# Is the following statement true or false? And why? "The system Ax=b, where A is a n x n matrix and b∈ℝ^n, admits a unique solution if and only if det(A)≠0."

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### 2 Answers

It's true. Have you learned that det(A) `ne` 0 if and only if the n columns of A, considered as n dimensional vectors, are linearly independent? If the columns are independent, then they span `RR^n,` and the key property of a spanning set is that *every *vector ` ``b in RR^n` can be written uniquely as a linear combination of the vectors in the spanning set.

If we write `A` as `[v_1 , v_2 , ... , v_n]`, where the `v`'s are column vectors and det(A) `ne` 0, then there exist unique scalars `x_1, x_2, ..., x_n` such that

`x_1v_1+x_2v_2+...+x_nv_n=b.`

Then `(x_1, x_2, ..., x_n)` is the unique solution to `Ax=b.`

If you prefer the language of linear transformations, this says that the linear transformation described by the matrix `A` is one to one and onto if and only if det(A) `ne` 0.

Graphically, although it's hard to visualize if `A` is bigger than 3x3, having nonzero determinant means that the range of the transformation described by `A` is `RR^n` . If the determinant is zero then the range is only a proper subspace of `RR^n` and so for any `b` not in this subspace, there will be no solution.

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### User Comments

Thanks, good described! :)