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Exercise 3.3.22

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If $A x_{1} = b$ and also $A x_{2} = b$ then $A (x_{1} - x_{2}) = 0$ and we can add $x_{1} - x_{2}$ to any solution of $Ax = B$ : the solution $x$ is not unique. But there will be no solution to $Ax = B$ if $B$ is not in the column space.

mathchakra

2022-01-23 15:06

Exercise 3.3.22

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