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Problem #21 from pg.175 and problem #5 from pg.195.

#21, pg.175: Use determinants to find out if the matrix is invertible:

$\left[ \begin{array}{ll} 2 & 3 & 0 \\ 1 & 3 & 4 \\ 1 & 2 & 1 \end{array} \right].$ Solution: Compute the determinant:

$\begin{array}{ll} \mathrm{det} \left[ \begin{array}{ll} 2 & 3 & 0 \\ 1 & 3 & 4 \\ 1 & 2 & 1 \end{array} \right] &= 2 (3-8) - 3 (1-4) + 0 \\ &=-10+9 \\ &= -1. \end{array}$ We see that the determinant is nonzero and hence we can conclude that

$A$ is invertible.

#5, pg.195: Determine if the given set is a subspace of

$\mathbb{P}_n$ (the set of polynomials of degree

$\leq n$ ): all polynomials of the form

$p(t)=at^2$ , where

$a \in \mathbb{R}$ .
Solution: Let

$S$ denote the set of polynomials of the form

$p(t)=at^2$ where

$a \in \mathbb{R}$ . Clearly

$S \subset \mathbb{P}_n$ . The zero vector of

$\mathbb{P}_n$ is the zero polynomial and that polynomial is in

$S$ -- to see this consider when

$a=0$ . If

$p_1,p_2 \in S$ and

$\alpha,\beta$ are scalars, then we know there are real numbers

$a_1,a_2$ such that

$p_1(t)=a_1t^2$ and

$p_2(t)=a_2t^2$ . Now compute

$\begin{array}{ll} \alpha p_1(t) + \beta p_2(t) &= \alpha a_1 t^2 + \beta a_2 t^2 \\ &= (\alpha a_1 + \beta a_2)t^2, \end{array}$ but this is a polynomial in

$S$ because

$\alpha a_1 + \beta a_2$ is a real number. Hence by the subspace criterion,

$S$ is a subspace of

$\mathbb{P}_n$ .

Note: many people solved this problem by noting the space in question is equal to $\mathrm{span}\{t^2\}$ and hence by Theorem 1 it is a subspace. This is a perfectly good proof!