In the previous section, we saw how a system of linear equations, like finding values for $x_1$ and $x_2$ that satisfy multiple conditions, can be neatly packaged into the matrix equation $Ax = b$ . Here, $A$ is the matrix of coefficients, $x$ is the vector of unknown variables we want to find, and $b$ is the vector of constants on the right-hand side.

But what does it actually mean to "solve" this equation $Ax = b$ ?

Think back to simple algebra. If you have an equation like $5x = 10$ , solving it means finding the value for the variable $x$ that makes the statement true. In this case, $x=2$ is the solution because $5 \times 2 = 10$ .

The concept is very similar for matrix equations, but now we are looking for a vector $x$ , not just a single number. A solution to the system $Ax = b$ is a specific vector $x$ whose components (the individual values like $x_1, x_2, \dots, x_n$ ) simultaneously satisfy all the linear equations represented within the matrix equation. When you perform the matrix multiplication $Ax$ , the result must exactly equal the vector $b$ .

Let's consider a concrete example. Suppose we have the following system of two linear equations with two unknowns:

$2x_1 + 3x_2 = 8$ $x_1 + x_2 = 3$

We can write this in matrix form $Ax = b$ as:

A = \begin{bmatrix} 2 & 3 \\ 1 & 1 \end{bmatrix}, \quad x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}, \quad b = \begin{bmatrix} 8 \\ 3 \end{bmatrix}

So the matrix equation is:

\begin{bmatrix} 2 & 3 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 8 \\ 3 \end{bmatrix}

Now, let's propose a potential solution: $x = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ . Is this the solution? To check, we substitute this vector $x$ back into the left side of the equation ( $Ax$ ) and perform the matrix multiplication:

Ax = \begin{bmatrix} 2 & 3 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} (2 \times 1) + (3 \times 2) \\ (1 \times 1) + (1 \times 2) \end{bmatrix} = \begin{bmatrix} 2 + 6 \\ 1 + 2 \end{bmatrix} = \begin{bmatrix} 8 \\ 3 \end{bmatrix}

Look at the result! The vector we calculated, $\begin{bmatrix} 8 \\ 3 \end{bmatrix}$ , is exactly the vector $b$ from our original equation. Since substituting $x = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ into $Ax$ gives us $b$ , we can confirm that $x = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ (meaning $x_1=1$ and $x_2=2$ ) is indeed a solution to this system.

Geometrically, for a system with two variables, each linear equation represents a line on a 2D plane. The solution to the system is the point where these lines intersect. Our solution $(x_1=1, x_2=2)$ corresponds to the coordinates $(1, 2)$ where the lines $2x_1 + 3x_2 = 8$ and $x_1 + x_2 = 3$ cross.

The solution $(1, 2)$ is the point where the two lines representing the equations intersect.

It's important to note that not all systems of linear equations have a single, unique solution like this one. Sometimes, there might be:

No solution: Imagine two parallel lines that never intersect. This happens if the equations are contradictory.
Infinitely many solutions: Imagine two equations that actually represent the exact same line. Any point on that line is a solution.

These cases correspond to specific properties of the matrix $A$ . For much of our work, especially when using methods like the matrix inverse (which we'll discuss next), we are interested in systems that have exactly one unique solution.

So, the "concept of a solution" boils down to finding that specific vector $x$ which, when multiplied by the matrix $A$ , produces the vector $b$ . The upcoming sections will explore methods for finding this vector $x$ , starting with the idea of the matrix inverse.