T$ with ordered bases $\mathcal B$ for its input and $\mathcal B'$ for its output. (In some sources you’ll see a similar notation that swaps the input and output bases, but I like this convention because it lets you visually “cancel” an adjacent input and output.) The symbol $\mathcal E$ represents the standard basis of the relevant vector space. In this notation we then have the change-of-basis formula $$[v]_{\mathcal B'} = [\operatorname{id}]_{\mathcal B'}^{\mathcal B} [v]_{\mathcal B}.\tag{*}$$ Notice how the upper $\mathcal B$ “cancels” the lower $\mathcal B$ to its right. I’ll also make use of the identity $[\operatorname{id}]_{\mathcal B'}^{\mathcal B} = \left([\operatorname{id}]_{\mathcal B}^{\mathcal B'}\right)^{-1}$.
Using this notation, your task is to find $[F]_{\mathcal E}^{\mathcal E}$. You’ve been given the values of $F$ for three different vectors. Happily, they’re linearly independent (verify this!), so you can use them as a basis $\mathcal B$ of $\mathbb R^3$. Recalling that the columns of a transformation matrix are the images of the basis vectors, you can form the matrix $$A = \begin{bmatrix} 3&-1&0 \\ 4&4&2 \\ 3&-1&-2 \end{bmatrix}.$$ Now take a look at what bases are being used for this matrix. The input basis is, of course, $\mathcal B$, but the given values of $F$ are in the standard basis, so $A=[F]_{\mathcal E}^{\mathcal B}$. So, the output is already in the correct basis for $[F]_{\mathcal E}^{\mathcal E}$, but the input needs to be adjusted. Applying the change-of-basis formula (*) gives $$[F]_{\mathcal E}^{\mathcal E} = [F]_{\mathcal E}^{\mathcal B} [\operatorname{id}]_{\mathcal B}^{\mathcal E} = [F]_{\mathcal E}^{\mathcal B} \left([\operatorname{id}]_{\mathcal E}^{\mathcal B}\right)^{-1}.$$ Now, $[\operatorname{id}]_{\mathcal E}^{\mathcal B}$ is the matrix with the elements of $\mathcal B$ as its columns, so in the end you need to compute $$[F]_{\mathcal E}^{\mathcal E} = \begin{bmatrix} 3&-1&0 \\ 4&4&2 \\ 3&-1&-2 \end{bmatrix} \begin{bmatrix} 1&1&0 \\ 1&0&1 \\ 0&1&1 \end{bmatrix}^{-1}.$$