Vector Fields — Question 7

Question 7 – Solution

Strategy. The values on the coordinate basis vectors are exactly the columns of the matrix representing a linear field.

Step 1: Recover the coefficients Substitution at $(1,0)$ gives $\langle a,c\rangle=\langle 2,-1\rangle$, while substitution at $(0,1)$ gives $\langle b,d\rangle=\langle 3,4\rangle$. Hence $$\boxed{\mathbf F(x,y)=\langle 2x+3y,-x+4y\rangle}.$$ Equivalently, $$\mathbf F(x,y)= \begin{pmatrix}2&3\\-1&4\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}.$$

Step 2: Evaluate $$\mathbf F(-1,2) =\langle-2+6,1+8\rangle =\boxed{\langle 4,9\rangle}.$$

Step 3: Find zero vectors Solve $$2x+3y=0,\qquad -x+4y=0.$$ The second equation gives $x=4y$; the first then gives $11y=0$. Thus $$\boxed{(x,y)=(0,0)}$$ is the only zero. Equivalently, the coefficient determinant is $2(4)-3(-1)=11\ne 0$, so the linear map has a trivial nullspace.

Verification The recovered formula returns both measured vectors when the two basis points are substituted, so all four coefficients satisfy the data.