Let f : \mathbb{r}^{2} \to \mathbb{r}^{2} be the linear transformation defined by f(\vec{x}) = \left[\begin{array}{cc} 4 & 3\cr 2 & 1 \end{array}\right] \vec{x}. let \begin{array}{lcl} \mathcal{b} & = & \lbrace \left< 1,-1\right> , \left< 2,-3\right> \rbrace, \\ \mathcal{c} & = & \lbrace \left< 1,1\right> , \left< -2,-1\right> \rbrace, \end{array} be two different bases for \mathbb{r}^{2}. find the matrix \lbrack f \rbrack_{\mathcal{b}}^{\mathcal{c}} for f relative to the basis \mathcal{b} in the domain and \mathcal{c} in the codomain.

Step-by-step explanation:

Remember that is the matrix whose columns are the images under f of the vectors of the basis written in the coordinates of the basis . Then we have to do the following:

For (1), note that any vector can be written as Therefore, the -- coordinates of are

For (2), we calculate:

Now we use the results obtained in steps (1) and (2) for finding and as requested in (3):

Therefore, the matrix for f relative to the basis in the domain and in the codomain is given by

answer: i think the answer is etheir c or d but i don't know which one but have a good day!

step-by-step explanation:

hello :

70y^8 + 30y^6 = 2 ( 35 y^8 + 15y^6)

the lenght is : 35 y^8

the width is : 15y^6