1 Answers
- The equations are (a+2b=7) and (2a-b=1), giving \[ \boxed{a=\frac95,\quad b=\frac{13}5}. \]
- The set must span the space and be linearly independent.
- The columns are (T(1,0)) and (T(0,1)), so \[ \boxed{A=\begin{bmatrix}1&2\\3&-1\end{bmatrix}}. \]
- \[ \boxed{\begin{bmatrix}4\\10\end{bmatrix}}. \]
- Adding the equations gives (3x=6), so (oxed{x=2,y=3}).
- Yes, since (1(2)+2(-1)+3(0)=0).
- \[ \boxed{\begin{bmatrix}-1\\2\\1\end{bmatrix}}. \]
- (oxed{10}).
- Since (A) is triangular, the eigenvalues are its diagonal entries: (oxed{4,2}).
- Write (c_1(1+x)+c_2x+c_3x2=2+5x-x2). Thus (oxed{[p]_{B}=[2,3,-1]^T}).
- Applying (B) and then (A) gives (A(Bx)=(AB)x), so (AB) represents composition.
- A zero determinant collapses volume into a lower-dimensional set, so distinct inputs can map to the same output; therefore no inverse exists.
2 Grading
Questions 1–10 are computational and worth 2 points each. Questions 11–12 are conceptual and worth 2 points each. Award 1 point for the correct method with an arithmetic or notation error, and 2 for a correct, explicit answer.