Linear Algebra - Basis for a row space

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To find a basis for the row space of matrix A, the matrix is transposed and converted to reduced row echelon form. The presence of leading ones in every column indicates that all row vectors from A are included in the basis. This confirms that the rows of A form a basis for R4. Consequently, every vector in R4 can be expressed as a linear combination of these row vectors. The approach taken is correct and effectively demonstrates the relationship between the row space and R4.
jinksys
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A =
Code:
 1  2 -1  3
 3  5  2  0
 0  1  2  1 
-1  0 -2  7

Problem: Find a basis for the row space of A consisting of vectors that are row vector of A.

My attempt:

I transpose the matrix A and put it into reduced row echelon form. It turns out that there are leading ones in every column. Therefore, the basis includes every row vector from A.

Is this the correct way to handle this problem?
 
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Yes, that works.
 
What you found means that the rows of A form a basis for R4, meaning that every vector in R4 can be written as a linear combination of the four vectors that are the rows of A.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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