Rank of a Matrix

The rank of a matrix with m rows and n columns is a number r with the following properties:

• r is less than or equal to the smallest number out of m and n.
• r is equal to the order of the greatest minor of the matrix which is not 0.

Determining the Rank of a Matrix

• We pick an element of the matrix which is not 0.
• We calculate the order 2 minors which contain that element until we find a minor which is not 0.
• If every order 2 minor is 0, then the rank of the matrix is 1.
• If there is any order 2 minor which is not 0, we calculate the order 3 minors which contain the previous minor until we find one which is not 0.
• If every order 3 minor is 0, then the rank of the matrix is 2.
• If there is any order 3 minor which is not 0, we calculate the order 4 minors until we find one which is not 0.
• We keep doing this until we get minors of an order equal to the smallest number out of the number of rows and the number of columns.

Example 42
$A=\begin{pmatrix} 1 & 2 & 4\\ 3 & 6 & 5 \end{pmatrix}$

The matrix has 2 rows and 3 columns, so the greatest possible value of its rank is 2. We pick any element which is not 0.

$\begin{pmatrix} \color{red}{1} & 2 & 4\\ 3 & 6 & 5 \end{pmatrix}$

We form an order 2 minor containing 1.
$\begin{pmatrix} \color{red}{1} & \color{red}{2} & 4\\ \color{red}{3} & \color{red}{6} & 5 \end{pmatrix}$

We calculate this minor.
$\begin{vmatrix} \color{red}{1} & \color{red}{2}\\ \color{red}{3} & \color{red}{6} \end{vmatrix}=6 - 6 = 0$

We form another order 3 minor containing 1. $A=\begin{pmatrix} \color{blue}{1} & 2 & \color{blue}{4}\\ \color{blue}{3} & 6 & \color{blue}{5} \end{pmatrix}$

We calculate this minor.
$\begin{vmatrix} \color{blue}{1} & \color{blue}{4}\\ \color{blue}{3} & \color{blue}{5} \end{vmatrix}= 5 - 12 = -7 \neq 0.$

The rank is 2.

Example 43
$B=\begin{pmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1\\ \end{pmatrix}$

We pick an element which is not 0.
$\begin{pmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & \color{red}{1} & 1 \end{pmatrix}$

We calculate order 2 minors containing this element. $\begin{pmatrix} 1 & 1 & 1\\ \color{red}{1} & \color{red}{1} & 1\\ \color{red}{1} & \color{red}{1} & 1 \end{pmatrix}$

$\begin{vmatrix} \color{red}{1} & \color{red}{1} \\ \color{red}{1} & \color{red}{1} \end{vmatrix}=0$ (because it has two equal rows)

Every other order 2 minor is 0 because it's the same as the others. In this case, the rank of the matrix is 1.

Example 44
$B=\begin{pmatrix} 3 & 8 & 2\\ 2 & 1 & 1\\ 5 & 3 & 4\\ 7 & 4 & 5 \end{pmatrix}$

The matrix has 4 rows and 3 columns, so the greatest possible value of its rank is 3.

We pick an element which is not 0.
$\begin{pmatrix} 3 & 8 & 2\\ 2 & 1 & 1\\ 5 & 3 & \color{red}{4}\\ 7 & 4 & 5 \end{pmatrix}$

We calculate an order 2 minor containing 4.
$\begin{pmatrix} 3 & 8 & 2\\ 2 & \color{red}{1} & \color{red}{1}\\ 5 & \color{red}{3} & \color{red}{4}\\ 7 & 4 & 5 \end{pmatrix}$

$\begin{vmatrix} \color{red}{1} & \color{red}{1}\\ \color{red}{3} & \color{red}{4}\\ \end{vmatrix} = 4 - 3 = 1$

We form an order 3 minor containing the previous minor. $\begin{pmatrix} 3 & 8 & 2\\ \color{red}{2} & \color{red}{1} & \color{red}{1}\\ \color{red}{5} & \color{red}{3} & \color{red}{4}\\ \color{red}{7} & \color{red}{4} & \color{red}{5} \end{pmatrix}$

We calculate this minor.
$\begin{pmatrix} \color{red}{2} & \color{red}{1} & \color{red}{1}\\ \color{red}{5} & \color{red}{3} & \color{red}{4}\\ \color{red}{7} & \color{red}{4} & \color{red}{5} \end{pmatrix}=0$ because $R_{1}+R_{2}=R_{3}$

We calculate another order 3 minor containing the previous minor.
$\begin{pmatrix} \color{blue}{3} & \color{blue}{8} & \color{blue}{2}\\ \color{blue}{2} & \color{blue}{1} & \color{blue}{1}\\ \color{blue}{5} & \color{blue}{3} & \color{blue}{4}\\ 7 & 4 & 5 \end{pmatrix}$

$\begin{vmatrix} \color{blue}{3} & \color{blue}{8} & \color{blue}{2}\\ \color{blue}{2} & \color{blue}{1} & \color{blue}{1}\\ \color{blue}{5} & \color{blue}{3} & \color{blue}{4} \end{vmatrix} =$ $12 + 12 +40 -10 -9 -64 =-19 \neq 0$
The rank of the matrix is 3.

Example 45
$D=\begin{pmatrix} 1 & 5 & 1 & 6\\ 2 & 3 & 2 & 5\\ 6 & 1 & 6 & 7 \end{pmatrix}$

D is a matrix with 3 rows and 4 columns, so the greatest possible value of the rank is 3.

We pick an element which is not 0.
$\begin{pmatrix} 1 & \color{red}{5} & 1 & 6\\ 2 & 3 & 2 & 5\\ 6 & 1 & 6 & 7 \end{pmatrix}$

We form an order 2 minor containing 5.
$\begin{pmatrix} \color{red}{1} & \color{red}{5} & 1 & 6\\ \color{red}{2} & \color{red}{3} & 2 & 5\\ 6 & 1 & 6 & 7 \end{pmatrix}$

$\begin{vmatrix} 1 & 5\\ 2 & 3 \end{vmatrix}= 3 - 10 = -7 \neq 0$

We form an order 3 minor containing the previous minor.
$\begin{pmatrix} \color{red}{1} & \color{red}{5} & \color{red}{1} & 6\\ \color{red}{2} & \color{red}{3} & \color{red}{2} & 5\\ \color{red}{6} & \color{red}{1} & \color{red}{6} & 7 \end{pmatrix}$

$\begin{vmatrix} \color{red}{1} & \color{red}{5} & \color{red}{1}\\ \color{red}{2} & \color{red}{3} & \color{red}{2}\\ \color{red}{6} & \color{red}{1} & \color{red}{6} \end{vmatrix} = 0$ (because it has 2 equal columns)

In this case, we form another order 3 minor containing the order 2 minor which is not 0.
$\begin{pmatrix} \color{blue}{1} & \color{blue}{5} & 1 & \color{blue}{6}\\ \color{blue}{2} & \color{blue}{3} & 2 & \color{blue}{5}\\ \color{blue}{6} & \color{blue}{1} & 6 & \color{blue}{7} \end{pmatrix}$

$\begin{pmatrix} \color{blue}{1} & \color{blue}{5} & \color{blue}{6}\\ \color{blue}{2} & \color{blue}{3} & \color{blue}{5}\\ \color{blue}{6} & \color{blue}{1} & \color{blue}{7} \end{pmatrix} = 0$ because $C_{1} + C_{2}=C_{3}$

Since all order 3 minors are 0, the rank of matrix D is 2.

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