Economics at Zero Marginal Cost: Rank of a Matrix

The rank of a matrix is equal to the order of the highest order non-singular matrix contained in a given matrix. If matrix A is given, then rank of A is written as: rank (A)

The rank of a matrix is the maximum number of linearly independent rows (columns) of a given matrix. The rank of a matrix would be zero only if the matrix has all zero elements, i.e., when it is a null matrix. If a matrix has only one element, its minimum rank would be one.

If A is m x n, then $rank\ (A_{m\times n})\leq min(m,n)$ , where min( m, n) denotes the smaller of the two numbers m and n (or their common value if m = n).

For example, the rank of a 3 x 5 matrix can be no more than 3, and the rank of a 4 x 2 matrix can be no more than 2.

If A is a 3 x 5 matrix, then $rank\ (A_{3\times 5})\leq min(3,5)\leq3$

1. Find the rank of

$A=\begin{bmatrix} 1\ \ \ \ \ \ \ \ \2\ \ \ \ \ \ \ \3\\ -2\ \ \ \ \ \ 4\ \ \ \ \ \ \6\\ -3\ \ \ -6\ \ \ -9 \end{bmatrix} \$

Solution:

Given:

$A=\begin{bmatrix} 1\ \ \ \ \ \ \ \ \2\ \ \ \ \ \ \ \3\\ -2\ \ \ \ \ \ 4\ \ \ \ \ \ \6\\ -3\ \ \ -6\ \ \ -9 \end{bmatrix} \$

The order of matrix A is 3x3

Using the formula: $rank\ (A_{m\times n})\leq min(m,n)$

The possible rank of matrix A is:

$rank\ (A_{3\times 3})\leq min(3,3)\leq 3$

The highest possible sub-matrix of the given matrix is 3x3 order :

$A=\begin{bmatrix} 1\ \ \ \ \ \ \ \ \2\ \ \ \ \ \ \ \3\\ -2\ \ \ \ \ \ 4\ \ \ \ \ \ \6\\ -3\ \ \ -6\ \ \ -9 \end{bmatrix} \$

$\\\left |A \right |=1\begin{vmatrix} 4\ \ \ \ \6\\ -6\ -9 \end{vmatrix}-2 \begin{vmatrix} 2\ \ \ \ \6\\ -3\ -9 \end{vmatrix}+3 \begin{vmatrix} 2\ \ \ \ \4\\ -3\ -6 \end{vmatrix}\\\\ \left |A \right |=1(-36-(-36))-2(-18-(-18))+3(-12-(-12))\\ \left |A \right |=1(-36+36)-2(-18+18)+3(-12+12)\\ \left |A \right |=1(0)-2(0)+3(0)\\ \left |A \right |=0-0+0\\ \left |A \right |=0\\$