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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!
