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One of my homework problems asks whether the set S = {(2, -1, 2), (1, 0, 3), (3, -2, 1)} is linearly independent.

My book lists out a bunch of steps for problems similar to this, which basically say that I should first form the following system of equations:

2c1 + c2 + 3c3 = 0

-c1 - 2c3 = 0

2c1 + 3c2 + c3 = 0

Then form the augmented matrix: \begin{bmatrix}2&1&3&0\\-1&0&-2&0\\2&3&1&0\end{bmatrix} And then reduce this using Gaussian elimination in order to tell whether the system has a unique solution.

My question is, couldn't I just form the coefficient matrix (rather than the augmented matrix), find its determinant, and if it's nonzero conclude that the system has only the trivial solution and is linearly independent?

Or do I really need to use Gaussian elimination?

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    You can use both methods because the matrix is square. However, it is better to ask your professor - maybe the goal of the exercise is to practice Gaussian elimination? – 2017-01-10
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    The determinant suffices to answer the question in the case of a square matrix. But even then, the Gaussian elimination gives you more information. In the case where they are _not_ independent, it allows you to find the coefficients of a linear combination of the columns that equals to the zero vector. – 2017-01-10
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    Further to quasi's remark, if you write down the 3 by 3 matrix with rows the vectors in S, then do Gaussian elimination, a row of zeros will imply dependence. Further, the remaining none zero rows, written as columns, will form a basis of S. – 2017-01-10

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