Question

what are the benefits of finding rank in a matrices?

Answer 1

Maybe of the most practical importance to a first year linear algebra student is that the rank gives you information about the number of solutions to a system of linear equations.

Example:
\[\begin{cases}x+y=1\\
y+z=1\\
x+2y+z=2\end{cases}\iff\begin{bmatrix}1&1&0\\0&1&1\\1&2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\1\\2\end{bmatrix}\]Upon inspection, you can see that the third row is a linear combination of the first and second rows, while the remaining rows are independent. Indeed, reducing the rows of the coefficient matrix, yields
\[\left[\begin{array}{ccc|c}
1&1&0&1\\0&1&1&1\\1&2&1&2\end{array}\right]\implies\left[\begin{array}{ccc|c}
\boxed1&0&-1&0\\0&\boxed1&1&1\\0&0&0&0\end{array}\right]\]where the boxes surround the pivots of the matrix. The number of pivots of a matrix is the matrix's rank. For this matrix, because the rank is smaller than the number of columns, this matrix has a non-trivial nullspace (at least one non-zero vector in the nullspace). What this means is that the equation \(\mathbf{Ax}=\mathbf{b}\), if it has a particular solution, then it also has an infinite number of additional solutions.

The particular solution to this system is \(\begin{bmatrix}0\\1\\0\end{bmatrix}\), but adding any scalar multiple of \(\begin{bmatrix}1\\-1\\1\end{bmatrix}\) also satisfies the system; that is, the solution set is \(\left\{\mathbf{x}=\begin{bmatrix}k\\1-k\\k\end{bmatrix}:k\in\mathbb{R}\right\}\).