Learning Lab

What Does the Coefficient Matrix Tell Us?

The coefficient matrix can be used to find information about the number of solutions to a system of equations. For any system of equations, if the coefficient matrix is:

1. Singular (that is the determinant \(=0\)) there will be an infinite number of solutions or no solutions.

2. Non-singular (determinant \(\neq0\)) there will be a unique solution.

Example 1

Consider the system of equations \[\begin{align*} x+2y-z & =3\\ x+3y & =4\\ -x-y+2z & =-2 \end{align*}\]

The augmented matrix for this system is \[\begin{align*} \left[\begin{array}{ccc} 1 & 2 & -1\\ 1 & 3 & 0\\ -1 & -1 & 2 \end{array}\left|\begin{array}{c} 3\\ 4\\ -2 \end{array}\right.\right]. \end{align*}\] Using elementary row operations we get: \[\begin{align*} \begin{array}{c} \\ R_{2}'=R_{2}-R_{1}\\ R_{3}'=R_{3}+R_{1} \end{array}\left[\begin{array}{ccc} 1 & 2 & -1\\ 0 & 1 & 1\\ 0 & 1 & 1 \end{array}\left|\begin{array}{c} 3\\ 1\\ 1 \end{array}\right.\right] \end{align*}\] \[\begin{align*} \begin{array}{c} \\ \\ R_{3}'=R_{3}-R_{2} \end{array}\left[\begin{array}{ccc} 1 & 2 & -1\\ 0 & 1 & 1\\ 0 & 0 & 0 \end{array}\left|\begin{array}{c} 3\\ 1\\ 0 \end{array}\right.\right] \end{align*}\] Row \(3\) tells us that \(0x+0y+0z=0.\) This is true for any value of \(x,y\) or \(z.\) So we are able to set any one of these variables to some number \(t\in\mathbb{R}\).11 It can only be one variable as there other equations to be satisfied in \(R_{2}\) and \(R_{1}.\) For this example,22 Even though we set \(z=t,\) we could set either \(x=t\) or \(y=t\). We still get a solution to the system of equations but it will look different. we set \(z=t.\) From \(R_{2}\) and substituting \(t\) for \(z\) we have \[\begin{align*} y+z & =1\\ y+t & =1\\ y & =1-t. \end{align*}\] Now from \(R_{1}\), substituting \(z=t,\,y=1-t\) we have \[\begin{align*} x+2y-z & =3\\ x+2\left(1-t\right)-t & =3\\ x+2-2t-t & =3\\ x & =1+3t. \end{align*}\] So the solution is \(x=1+3t,\,y=1-t\) and \(z=t\) where \(t\in\mathbb{R}\).33 As there are an infinite number of choices for \(t,\) there are an infinite number of solutions to the system of equations.

In general if the reduced augmented matrix has one or more rows of zeros, there will be an infinite number of solutions to the system of equations.44 You may recall that the determinant of a matrix with a row of zeros is zero. Hence the matrix is singular.

Example 2

Consider the system of equations \[\begin{align*} x+2y-z & =3\\ 2x+4y-2z & =6\\ -5x-10y+5z & =-15 \end{align*}\]

The augmented matrix for this system is \[\begin{align*} \left[\begin{array}{ccc} 1 & 2 & -1\\ 2 & 4 & -2\\ -5 & -10 & 5 \end{array}\left|\begin{array}{c} 3\\ 6\\ -15 \end{array}\right.\right]. \end{align*}\] Using elementary row operations we get \[\begin{align*} \begin{array}{c} \\ R_{2}'=R_{2}-2R_{1}\\ R_{3}'=R_{3}+5R_{1} \end{array}\left[\begin{array}{ccc} 1 & 2 & -1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\left|\begin{array}{c} 3\\ 0\\ 0 \end{array}\right.\right]. \end{align*}\] We have only one equation and three unknowns. In this case we must let two of the three variables be free. For example, we set \(z=t\), where \(t\in\mathbb{R}\) and \(y=s,\) where \(s\in\mathbb{R}\) .55 It doesn’t matter which two variables are selected to be a parameter \(s\) or \(t.\) We could set \(x=s\) and \(y=t.\) Substituting in row \(1\) gives: \[\begin{align*} s+2t-z & =3\\ z & =s+2t-3. \end{align*}\] While this looks different to the solution at left, it is still a valid solution to the set of equations. Substituting into \(R_{1}\) we get \[\begin{align*} x+2y-z & =3\\ x+2s-t & =3\\ x & =3-2s+t. \end{align*}\] So the solution is \(x=3-2s+t,\,y=s,\,z=t\) where \(s\in\mathbb{R}\) and \(t\in\mathbb{R}\).