For $α ∈ R$, let $q(x_1, x_2) =x_1^2 + 2αx_1x_2 + \frac{1}{2}x_2^2$ for $(x_1,x_2) \in R^2$
(a) Take $α = \frac14$ and let B be the symmetric matrix of q with respect to the basis{${(1, 0),(1, 1)} $}of $R^2$. Then the entry in row $1$, column $2$ of B equals____????
my attempst : $q = \begin{bmatrix} 1& \frac{1}{4}\\\frac{1}{4}&\frac{1}{2}\end{bmatrix}$
now the entries in row 1 is $1$ and $\frac 14$ and column is $\frac 14$ and $\frac{1}{2}$
Is its corrects ??????
Respect to the standard basis
$$q=x^TAx$$
with
$$A = \begin{bmatrix} 1& \frac{1}{4}\\\frac{1}{4}&\frac{1}{2}\end{bmatrix}$$
let $M = \begin{bmatrix} 1& 1\\0&1\end{bmatrix}$ the matrix for the change of basis $x=My$ then we have
$$q=x^TAx=(My)^TAMy=y^TM^TAMy$$