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I am reading about how to find the rank of matrices. And one way to find this is to see number of independent rows. As number of independent rows equals to the rank. If matrix is small i.e with 4 elments or 9. We can check linear independence. But if we have matrices with many elements then how can we check linear dependent rows. Is there any easy way?

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I know gaussian elimination method. But i want to know using linear depedence it is possible? If yes then easy method?

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    If you want to do it by hand, then Gaussian elimination is an obvious approach. This being said, what do you understand under "many elements"? – 2017-01-09
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    Usually it's done by Gaussian Elimination. If the number of rows is very large, one can use a Singular Value Decomposition, which is hard to determine by hand but more stable, when it comes to numerical errors – 2017-01-09
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    If you don't want to do it by hand, then use something like the [$QR$](https://en.wikipedia.org/wiki/QR_decomposition#Connection_to_a_determinant_or_a_product_of_eigenvalues) decomposition. – 2017-01-09

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One method is the Gaussian elimination. For example:

$$\begin{aligned} A=&\begin{bmatrix}{\boxed{1}}&{2}&{-1}&2\\{4}&{1}&{0}&2\\{2}&{-3}&{2}&-2\\ 1&0&-1&3\\2&1&2&-4\end{bmatrix}\begin{matrix}\sim \\R_2-4R_1\\R_3-2R_1\\R_4-R_1\\R_5-RF_1\end{matrix}\\ &\begin{bmatrix}{1}&{2}&{-1}&2\\{0}&{\boxed{-7}}&{4}&-6\\{0}&{-7}&{4}&-6\\ 0&-2&0&1\\0&-3&4&-8\end{bmatrix}\begin{matrix}\sim \\R_3-R_2\\7R_4-2R_2\\7R_5-3R_2\end{matrix}\\ &\begin{bmatrix}{1}&{2}&{-1}&2\\{0}&{-7}&{4}&-6\\{0}&{0}&{0}&0\\ 0&0&-8&19\\0&0&16&-38\end{bmatrix}\begin{matrix}\sim \\R_3\leftrightarrow R_5 \end{matrix}\\ &\begin{bmatrix}{1}&{2}&{-1}&2\\{0}&{-7}&{4}&-6\\{0}&{0}&{\boxed{16}}&-38\\ 0&0&-8&19\\0&0&0&0\end{bmatrix}\begin{matrix}\sim \\2R_4+R_3 \end{matrix}\\ &\begin{bmatrix}{1}&{2}&{-1}&2\\{0}&{-7}&{4}&-6\\{0}&{0}&{16}&-38\\ 0&0&-8&19\\0&0&0&0\end{bmatrix}\begin{matrix}\sim \\2R_4+R_3 \end{matrix}\\ &\begin{bmatrix}{\boxed1}&{2}&{-1}&2\\{0}&{\boxed{-7}}&{4}&-6\\{0}&{0}&{\boxed{16}}&-38\\ 0&0&0&0\\0&0&0&0\end{bmatrix}\Rightarrow \text{rank }A=3. \end{aligned}$$