Solution to Linear Algebra Hoffman & Kunze Chapter 9.2
Exercise 9.2.1 Solution: (a) No. Since $f(0,\beta)\ne 0$. (b) No. Since $f((0,0),(1,0))\ne 0$. (c) Yes. Since $f(\alpha,\beta)=4x_1\bar y_1$. (d) No. Because of $\bar x_2$ there, it is not linear on…