Science:Math Exam Resources/Courses/MATH221/April 2010/Question 01 (a)

MATH221 April 2010

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Question 01 (a)
Let $\displaystyle A={\begin{bmatrix}1&2&3&-2&4\\2&5&8&-1&6\\1&4&7&5&2\\1&3&5&1&2\\\end{bmatrix}}$ and $\displaystyle U={\begin{bmatrix}1&2&3&-2&4\\0&1&2&3&-2\\0&0&0&1&2\\0&0&0&0&0\\\end{bmatrix}}$ Then the matrix U is an echelon form for A (you may assume this, you don't have to do the row reduction again.) Find the rank 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
One way to find the rank of the matrix $A$ is determining the number of linearly independent rows.

Hint 2
Since the echelon form is obtained by using row operations only, the row space of $U$ is the same as the row space of $A$.

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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 rank of the matrix $A$ is found by determining the number of linearly independent rows or, equivalently, the number of linearly independent columns. To obtain the echelon form one only performs row operations (adding, subtracting, swapping), hence the echelon form $U$ has the same row space as $A$. For $U$ we easily read off three linearly independent vectors, namely $[1,2,3,-2,4]^{T}$, $[0,1,2,3,-2]^{T}$, $[0,0,0,1,2]^{T}$. Since the fourth row of $U$ is the zero vector we know that there is no fourth linearly independent vector in the row space of $U$. Hence the number of linearly independent rows of $U$ is 3. This means that the number of linearly independent rows of $A$ is also 3. Finally, we obtain that $A$ has rank 3.

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