Volume as the Determinant of a Square Linear System

We have seen that the space of linear equations is actually a vector space, and that the linear equations in any particular system span a subspace of this vector space. However, there is also another way to interpret linear systems in terms of vectors: a linear system can be interpreted as a single vector equation stating that some multiples of particular vectors add up to another particular vector.

For example, we can write the system below as a vector equation by interpreting each side of the equation as a vector:

$\begin{align*} x+2y+3z = 2 \\ −x−y−2z = 1 \\ 2x+4y+7z = 5 \\ \text{ } \\ \begin{pmatrix} x+2y+3z \\ −x−y−2z \\ 2x+4y+7z \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} \\ \text{ } \\ \begin{pmatrix} x \\ −x \\ 2x \end{pmatrix} + \begin{pmatrix} 2y \\ −y \\ 4y \end{pmatrix} + \begin{pmatrix} 3z \\ −2z \\ 7z \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} \\ \text{ } \\ \begin{pmatrix} 1 \\ −1 \\ 2 \end{pmatrix} x + \begin{pmatrix} 2 \\ −1 \\ 4 \end{pmatrix} y + \begin{pmatrix} 3 \\ −2 \\ 7 \end{pmatrix} z = \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} \end{align*}$


This equation states that some multiples $x$, $y$, and $z$ of the coefficient vectors $\left< 1,-1,2 \right>$, $\left< 2,-1,4 \right>$, and $\left< 3,-2,7 \right>$ sum to the constant vector $\left< 2,1,5 \right>$. You might recall that we solved this system earlier using reduction, and we found that the solution was $x=-5$, $y=2$, and $z=1$. Now, we see that these are simply the multiples of the coefficient vectors that sum to the constant vector.

$\begin{align*} \begin{pmatrix} 1 \\ −1 \\ 2 \end{pmatrix} (−5) + \begin{pmatrix} 2 \\ −1 \\ 4 \end{pmatrix} (2) + \begin{pmatrix} 3 \\ −2 \\ 7 \end{pmatrix} (1) = \begin{pmatrix} −5 \\ 5 \\ −10 \end{pmatrix} + \begin{pmatrix} 4 \\ −2 \\ 8 \end{pmatrix} + \begin{pmatrix} 3 \\ −2 \\ 7 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} \end{align*}$


For the linear system to have a solution, there must be some multiples of the coefficient vectors that add to the constant vector. In other words, for the linear system to have a solution, the constant vector must be in the span of the coefficient vectors.

Thinking about linear systems in terms of coefficient vectors can provide useful intuition. For example, we can tell that the linear system below has a solution because its coefficient vectors span the full 2-dimensional plane.

$\begin{align*} \begin{matrix} x&+&y&+&2z &=& 5 \\ x & + & 2y & + & 2z &=& −3 \end{matrix} \hspace{.5cm} \rightarrow \hspace{.5cm} x \begin{pmatrix} 1 \\ 1 \end{pmatrix} + y \begin{pmatrix} 1 \\ 2 \end{pmatrix} + z \begin{pmatrix} 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ −3 \end{pmatrix} \end{align*}$


Moreover, since there are 3 coefficient vectors spanning a 2-dimensional plane, there must be a dependent vector, so there must be infinitely many solutions. For example, $\left< 2,2 \right>$ is a multiple of $\left< 1,1 \right>$, so in any solution we can increase $z$ by some amount and decrease $x$ by twice that amount to yield another solution. Thus since $x=13, \, y=-8, \, z=0$ is a solution, so is $x=11, \, y=-8, \, z=1$, and $x=9, \, y=-8, \, z=2$, and so on.

When there are exactly N coefficient vectors that form an N-dimensional parallelepiped, we can also extend this intuition to relate to the volume of the coefficient vectors. Such linear systems are called square linear systems because they consist of N rows of equations and N columns of variables. In a square system, the volume of the coefficient vectors is called the determinant, because it determines much about the solutions of the system.

When the determinant is nonzero, there is exactly one solution. When the determinant is nonzero, the N coefficient vectors form a parallelepiped which extends some nonzero amount in all N dimensions, and consequently the coefficient vectors span the full N-dimensional space, guaranteeing a solution. Moreover, the solution must be unique. For N vectors to span N dimensions, the vectors must be independent – meaning that no vector can be written in terms of the others, and thus guaranteeing that there is only one solution.

For example, the following linear system has a nonzero determinant, and a single solution $x=0, \, y=1, \, z=1$.

$\begin{align*} \begin{matrix} x&+&y&+&2z &=& 3 \\ x & + & y & + & z &=& 2 \\ x & + & 2y & + & 3z & = & 5 \end{matrix} \hspace{.5cm} &\rightarrow \hspace{.5cm} x \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + y \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} + z \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 5 \end{pmatrix} \\ \mbox{determinant} = \mbox{det} \begin{pmatrix} \left< 1,1,1 \right> \\ \left< 1,1,2 \right> \\ \left< 2,1,3 \right> \end{pmatrix} &= V \begin{pmatrix} \left< 1,1,1 \right> \\ \left< 1,1,2 \right> \\ \left< 2,1,3 \right> \end{pmatrix} = 1 \end{align*}$


On the other hand, when the determinant is zero, there are either no solutions or infinitely many solutions. When the determinant is zero, the coefficient vectors form a parallelepiped that is flat in at least one dimension, and consequently the coefficient vectors span only a smaller subspace of N-dimensional space, which may or may not contain the constant vector. If the subspace does not contain the constant vector, then there is no solution. If the subspace does contain the constant vector, then there is a solution, and moreover, since a set of N vectors spanning fewer than N dimensions must contain at least one dependent vector, there must be infinitely many solutions.

For example, the following linear system has a zero determinant and no solutions.

$\begin{align*} \begin{matrix} x&+&y&+&2z &=& 3 \\ x & + & y & + & 2z &=& 4 \\ x & + & 2y & + & 3z & = & 5 \end{matrix} \hspace{.5cm} &\rightarrow \hspace{.5cm} x \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + y \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} + z \begin{pmatrix} 2 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix} \\ \mbox{determinant} = \mbox{det} \begin{pmatrix} \left< 1,1,1 \right> \\ \left< 1,1,2 \right> \\ \left< 2,2,3 \right> \end{pmatrix} &= V \begin{pmatrix} \left< 1,1,1 \right> \\ \left< 1,1,2 \right> \\ \left< 2,2,3 \right> \end{pmatrix} = 0 \end{align*}$


On the other hand, the following linear system has a zero determinant, and infinitely many solutions.

$\begin{align*} \begin{matrix} x&+&y&+&2z &=& 3 \\ x & + & y & + & 2z &=& 3 \\ x & + & 2y & + & 3z & = & 5 \end{matrix} \hspace{.5cm} &\rightarrow \hspace{.5cm} x \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + y \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} + z \begin{pmatrix} 2 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 5 \end{pmatrix} \\ \mbox{determinant} = \mbox{det} \begin{pmatrix} \left< 1,1,1 \right> \\ \left< 1,1,2 \right> \\ \left< 2,2,3 \right> \end{pmatrix} &= V \begin{pmatrix} \left< 1,1,1 \right> \\ \left< 1,1,2 \right> \\ \left< 2,2,3 \right> \end{pmatrix} = 0 \end{align*}$


One solution, for example, is $x=0, \, y=1, \, x=1$. But since $\left< 2,2,3 \right>$ is the sum of $\left< 1,1,1 \right>$ and $\left< 1,1,2 \right>$, we can obtain another solution by increasing $x$ and $y$ by some amount, and decreasing $z$ by that same amount. For example, another solution is $x=1, \, y=2, \, x=0$, and yet another solution is $x=2, \, y=3, \, x=-1$.

Later, we will see that the determinant plays a fundamental role in understanding transformations of vectors, which are called matrices. For now, though, we will just get in the habit of writing volume using the determinant operator $\det$ in place of $V$.


Practice Problems


Determine whether the system has A) exactly one solution, or B) no solutions or infinitely many solutions. (You can view the solution by clicking on the problem.)

$\begin{align*} 1) \hspace{.5cm} x+y &= 1 \\ 2x+y &= 3 \end{align*}$
Solution:
$\begin{align*} \text{A) exactly one solution} \end{align*}$


$\begin{align*} 2) \hspace{.5cm} 2x+4y &= 0 \\ −x−2y &= 3 \end{align*}$
Solution:
$\begin{align*} &\text{B) no solutions or} \\ &\text{infinitely many solutions} \end{align*}$


$\begin{align*} 3) \hspace{.5cm} x−y+z &= 10 \\ 2x+y+z &= 12 \\ 4x−y+3z &= 13 \end{align*}$
Solution:
$\begin{align*} &\text{B) no solutions or} \\ &\text{infinitely many solutions} \end{align*}$


$\begin{align*} 4) \hspace{.5cm} x+y+z &= 27 \\ x+2y &= 31 \\ y+2z &= 42 \end{align*}$
Solution:
$\begin{align*} \text{A) exactly one solution} \end{align*}$


$\begin{align*} 5) \hspace{.5cm} x−y &= 53 \\ x+y+z &= 11 \\ x+z &= 16 \end{align*}$
Solution:
$\begin{align*} \text{A) exactly one solution} \end{align*}$


$\begin{align*} 6) \hspace{.5cm} 7x+3y−2z &= 11 \\ x−y−z &= 12 \\ 8x−2y−3z &= −7 \end{align*}$
Solution:
$\begin{align*} \text{A) exactly one solution} \end{align*}$


$\begin{align*} 7) \hspace{.5cm} 2x+2y−z &= −27 \\ x+y+z &= 11 \\ 7x+5y &= 83 \end{align*}$
Solution:
$\begin{align*} \text{A) exactly one solution} \end{align*}$


$\begin{align*} 8) \hspace{.5cm} x+y−3z &= 12 \\ x−y+3z &= −12 \\ y−3z &= 4 \end{align*}$
Solution:
$\begin{align*} &\text{B) no solutions or} \\ &\text{infinitely many solutions} \end{align*}$


$\begin{align*} 9) \hspace{.5cm} w + x + y + z &= 42 \\ w−x−y−z &= 27 \\ x+z &= −11 \\ w+2x+z &= −14 \end{align*}$
Solution:
$\begin{align*} \text{A) exactly one solution} \end{align*}$


$\begin{align*} 10) \hspace{.5cm} w+x+y−z &= 12 \\ 3w+2x+y &= −7 \\ x+2y+3z &= 11 \\ w+2x+3y+2z &= 14 \end{align*}$
Solution:
$\begin{align*} &\text{B) no solutions or} \\ &\text{infinitely many solutions} \end{align*}$


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