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Exercise 3.4.39

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If the 5 by 5 matrix $[A b]$ is invertible, $b$ is not a combination of the columns of $A :$ no solution to $Ax = b$ . If $[A b]$ is singular, and the 4 columns of $A$ are independent $($ rank 4 $), b$ is a combination of those columns. In this case $Ax = b$ has a solution.

mathchakra

2022-01-23 16:00

Exercise 3.4.39

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