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$M$ is a $2\times{4}$ matrix, and from its row reduces form, we have found the spanning vectors $$v_{1}= \begin{bmatrix}2\\1\\0\\0\end{bmatrix}\quad\text{and}\quad v_{2}= \begin{bmatrix}-1\\0\\2\\1\end{bmatrix}.$$ For what value of $T$ is $u= \begin{bmatrix}1&-2&3&T\end{bmatrix} $ in the row space of $M$?

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  • $\begingroup$ Should $v_1, v_2$ be row vectors instead? Are they the rows of $M$? $\endgroup$ – Arthur Jun 14 '16 at 0:18
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If $u$ would be in the row space, then you could obtain it as a linear combination of $v_1$ and $v_2$ as a basis of row space. Now, assume that $u=av_1+bv_2$. From the second component, we have $a=-2$, and from the second one, we have $b=3/2$ while these two values do not satisfies the first component. So, there is no such a $T$.

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  • $\begingroup$ I know about this, but what if $v_{1}$ and $v_{2}$ are the vectors that span the column space? How can you be sure that these are the row spacing vectors? $\endgroup$ – kaboommath Jun 14 '16 at 0:27
  • $\begingroup$ Your matrix is $2\times 4$. So, it has two rows which are $4-$tuple vectors and four columns which are $2-$tuple vectors. Now, how $v_1$ and $v_2$ can generate the column space which comprises $2-$tuple vectors? I think you may need to revise the question. $\endgroup$ – Majid Jun 14 '16 at 0:47
  • $\begingroup$ OK, I get it now. $\endgroup$ – kaboommath Jun 14 '16 at 0:49

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