Let $b : \Bbb{R}^2 \times \Bbb{R}^2 \to \Bbb{R}$ be the bilinear form defined by $$b((x_1,x_2),(y_1,y_2)) = x_1 y_1 - 2x_1 y_2 + x_2 y_1 + 3x_2 y_2.$$ Find the $2 \times 2$ matrix $B$ of $b$ relative to the basis $U=\{u_1,u_2\} = \{(0,1),(1,1)\}$
From what I understand, if $B=[b_{ij}]$ is the matrix representation, then $b_{ij} = b(u_i,u_j)$. If this is so, then I should get $$b_{11} = b((0,1),(0,1)) = 0 \cdot 0 - 2 \cdot 0 \cdot 1 + 1 \cdot 0 + 3 \cdot 1 \cdot 1 =3,$$ $$b_{12} = b((0,1),(1,1)) = 4,$$ $$b_{21} = b((1,1),(0,1)) = 1,$$ and $$b_{22} = b((1,1),(1,1)) = 3$$
However, the answer is $B = \begin{bmatrix} 0 & 4 \\ -1 & 3 \\ \end{bmatrix}$. Where did I go wrong?
If a vector has coordinates $[a,b]$ in $U$, then the vector is $a(0,1)+b(1,1)=(b,a+b)$.
Let's compute $b$ in coordinates in $U$. I will be using $[\cdot,\cdot]$ to denote coordinates in $U$ and $(\cdot,\cdot)$ to denote the vectors themselves or, since they look the same, coordinates in $\{(1,0),(0,1)\}$
\begin{align}b([a_1,a_2],[b_1,b_2])&=b((a_2,a_1+a_2),(b_2,b_1+b_2))\\ &=a_2b_2-2a_2(b_1+b_2)+(a_1+a_2)b_2+3(a_1+a_2)(b_1+b_2)\\ &=3a_1b_1+4a_1b_2+2a_2b_1+3a_2b_2\\ &=a_1(3b_1+4b_2)+a_2(2b_1+3b_2)\\ &=\left[\begin{matrix}a_1&a_2\end{matrix}\right]\left[\begin{matrix}3b_1+4b_2\\2b_1+3b_2\end{matrix}\right]\\ &=\left[\begin{matrix}a_1&a_2\end{matrix}\right]\left(\begin{matrix}3&4\\2&3\end{matrix}\right)\left[\begin{matrix}b_1\\b_2\end{matrix}\right] \end{align}