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If a matrix is not invertible, can I say the matrix has many solutions? Or is the matrix has many solutions as well as the possibility or no solutions? And so if a matrix is invertible, the system of equations will definitely only have one unique solution?

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Well, think about it. If the matrix is not invertible, what does that say about it's determinant?
If it's invertible, we know the only solution to Ax = 0 is the trivial solution
And are you talking about homogeneous or non homogeneous?

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If a matrix is not invertible, the determinant is zero. But how does this tell about the number of solutions? If a matrix is invertible, there are many solutions to Ax=0, isn't it? I am referring to both homogeneous and non-homogeneous types.
If the coefficient matrix is a singular square matrix A, then the system A x = b can have no solutions of an infinite number of solutions, depending on what column matrix c is.
If column matrix c is all 0's, then Ax = c only the trivial solution, no matter what numbers are in singular sqare matrix A.
Is column matrix c the last column of the matrix? Last column in the sense that if it is a 3x3 matrix, the 3 column is the column matrix c?
This is the only time A x = c has exactly one solution. For nonzero C there are an infinite number of solutions or no solutions. Some systems may have uneque solutions for some, but not all, unknowns. The rest of the unknowns are still undeterminable.
If a matrix is singular, that's it's not invertible, the system of equations can have either no solutions or many solutions. Am I right about this? Is the column matrix C you refer to the last column of a matrix? Like in a 3x3 matrix, is column matrix C the column number 3? Or does the column matrix C referring to the "result column" in an augmented matrix? So by knowing that a matrix is not invertible is still not enough to tell if the system of equations has unique solution, many solutions or no solution? thanks.
Right. You need to augment the coefficient matris with column matrix c and then get it to reduced row-echelon form. That means using the Gauss-Jordan method.
Are you familiar with those terms?
Yes, I understand the terms. Thanks a lot! So I can also say that if a matrix is invertible, it is a good matrix that surely has only one unique solution, am I right about this?
Assume your 2x2 matrix represents lines. Then they cross or are parallel or are the same line. The first case will form an invertible matrix (unique solution). Second and third is no solution/infinite number of solutions. You can think of 3x3 matrices/planes in much the same way.

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