## problem 709

Let $S=\\mathbfv_1,\mathbfv_2,\mathbfv_3,\mathbfv_4,\mathbfv_5\$ where\<\mathbfv_1=\beginbmatrix1 \\ 2 \\ 2 \\ -1\endbmatrix,\;\mathbfv_2=\beginbmatrix1 \\ 3 \\ 1 \\ 1\endbmatrix,\;\mathbfv_3=\beginbmatrix1 \\ 5 \\ -1 \\ 5\endbmatrix,\;\mathbfv_4=\beginbmatrix1 \\ 1 \\ 4 \\ -1\endbmatrix,\;\mathbfv_5=\beginbmatrix2 \\ 7 \\ 0 \\ 2\endbmatrix.\>Find a basis because that the expectations $\Span(S)$.

You are watching: Find a basis for the space spanned by the given vectors We will provide two solutions.

## Solution 1.

We use the leading 1 method.Let $A$ it is in the procession whose tower vectors space vectors in the collection $S$:\Applying the elementary heat operations to $A$, we obtain\beginalign*A=\beginbmatrix1 & 1 & 1 & 1 & 2 \\2 & 3 & 5 & 1 & 7 \\2 & 1 & -1 & 4 & 0 \\-1 & 1 & 5 & -1 & 2\endbmatrix\xrightarrow\substackR_2-2R_1 \\ R_3-2R_1\beginbmatrix 1 & 1 & 1 & 1 & 2 \\ 0 & 1 & 3 & -1 & 3 \\ 0 & -1 & -3 & 2 & -4 \\ 0 & 2 & 6 & 0 & 4 \endbmatrix\\<6pt> \xrightarrow\substackR_1-R_2 \\ R_3+R_2 \beginbmatrix 1 & 0 & -2 & 2 & -1 \\ 0 & 1 & 3 & -1 & 3 \\ 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 2 & -2 \endbmatrix \xrightarrow\substackR_1-2R_3 \\ R_2+R_3 \beginbmatrix 1 & 0 & -2 & 0 & 1 \\ 0 & 1 & 3 & 0 & 2 \\ 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 \endbmatrix=\rref(A).\endalign*Observe the the first, second, and also fourth tower vectors the $\rref(A)$ save the leading 1 entries.Hence, the first, second, and also fourth column vectors the $A$ kind a basis of $\Span(S)$.Namely,\<\left\ \beginbmatrix 1 \\ 2 \\ 2 \\ -1 \endbmatrix, \beginbmatrix 1 \\ 3 \\ 1 \\ 1 \endbmatrix, \beginbmatrix 1 \\ 1 \\ 4 \\ -1 \endbmatrix\right \\> is a basis for $\Span(S)$.

## Solution 2.See more: Songs Like Can T Help Falling In Love, Can'T Help Falling In Love By Elvis Presley

Let\Then $\Span(S)$ is the column an are of $A$, i m sorry is the row space of $A^T$. Making use of row operations, we have\\<\to\beginbmatrix1 & 0 & 0 & -13 \\0 & 1 & 0 & 4 \\0 & 0 & 1 & 2 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0\endbmatrix.\>Therefore, the set of nonzero rows\<\left\\beginbmatrix1 \\ 0 \\ 0 \\ -13\endbmatrix,\beginbmatrix0 \\ 1 \\ 0 \\ 4\endbmatrix,\beginbmatrix0 \\ 0 \\ 1 \\ 2\endbmatrix\right\\>is a basis because that the row room of $A^T$, which equates to $\Span(S)$.

Click right here if resolved 137 