Problem 709

Let $S=\mathbfv_1,mathbfv_2,mathbfv_3,mathbfv_4,mathbfv_5$ whereFind a basis for the span $Span(S)$.

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We will give 2 solutions.

Equipment 1.

We apply the leading 1 technique.Let $A$ be the matrix whose column vectors are vectors in the collection $S$:Applying the elementary row operations to $A$, we obtaineginalign*A=eginbmatrix1 & 1 & 1 & 1 & 2 \2 & 3 & 5 & 1 & 7 \2 & 1 & -1 & 4 & 0 \-1 & 1 & 5 & -1 & 2endbmatrixxrightarrowsubstackR_2-2R_1 \ R_3-2R_1eginbmatrix 1 & 1 & 1 & 1 & 2 \ 0 & 1 & 3 & -1 & 3 \ 0 & -1 & -3 & 2 & -4 \ 0 & 2 & 6 & 0 & 4 endbmatrix\<6pt> xrightarrowsubstackR_1-R_2 \ R_3+R_2 eginbmatrix 1 & 0 & -2 & 2 & -1 \ 0 & 1 & 3 & -1 & 3 \ 0 & 0 & 0 & 1 & -1 \ 0 & 0 & 0 & 2 & -2 endbmatrix xrightarrowsubstackR_1-2R_3 \ R_2+R_3 eginbmatrix 1 & 0 & -2 & 0 & 1 \ 0 & 1 & 3 & 0 & 2 \ 0 & 0 & 0 & 1 & -1 \ 0 & 0 & 0 & 0 & 0 endbmatrix= ref(A).endalign*Observe that the first, second, and fourth column vectors of $ ref(A)$ contain the leading 1 entries.Hence, the initially, second, and also fourth column vectors of $A$ create a basis of $Span(S)$.Namely, is a basis for $Span(S)$.

Systems 2.

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LetThen $Span(S)$ is the column space of $A$, which is the row room of $A^T$. Using row operations, we have< oeginbmatrix1 & 0 & 0 & -13 \0 & 1 & 0 & 4 \0 & 0 & 1 & 2 \0 & 0 & 0 & 0 \0 & 0 & 0 & 0endbmatrix.>Therefore, the collection of nonzero rowsis a basis for the row room of $A^T$, which equals $Span(S)$.

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