Science:Math Exam Resources/Courses/MATH307/April 2012/Question 03

MATH307 April 2012

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Question 03
Suppose that A is a 3 x 4 matrix and $\mathrm {rref} (A)={\begin{bmatrix}1&2&0&2\\0&0&1&2\\0&0&0&0\end{bmatrix}}$ For each of the following, either find what is asked for, or indicate that there is insufficient information to determine it: (i) the rank of A, (ii) a basis for N(A), (iii) a basis for R(A), (iv) a basis for N(AT), (v) a basis for R(AT).

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

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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. (i) The rank of A is determined by the number of pivots in rref(A). Therefore, the rank of A is 2. (ii) A basis for N(A) is determined by inspecting the free variables of rref(A), which are 2 and 4. Let $x_{2}=s$ and $x_{4}=t$. So, $x_{1}=-2x_{2}-2x_{4}$ and $x_{3}=-2x_{4}$. Then we can write: ${\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{bmatrix}}=s{\begin{bmatrix}-2\\1\\0\\0\end{bmatrix}}+t{\begin{bmatrix}-2\\0\\-2\\1\end{bmatrix}}$. Therefore, a basis for N(A) is ${\begin{bmatrix}-2\\1\\0\\0\end{bmatrix}}$ and ${\begin{bmatrix}-2\\0\\-2\\1\end{bmatrix}}$. (iii) A basis for R(A) are the columns of A corresponding to the pivot columns of rref(A). Since we don’t know A, we cannot determine a basis for R(A). (iv) A basis for N(AT) is determined by inspecting the free variables of rref(AT). Since we don’t know rref(AT), we cannot determine a basis for N(AT). (v) A basis for R(AT) are the rows of rref(A) that correspond to the pivots. Therefore a basis for R(AT) is ${\begin{bmatrix}1\\2\\0\\2\end{bmatrix}}$ and ${\begin{bmatrix}0\\0\\1\\2\end{bmatrix}}$.