Linear Algebra - There is system Ax=b of n linear equations inan unknowns. Q: If it has
Question:
Linear Algebra - There is system Ax=b of n linear equations inan 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 whichmeans then A can be reduced to triangular matrix (eitherrow-echelon form or just triangular) with a_ij (i and j not being0). This means determinant exists and A is invertible. which theymarked it wrong.
The correct answer, they said would be below.
Conversely, if Ax = b has a unique solution x, then weclaim that A
cannot be singular. Indeed, if A were singular, then the equationAx = 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 twosolutions 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?
Financial accounting
ISBN: 978-1118285909
IFRS Edition
Authors: Jerry J. Weygandt, Donald E. Kieso, Paul D. Kimmel