The an initial question "The vector b is in the kernel that T" I obtained as false since I go A*b and got (-7, 24, -25) i beg your pardon is no a 0 vector for this reason I placed it as false.

You are watching: Determine if the given vector is in the range of t(x) = ax, where

The 2nd question "The vector c is in the variety of T" I placed as false due to the fact that I was unable to execute an augmented procession of Ax = c because the dimensions room not the same.

I obtained the concern right due to the fact that they both turned out to it is in false however I am not certain if my thinking are correct. Deserve to somebody verify if i am interpreting this trouble correctly? thank you

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· 7 yr. Ago

Both room correct reasoning. The kernel of a linear revolution A is by definition the collection of every vectors which acquire mapped to 0; it is as an easy as calculating ab and compare it come 0 to see whether or not b is in the kernel. The vector c have the right to only be in the variety of A just if it lies in the codomain the A, i beg your pardon is R3 for her example; together c lies in R2, it couldn't perhaps be in the variety of A.

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Op · 7 yr. Ago · edited 7 yr. Ago

Okay, thank you. I just wanted come make certain I construed the concepts for the upcoming exam.

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Edit: If the c vector remained in the exact same dimensions would certainly that automatically make the vector in the range? Or would I need to do RREF to determine if it's in the range?

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