Science:Math Exam Resources/Courses/MATH221/December 2008/Question 01 (a)

MATH221 December 2008

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Question 01 (a)
Let $A$ denote the matrix $A={\begin{bmatrix}3&0&-2\\1&-2&3\\4&-2&1\end{bmatrix}}$ Find a basis for the column space of $A$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Perform row reduction and identify the pivot columns.

Hint 2
It may be easier to perform row reduction after interchanging the first and second row.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We perform row reduction, interchanging the first and second rows in the first step to make calculations simpler: ${\begin{bmatrix}3&0&-2\\1&-2&3\\4&-2&1\end{bmatrix}}\rightarrow {\begin{bmatrix}1&-2&3\\3&0&-2\\4&-2&1\end{bmatrix}}\rightarrow {\begin{bmatrix}1&-2&3\\0&6&-11\\0&6&-11\end{bmatrix}}\rightarrow {\begin{bmatrix}1&-2&3\\0&6&-11\\0&0&0\end{bmatrix}}.$ The first and second columns are pivot columns, so a basis for the column space is $\color {blue}\left\{{\begin{bmatrix}3\\1\\4\end{bmatrix}},\,{\begin{bmatrix}0\\-2\\-2\end{bmatrix}}\right\}.$

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