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ya

Therefore, \(x_1 = y_1\); \(x_2 = y_2\) and \(x_3 = y_3\) making this a one-one function.

the matrix is linearly independant

For instance, \(x_3 = x - y\) from the first two equations.

Similarly, \(x_1 = \dfrac{y+z -3x_3}2 = \dfrac{y + z - 3x + 3y}{2} =\dfrac{4y - 3x + z}{2}\)

If you know basic linear algebra, Dan's solution is actually WAYYYY simpler than mine is.

ok so a simple answer would be if its linearly independent its onto and if not 1-1?