We begin with a familiar example, performed in a novel way.
The matrix $B$ of the previous example is called the inverse of $A$. When $A$ and $B$ are combined via matrix multiplication, the result is the identity matrix, which can be inserted “in front” of $\vect{x}$ as the first step in finding the solution. This is entirely analogous to how we might solve a single linear equation like $3x=12$. \begin{equation*} x=1x=\left(\frac{1}{3}\left(3\right)\right)x=\frac{1}{3}\left(3x\right)=\frac{1}{3}\left(12\right)=4 \end{equation*}
Here we have obtained a solution by employing the “multiplicative inverse” of $3$, $3^{-1}=\frac{1}{3}$. This works fine for any scalar multiple of $x$, except for zero, since zero does not have a multiplicative inverse. Consider separately the two linear equations, \begin{align*} 0x&=12 & 0x&=0 \end{align*}
The first has no solutions, while the second has infinitely many solutions. For matrices, it is all just a little more complicated. Some matrices have inverses, some do not. And when a matrix does have an inverse, just how would we compute it? In other words, just where did that matrix $B$ in the last example come from? Are there other matrices that might have worked just as well?