On the other hand, since $A,B$ being nonsingular gives us that $AB$ is nonsingular, then we also have that 2) if $AB$ is singular, then at least one of $A$ and $B$ must be singular as well. x np.linalg.lstsq (E, D, rcondNone) 0 We cant confirm this is one solution with. For example, a circuit containing a floating. Im looking at the problem but do not know any more than when I did at 6AM when I posted earlier. Instead of solving Ex D the method of least squares finds an x that minimizes the squared norm-2 of Ex - D. A singular matrix error occurs when the circuit does not have a unique and finite solution. Notice, by the way, that we also showed that 1) $AB$ is singular if $B$ is the one assumed singular. A common way to do that is using the method of least squares. But now we have that $(AB)a=A(Ba)=Ab=0$, and we conclude (again) that $AB$ is singular. Now, use that $B$ is nonsingular to find $a$ such that $Ba=b$. If, on the other hand, $B$ is nonsingular, use that $A$ is singular to find $b\ne 0$ such that $Ab=0$. If $B$ is also singular, then for some $x\ne 0$ we have $Bx=0$, but then $(AB)x=A(Bx)=A0=0$, and we conclude that $AB$ is also singular. The Normal equations can still be solved even when the determinant of X X is zero. If the determinant is zero, it is singular if not, it is non-singular. That $C$ is nonsingular, on the other hand, gives us in particular that the column space of $C$ has full rank, that is, for any vector $b$ there is a vector $a$ such that $Ca=b$. To answer the title question, all you need to do is to calculate the determinant of the matrix. One approach is this: That a matrix $C$ is singular gives us in particular that its null space is non-trivial, that is, for some vector $x\ne0$ we have $Cx=0$.
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