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Hello.
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Today we're going to find the null space of some matrix, and we're also going to verify the rank nullity theorem.
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So our given matrix today is a equals 1, 4, negative 1, 3, 2, 9, negative 1, 7, and 2, 8, negative 2, and 6.
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So we're going to want to put this into reduced row echelon form.
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So right from the get go, we can see that we want to preserve the first row because we already have that leading one in this initial operation.
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And we can actually do row two plus negative two times row one, which gives us zero, nine plus 4 times negative 2, negative 8, so that's 1.
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And then negative 1 plus negative 1 times negative 2.
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So negative 1 plus 2 is again 1.
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And then 7 minus 6, which is again 1.
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And then for the third row, we can actually also do row 3 plus negative 2 times row 1.
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And as you see straight across the board, basically if you just multiply by negative 2, it cancels out everything in row 3.
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So we're actually going to get 0, 0, 0, and 0.
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So let's rewrite that again.
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1, 4, negative 1, 3, 0, 1, 1, 1, and then all 0.
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So we know that we are preserving the bottom row because that is all zeros, which is already in its most reduced form.
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And we have a leading one and we have a leading one in this column as well.
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So that's already good.
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But we have, if we look here, we have a number over the one in the second row.
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So then when i get rid of that...