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I need help finding a value of h so that $\begin{bmatrix} 0\\1\\h+2\end{bmatrix}$ is in the span of the two matrices:$\begin{bmatrix} 2\\1\\1 \end{bmatrix}, \begin{bmatrix} 3\\1\\2 \end{bmatrix}.$

My approach was to write the augmented matrix and reduce it like this: $\begin{bmatrix} 2&3&0\\1&1&1\\1&2&h+2\end{bmatrix} $

add $-1/2$ times row $1$ to row $2$, and $-1/2$ times row $1$ to row $3$: $\begin{bmatrix} 2&3&0\\0&-1/2&1\\0&1/2&h+2\end{bmatrix}.$

Now we have that $h + 2 = \frac{1}{2} \cdot x_2$, and $x_2 = -2$. So, $\hbox{h = -3}$.

However, I do not think this is correct. Can someone please check my work?

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  • $\begingroup$ Why do you not think this is correct? $\endgroup$
    – Wolfram
    Jan 29, 2017 at 17:37
  • $\begingroup$ Ok, note that initial vectors are the columns of the matrix, but you use the transformations of the rows, not columns. Do you understand why this is legal? $\endgroup$
    – Wolfram
    Jan 29, 2017 at 17:41
  • $\begingroup$ Yes it is an augmented matrix. $\endgroup$
    – user400359
    Jan 29, 2017 at 17:41
  • $\begingroup$ Using @Upstart's approach, I got the same answer. $\endgroup$
    – user400359
    Jan 29, 2017 at 17:43
  • $\begingroup$ I don't see any calculation errors, so I think the answer is indeed correct. Is that all you wanted to know? $\endgroup$
    – Wolfram
    Jan 29, 2017 at 17:44

1 Answer 1

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Find the determinant of the matrix and see where the determinant is $0$. That value of $h$ will be your answer. The det can be zero even if the first two columns are linearly independent but in your case the first two columns are independent so if the matrix determinant is $0$ then it is because of the third column.

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  • $\begingroup$ What did you get from this approach? $\endgroup$
    – user400359
    Jan 29, 2017 at 17:42
  • $\begingroup$ I got the same answer as I originally did. $\endgroup$
    – user400359
    Jan 29, 2017 at 17:43
  • $\begingroup$ then it's correct $\endgroup$
    – Upstart
    Jan 29, 2017 at 17:45

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