Why does a singular matrix imply that it does not have a solution?
(Trying to learn linear algebra over here)
The augmented matrix in question:
$$\begin{bmatrix}0 & 1 &5 & -4\\1 & 4 & 3 & 2\\2 & 7 & 1 & -2\end{bmatrix}$$
So I tried to solve the matrix above but I couldn't. I decided to see what happened when I pushed it through Numpy (Python):
numpy.linalg.linalg.LinAlgError: Singular matrixSo I went back to the definition for a singular matrix:
A square matrix that is not invertible is called singular or degenerate.
The book simply says it is inconsistent. So, now I'm wondering: How does a singular matrix relate to it not having a solution?
$\endgroup$5 Answers
$\begingroup$Reducing an augmented matrix $[\mathbf{A}|\mathbf{b}]$ is equivalent to solving the system $\mathbf{Ax} = \mathbf{b}$. However, there is a unique solution to this system if and only if $\mathbf{A}$ is invertible. The solution is $\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}$.
If the matrix $\mathbf{A}$ is not invertible, there are either zero or an infinite number of solutions.
$\endgroup$ $\begingroup$The system has a solution (not necessarily a unique one) iff: $$rank(A) = rank(A\mid b).$$
We get that $rank (A) =2$, since $\det(A) = 0$, plus there exists a $2\times 2$ minor which is $\neq 0$.
However, $rank(A\mid b) = 3$, since the determinant $\begin{vmatrix} 1 &5 & -4\\ 4 & 3 & 2\\ 7 & 1 & -2\end{vmatrix}=170\neq 0.$
$\endgroup$ $\begingroup$Assume $A$ is singular.One explanation is by using determinants, using the fact that determinants are multiplicative, i.e., given square matrices $A,B$, we have $\det(AB)=\det{A}\det{B}$. Now, if $A$ is invertible, we would have $AC=I$, for some square matrix $C$. But then we need $$ \det(AC)=\det{A}\det{C}=\det{I}=1 $$ , so we would theen need to have $$\det{A}\det{C}=0(\det{C})=1 $$ , i.e., we would end u with $0. c=1$, which is not possible.
If $A$ were invertible, the the system $Ax=B$ would have the solution $$ x=C^{-1}B $$ , for $C$ as above the inverse of $A$. This solution $x$ is unique, and exists only if $C^{-1}$ exists, i.e. if $A$ is nonsingular. Still, if the system is consistent, you will still have infinitely many solutions.
$\endgroup$ $\begingroup$If the given system is homogeneous i.e. AX=0 then the problem have infinite number of solution. So the given system always have no solution; is not true.
$\endgroup$ $\begingroup$You can also think of this in terms of Cramer's Rule, since you have n equations and n unknowns. Cramer's rule gives a simple formula to calculate the unique solution of n equations with n unknows
The formula has det(A) in the denominator of the unique solution values, where A is the coefficient matrix (only the first 3 columns of your augmented matrix).
Clearly if det(A) is zero, then your solution can't exist. Hence, if the A is non singular, the solution can't exist. Your matrix is non singular, hence a real solution does not exist.
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