Linear Algebra - There is system Ax=b of n linear equations in an unknowns. Q: I

Question

Linear Algebra - There is system Ax=b of n linear equations in an unknowns.

Q: If it has a unique solution, then A is nonsingular

I have answered correctly the other way. but for question above, I said "unique solution means no existence of free variables which means then A can be reduced to triangular matrix (either row-echelon form or just triangular) with a_ij (i and j not being 0). This means determinant exists and A is invertible. which they marked it wrong.

The correct answer, they said would be below.

Conversely, if Ax = b has a unique solution x, then we claim that A
cannot be singular. Indeed, if A were singular, then the equation Ax = 0 would have
a solution z 6= 0. Let y = x + z, then Ay = A(x + z) = A^x + Az = b + 0 = b. This
means y = x + z is also the solution for Ax = b. Then we have two solutions x and
y = x + z. This happens since we assume that A is singular. Thus, if Ax = b has a
unique solution, then A must be nonsingular.

Could you explain why my answer would be wrong? Thank you

Explanation / Answer

If the system of linear equations has unique solution then X=(inverse of A)*b.

Inverse of A will exist only if determinant of matrix A is a non-zero value. Hence A must be non-singular.

Another way of explaining this is(I hope this is the way to extend your explanation little bit): If we take aumented form to get unique solution rank[A |b]=rank(A) = n

where n is the number of variables or size of the matrix A. Then to get full rank A must be invertible which implies A must be non-singular.

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