GO Classes | DA Course| Linear Algebra

Answer 1

The $\operatorname{rref}(\mathbf{A})$ is
$$
\left[\begin{array}{ccc}
1 & 2 & -1 \\
-3 & -4 & 2 \\
5 & 2 & 3
\end{array}\right] \sim\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
$$

Therefore, $r=\operatorname{rank}(\mathbf{A})=3$. The dimension of the co-domain is $m=3$ and therefore $\mathrm{T}_{\mathbf{A}}$ is onto. Therefore, the columns of $\mathbf{A}$ span all of $\mathbb{R}^{3}$, that is, every $\mathbf{b} \in \mathbb{R}^{3}$ can be written as a linear combination of the columns of $\mathbf{A}$ :
$$
\operatorname{span}\left\{\left[\begin{array}{c}
1 \\
-3 \\
2
\end{array}\right],\left[\begin{array}{c}
2 \\
-4 \\
2
\end{array}\right],\left[\begin{array}{c}
-1 \\
2 \\
3
\end{array}\right]\right\}=\mathbb{R}^{3}
$$

Answer 2

|A| = 8   

hence cols of A are LI

hence onto

Answer 3

Square matrix is the special case where there is a possibility of both one-one and onto. This matrix is both one one and onto as it's reduced form contains pivot in every row as well as in every column.