Solving Systems in Linear Algebra: Elimination vs. Matrix Inverse

In first-year linear algebra, you're often told you can solve a system of two equations:

2x + 3y = 7

4x + 6y = 14

by writing

A = $\begin{bmatrix}2 & 3 \\ 4 & 6\end{bmatrix}$, $x = \begin{bmatrix}x \\ y\end{bmatrix}$, $b = \begin{bmatrix}7 \\ 14\end{bmatrix}$

and then, in principle, using $x = A^{-1}b$.

But what actually happens?

Matrix Inverse Attempt (Wrong Path)

1. Compute $\det(A) = 2 \times 6 - 3 \times 4 = 12 - 12 = 0$
2. Try to write the inverse:
   • $A^{-1} = \frac{1}{\det(A)} \begin{bmatrix}6 & -3 \\ -4 & 2\end{bmatrix}$
   • But $\det(A) = 0$, so you're dividing by zero.
3. Attempting to multiply by this "inverse" produces undefined or infinite values—nothing meaningful.

The calculation cannot continue. The "answer" does not exist by this route. That's the check: division by zero means the method fails here.

Switch to the elimination method—row operations or substitution, which rely on linear combinations, not the invertibility of A.

Take the original system:

2x + 3y = 7 (1) 4x + 6y = 14 (2)

Multiply (1) by 2:

(1) x 2: 4x + 6y = 14

Now compare to (2):

(2): 4x + 6y = 14

So both equations are the same. That is, every solution to (1) is also a solution to (2). There is NOT just one solution; the entire line defined by 2x + 3y = 7 is the solution set.

General solution: Solve (1) for one variable in terms of the other.

2x + 3y = 7 → 2x = 7 - 3y → x = (7 - 3y)/2

So all points $\left(\frac{7-3y}{2},\ y\right)$ for real y are solutions.

This method works whether or not the matrix is invertible, as long as the system is consistent. The method’s cost is a few steps of algebra, but it always reveals whether the system has one solution, infinitely many, or none.

The matrix inverse method depends on the Invertible Matrix Theorem: a square matrix A has an inverse if and only if it is invertible, which is true exactly when $\det(A) \neq 0$. This is why the method failed above: a zero determinant signals the matrix is singular (not invertible), so the system cannot have exactly one solution.

Elimination, on the other hand, is just algebraic manipulation—it works for any system, square or not, and will reveal no solution (contradiction), one solution, or infinitely many (dependence).

The determinant check is non-negotiable. If $\det(A) = 0$, the inverse does not exist: no amount of rearrangement will save the method.

The matrix inverse method is fast and clean for small systems with exactly one solution and a nonzero determinant. For example, try:

x + y = 3

2x - y = 0

A = $\begin{bmatrix}1 & 1 \\ 2 & -1\end{bmatrix}$, b = $\begin{bmatrix}3 \\ 0\end{bmatrix}$

$\det(A) = 1 \times -1 - 1 \times 2 = -1 - 2 = -3 \neq 0$

Inverse exists:

$A^{-1} = \frac{1}{-3} \begin{bmatrix}-1 & -1 \\ -2 & 1\end{bmatrix}$

$x = A^{-1}b = \frac{1}{-3}\begin{bmatrix}-1 & -1 \\ -2 & 1\end{bmatrix}\begin{bmatrix}3 \\ 0\end{bmatrix} = \frac{1}{-3}\begin{bmatrix}-3 \\ -6\end{bmatrix} = \begin{bmatrix}1 \\ 2\end{bmatrix}$

Check: (1) 1+2=3, (2) 2-2=0. The seconds cancel. That's the check.

In this boundary case—square, unique solution, nonzero determinant—the inverse method is quick and gives the exact answer.

But for larger systems or those with repeated or dependent equations, elimination is both more general and more transparent. The inverse method is blind to infinite or zero solutions: it just fails quietly. Elimination tells you why.

Try this system:

3x + 6y = 9

x + 2y = 3

Write as A = [[3,6],[1,2]], b = [9,3]

Determinant: 3x2 - 6x1 = 6-6 = 0.

Again, the inverse breaks: division by zero. But elimination quickly shows these are the same line (since the first equation is three times the second), so the solution is infinite: any (x, y) with x + 2y = 3.

When the system is not square (e.g., more variables than equations), the inverse method is not even defined, but elimination still works.

Solve the system:

2x - y = 1

5x + 3y = 14

1. Write as a matrix equation and check the determinant.
2. Attempt the inverse method and elimination.

Answer: $\det = 2 \times 3 - (-1) \times 5 = 6+5=11 \neq 0$ so both methods work. Solution: x = 2, y = 3.

When the determinant is nonzero, either method is valid, but elimination is safer when the matrix might be singular or the system is not square.