Given a linear system $\begin{cases}2x - 3y &= 7\\x + 5y &= -3\end{cases}$

This system can be expressed as a matrix-vector product

$$\underbrace{\begin{bmatrix}2 & -3\\1 & 5\end{bmatrix}}_{A}\times\underbrace{\begin{bmatrix}x\\y\end{bmatrix}}_{\vec{x}} = \underbrace{\begin{bmatrix}7\\-3\end{bmatrix}}_{\vec{v}}$$

Or as two line equations $\begin{cases}2x - 3y - 7 &= 0\\x + 5y + 3 &= 0\end{cases}$

The two previous expressions for the system allow us to verify if a solution exists in two different ways.

Using the transformation matrix, we can deduce that an unique solution exists if the determinant of the matrix is different from zero. This makes sense as the determinant is the scaling factor for the area (in the 2D plane) of the parallelogram formed by the two basis vectors $\left(O,\vec{i},\vec{j}\right)$. If the determinant happens to be zero, the parallelogram has been flattened to a single line and $\vec{v}$ cannot be expressed in terms of $\left(O,\vec{i},\vec{j}\right)$, thus the system has no solution. The system has an infinite number of solutions if $\vec{v}$ is collinear and confounded to $\left(O,\vec{i},\vec{j}\right)$ after the transformation.

The inverse matrix will revert the transformation, expressing $\vec{v}$ in the basis represented by the inverse matrix is equivalent to $\vec{x}$ in the standard basis

$$A^{-1} \vec{v} = \vec{x}$$

The vector $\vec{v}\begin{pmatrix}7\\-3\end{pmatrix}$ is transformed by the inverse matrix and its tip is the solution of the system (and the intersection of the two lines).


Using the two line equations, it is trivial to see that the intersection point means there is a single unique solution, and that if the lines are parallel to each other there are no solutions. If the lines are parallel to each other and confounded, there is an infinite number of solutions.

But for non-invertible matrices, consider the linear system $\begin{cases}2x + \frac{1}{3}y &= 7\\8x + \frac{4}{3}y &= 28\end{cases}$

We see that the lines are parallel to each other and confounded, so there is an infinite number of solutions. But from the matrix viewpoint, we only see that the determinant is zero thus the matrix cannot be inverted. We have

$$\underbrace{\begin{bmatrix}2 & \frac{1}{3}\\8 & \frac{4}{3}\end{bmatrix}}_{T}\times\underbrace{\begin{bmatrix}x\\y\end{bmatrix}}_{\vec{x}} = \underbrace{\begin{bmatrix}7\\28\end{bmatrix}}_{\vec{v}}$$

However, if we keep $\vec{j}$ fixed, we can move freely $\vec{i}$ and the vector $\vec{v}$ is transformed so that its tip is touching both confounded lines, so $\left(O,\vec{i},\vec{j}\right)$ is one of all the possible inverse bases for the system.


We thus have an infinite number of inverse matrices such as

$$\underbrace{\begin{bmatrix}a & b\\c & d\end{bmatrix}}_{T^{-1}}\times\underbrace{\begin{bmatrix}7\\28\end{bmatrix}}_{\vec{v}} = \underbrace{\begin{bmatrix}x\\y\end{bmatrix}}_{\vec{x}}$$

where $x,y$ are the coordinates for $\vec{x}$ when the tip of the vector is touching both confounded lines.

How can we determine, from the matrix viewpoint, if a system has no solution or an infinite number of solutions, since the determinant is zero in both cases?

  • $\begingroup$ Infinite solutions or infinitely many solutions? $\endgroup$ Jan 22 '20 at 21:43
  • $\begingroup$ @RodrigodeAzevedo you are right, infinitely many solutions. $\endgroup$
    – explogx
    Jan 22 '20 at 21:46

Suppose we had the linear system represented by the matrix equation $Ax=b$. If $A$ is singular (that is, $\det A=0$), then whether there is a unique solution or infinitely many solutions depends on whether or not a single solution exists. If there exists a single solution to $Ax=b$, then there must be infinitely many.

In particular, if $x_0$ is any of the infinitely many solutions to $Ax_0=0$, and $x_p$ is a solution to $Ax_p=b$, then the general solution is given by $x=x_0+x_p$.

  • 1
    $\begingroup$ How do you determine those solutions? $\endgroup$
    – explogx
    Jan 21 '20 at 2:27
  • $\begingroup$ @gatosec often reduction to reduced row echelon form using elementary row operations (Gaussian elimination) is the simplest way. $\endgroup$
    – YiFan
    Jan 21 '20 at 2:40
  • $\begingroup$ Cramer's rule cannot be used I guess since the determinant is zero. $\endgroup$
    – explogx
    Jan 21 '20 at 2:50
  • $\begingroup$ Yes, that's right. $\endgroup$
    – YiFan
    Jan 21 '20 at 2:50
  • 1
    $\begingroup$ @gatosec you might want to look up the concept of a pseudo-inverse, and see if that's what you're looking for. $\endgroup$
    – YiFan
    Jan 21 '20 at 21:21

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