Science:Math Exam Resources/Courses/MATH221/December 2011/Question 08 (a)

MATH221 December 2011

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Question 08 (a)
Consider the matrix $A=\left({\begin{array}{ccccc}1&0&-1&2&4\\0&1&-2&0&-1\\-1&-2&5&1&4\\1&0&-1&1&2\end{array}}\right)$ a) 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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
How can the row reduced version of A help you find the column space?

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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. The basis of the column space of A are the columns of A corresponding to the linearly independent columns of the row-reduced version of A. We will find the row reduced matrix for A by performing Gaussian elimination as follows: $\left[{\begin{array}{ccccc}1&0&-1&2&4\\0&1&-2&0&-1\\-1&-2&5&1&4\\1&0&-1&1&2\end{array}}\right]$ Add row 1 to row 3; subtract row 1 from row 4. $\left[{\begin{array}{ccccc}1&0&-1&2&4\\0&1&-2&0&-1\\0&-2&4&3&8\\0&0&0&-1&-2\end{array}}\right]$. Add twice row 2 to row 3. $\left[{\begin{array}{ccccc}1&0&-1&2&4\\0&1&-2&0&-1\\0&0&0&3&6\\0&0&0&-1&-2\end{array}}\right]$. Multiply row 3 by 1/3. $\left[{\begin{array}{ccccc}1&0&-1&2&4\\0&1&-2&0&-1\\0&0&0&1&2\\0&0&0&-1&-2\end{array}}\right]$. Subtract twice row 3 from row 1; add row 3 to row 4 $\left[{\begin{array}{ccccc}1&0&-1&0&0\\0&1&-2&0&-1\\0&0&0&1&2\\0&0&0&0&0\end{array}}\right]$. The row reduced version of A has columns 1, 2, and 4 as linearly independent vectors. Therefore, the basis of the column space will be the first, second and fourth column of the original matrix A. Hence a basis for the column space is: $\left\{\left[{\begin{array}{c}1\\0\\-1\\1\end{array}}\right],\left[{\begin{array}{c}0\\1\\-2\\0\end{array}}\right],\left[{\begin{array}{c}2\\0\\1\\1\end{array}}\right]\right\}$.

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