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Let $A = \left(\begin{array}{rrr|r} 1&1&-15&36\\ 1&2&-10&41\\ 1&2&-9&42 \end{array}\right)$.

Here is the row reduction: $A = \left(\begin{array}{rrr|r} 1&0&0&51\\ 0&1&0&0\\ 0&0&1&1 \end{array}\right)$.

a) Determine all vectors $b$ for which $Ax = b$ has a solution. Write your answer as the span of a set of linearly independent vectors. (I have no idea what this is asking)

b) Do the columns of $A$ span $\mathbb{R}^3$? (How do can you tell?)

I'm lost in class. Please refrain form using rank and im., etc. because I haven't learn those yet. Please show steps and explanations, not only answers, so that I can learn. Thank you

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You need more help than we can give here (we can give answers, but you need much more than that). Make an appointment to see your teacher, to ask what you can do to catch up. Do they have any kind of tutoring services where you are? –  Gerry Myerson Mar 11 '13 at 2:28
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Manny, just giving answers won't help you learn, and we're not in a position to give full lessons. –  amWhy Mar 11 '13 at 2:35
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@Manny It would do you well here to remember that people are volunteering their time. –  John Moeller Mar 11 '13 at 2:40
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Manny-turned-Walter: Is the matrix an augmented matrix? That is, is A a $3\times 3$ matrix augmented with the vector b? Or is a as you've written, a $3 \times 4$ matrix? Whatever the case, please edit your post to include those details. Have you tried row reduction? You've got to dig in and get your "hands dirty". –  amWhy Mar 11 '13 at 2:54
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Could you please show your row reduction? edit the post to include the row reduced matrix. You need that to answer your questions. And is the question find all vectors $x$ for which $Ax = b$ is a solution? Since $b$ seems to be the vector to the right of the matrix: given? –  amWhy Mar 11 '13 at 4:16

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