Science:Math Exam Resources/Courses/MATH307/December 2010/Question 02 (b)

MATH307 December 2010

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Question 02 (b)
Suppose that $A={\begin{bmatrix}1&1&0&1&0\\2&2&1&3&0\\3&3&1&4&1\\4&4&1&5&1\end{bmatrix}}\quad {\text{rref}}(A)={\begin{bmatrix}1&1&0&1&0\\0&0&1&1&0\\0&0&0&0&1\\0&0&0&0&0\end{bmatrix}}$ (b) Write down a basis for N(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
The null space N(A) is the set of vectors x such that Ax = 0. To find a basis for N(A) rewrite the matrix rref(A) as a system of linear equations and solve for the unknowns.

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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. Considering $A{\vec {x}}={\vec {0}}$ with ${\vec {x}}={\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\end{bmatrix}}$. From the above rref, with $x_{2}$ and $x_{4}$ free, the following equations can be derived: ${\begin{aligned}x_{1}&=-x_{2}-x_{4}\\x_{2}&=x_{2}\\x_{3}&=-x_{4}\\x_{4}&=x_{4}\\x_{5}&=0\end{aligned}}$ Therefore ${\vec {x}}={\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\end{bmatrix}}=x_{2}{\begin{bmatrix}-1\\1\\0\\0\\0\end{bmatrix}}+x_{4}{\begin{bmatrix}-1\\0\\-1\\1\\0\end{bmatrix}}$ If we for example let $x_{2}$ = $x_{4}$ = 1, then a basis for N(A) is given by ${\text{span}}\left\{{\begin{bmatrix}-1\\1\\0\\0\\0\end{bmatrix}},{\begin{bmatrix}-1\\0\\-1\\1\\0\end{bmatrix}}\right\}$

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