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I was wondering how can I determine if there is a line of intersection with any matrix?

For example, if I have the following matrix:

$$\left(\begin{array}{rrr|r} 1 & -3 & -2 & -9 \\ 2 & -5 & 1 & 3 \\ -3 & 6 & 2 & 8 \\ \end{array} \right)$$

What does the solution have to look like for me to conclude that there is a line of intersection?

P.S. I know this matrix has a point of intersection but I used this as an example because I didn't know an example for a matrix that had a line of intersection.

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    $\begingroup$ If there are infinitely many solutions, then you are guaranteed to find a line (more generally subspace) of solutions. An equivalent condition is if the rank of the matrix is not full, and at least 1 solution exists. $\endgroup$
    – Calvin Lin
    Jun 11, 2013 at 14:03

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The existence of a "point of intersection" is the existence of a point $(x, y, z)$ that satisfies the system of equations: a point lying on each line represented by the corresponding system of equations:

$$\begin{align} x - 3y -2z & = -9 \\ 2x - 5y + z & = 3 \\ -3x + 6y + 2z & = 8 \end{align}$$

A unique solution exists (a single point of intersection exists) if the augmented matrix does not reduce to a row of all zeros, and no row has all zeros, augmented by a non-zero entry.

If you obtain a row of all zeros, through row reduction, then infinitely many points of intersection occur (two or more lines will be concurrent): the entries in one or more lines will be a scalar multiple of the entries of another. Put differently, there will be a line of intersection.

If the matrix reduces to a row of three zeros, with a non-zero entry in the last column of that row, no solution exists (i.e., no point of intersection exists.)

All these possibilities can be determined by reducing the matrix to row echelon form.

We can reduce your example matrix:

$$\left(\begin{array}{rrr|r} 1 & -3 & -2 & -9 \\ 2 & -5 & 1 & 3 \\ -3 & 6 & 2 & 8 \\ \end{array}\right)$$

$$\left(\begin{array}{rrr|r} 1 & -3 & -2 & -9 \\ 0 & 1 & 5 & 21 \\ 0 & 0 & 11 & 42 \\ \end{array}\right)$$

If reduced further, you'd see that the system represented by the system of equations has a unique point of intersection, hence no common line of intersection.

Note that if the last row were $(0\;\;2\;\;10\;\;42)$, we could "zero it" by taking $-2R_2 + R_3 \rightarrow R_3$, and obtain a row of all zeros, since the third row would be a scalar multiple of the second row. In that event, there would indeed be a line of intersection. If any row is a linear combination of the other rows, we have a linearly dependent system of equations: this shows when a row-reduced matrix has a row of all zeros. And in that case, there is at least a line of intersection.

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  • $\begingroup$ So if the third row of the system is a multiple of the second row or the first row then there is a line of intersection? I'm confused about how it would look in terms of $x$'s and $y$'s and $z$'s. Because for a the final answer for a P.O.I I would just write the solution but what would I write for a L.O.I? $\endgroup$
    – Jeel Shah
    Jun 11, 2013 at 14:43
  • $\begingroup$ Very nice Amy and good morning $\ast$ $\endgroup$
    – Mikasa
    Jun 11, 2013 at 14:44
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    $\begingroup$ @gekkostate What one typically does is set, say, $z = \alpha$, where $\alpha$ is any real number. Then express $x, y$ as values in terms of $z = \alpha$. So say you reduce the matrix to the rows the first two rows in the above row-reduced matrix, but the last row is all zeros. Then the solution is given by $z = \alpha, y = 21 - 5 \alpha, x = -9 + 2\alpha + 3(21 - 5\alpha)$, simplifying x, of course. $\alpha$ is then called a parameter. $\endgroup$
    – amWhy
    Jun 11, 2013 at 14:56
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    $\begingroup$ Some use the parameter $t$ instead of $\alpha$. This post gives an example of such a system: infinite number of solutions. $\endgroup$
    – amWhy
    Jun 11, 2013 at 15:02
  • $\begingroup$ @amWhy, however, in a similar question to this, I also let $\alpha$ to be the parameter, however, the solution of the line of intersection is different from the one when I just do cross product between two of the normal vectors of the plane. Why is that? $\endgroup$
    – CCC
    Jan 9, 2016 at 10:57
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If you are familiar with ranks and Rouché-Capelli (RC) theorem, the following reasoning can be used to answer. The system $A\mathbf x=\mathbf b$, where $A$ is an $n\times m$ matrix, has solutions if and only if $$ r(A)=r(A|\mathbf b)=r, $$ and in this case you have "$\infty^{n-r}$" solutions. This really means that your solutions depend on $n-r$ free parameters. So, if the rank $r=n-1$, your solution set depends on $n-(n-1)=1$ free parameter or, in other words, it's monodimensional. This helps the intuition that the solution set is indeed a line in this case. Hence, you'll get a line of solutions when the rank of the complete matrix $r$ is equal to the number of unknowns $n$ less 1. As an example, consider the system $$ \begin{align} x - y & = 2 \\ -2x +2y &=-4 \end{align} $$ The rank $r(A)=r(A|\mathbf b)=1$ and you can indeed write the solutions set as $x=2+t,y=t$, which is equal to the set of points on the line $y=x-2$.

A more interesting example (I like a lot $A$!) is $$ \begin{align} x +y +z& = 6 \\ 4x +5y +6z&=15\\ 7x+8y+10z&=24 \end{align} $$ The reduced complete matrix is $$ \left(\begin{array}{rrr|r} 1 & 2 & 3 & 6 \\ 0 & -3 & -6 & -9 \\ 0 & 0 & 0 & 0 \\ \end{array}\right) $$ from which you get the solutions $(x=t,y=3-2t,z=t), \forall t$. The solution set can be described in terms of a unique parameter $t$ and it is a straight line in $\mathbf R^3$.

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