Exercise 1.3.5

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$r = m = n$ , $A_{1} = [\begin{matrix} 1 & 3 \\ 2 & 4 \end{matrix}]$ . $A_{1}$ is invertible, so $A_{1} x = b$ has 1 solution for every $b$ . Or we can say that the column space is the full $R^{2}$ , so for every $b$ in $R^{2}$ , it’s in the column space, and we have a solution.
$r = m < n$ , $A_{2} = [\begin{matrix} 1 & 2 & 3 \\ 2 & 4 & 1 \end{matrix}]$ , since the column 1 and column 2 are dependent, we have $rank (A_{2}) = m = 2$ . For $A_{2} x = b$ , we have two equations and 3 unknowns, which generally gives us infinitely many solutions. Or we can say that the column space is $R^{2}$ in $R^{2}$ , so every $b$ from $R^{2}$ is in the column space and we always have a solution corresponding to the basis matrix. Since column 2 is a multiple of column 1, there are infinitely many solutions. However, if one of the columns is 0, e.g. $A_{2} = [\begin{matrix} 0 & 2 & 3 \\ 0 & 4 & 1 \end{matrix}]$ , we still have $rank (A_{2}) = 2$ , but there’s only one solution for $A_{2} x = b$ now.
$r = n < m$ , $A_{3} = [\begin{matrix} 1 & 3 \\ 2 & 4 \\ 3 & 6 \end{matrix}]$ , so $rank (A_{3}) = 2$ , for $A_{3} x = b$ , If $b$ is in the column space. For example, $b = [\begin{matrix} 4 \\ 6 \\ 9 \end{matrix}]$ , we’ll have a solution, but if $b$ is not in the column space, there’s no solution.
$r < m, r < n$ , $A_{4} = [\begin{matrix} 1 & 2 \\ 2 & 4 \\ 3 & 6 \end{matrix}]$ , $rank (A_{4}) = 1 < m, n$ . for $A_{4} x = b$ , we have 3 equations and 1 unknown. If $b$ is in the column space (a line along $[\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]$ ), we have infinitely many solutions, otherwise there’s no solution at all.

niuers

2020-03-20 00:00

Exercise 1.3.5

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