Another method? - Matrix & Linear Independence

RajdeepSingh7
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Question :
Let A be a 7 × 4 matrix. Show that the set of rows of A is linearly dependent.



Answer:
The row vectors of a matrix are linearly independent if and only if the rank of the matrix is equal to the number of rows in the matrix.
Since rank (A) = 4 , and the number of rows in the matrix is 7, the row vectors are linearly dependent.



I am sure that my answer is right, but I was considering or wondering if there was an alternative method, because the actual answer is worth 4 marks, so was wondering if there was a more mathematical sound proof possible to prove the answer?
 
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The rows of a 7 by 4 matrix are members of R4 which has dimension 4- any set of more than 4 vectors must be dependent.

Your basic statement that a 7 by 4 matrix has rank 4 is NOT correct. The matrix
\begin{bmatrix}1& 1 & 1 & 1\\1& 1 & 1 & 1 \\1& 1 & 1 & 1 \\1& 1 & 1 & 1 \\1& 1 & 1 & 1 \\1& 1 & 1 & 1 \\1& 1 & 1 & 1 \\1& 1 & 1 & 1 \end{bmatrix}
does not have rank 4.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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