Economics at Zero Marginal Cost: Rank of a Matrix

2. Find the rank of

$A=\begin{bmatrix} 1\ \ \3\ \ \4\ -2\\ 2\ \ \6\ \ \8\ -4\\ 3\ \ \0\ \ \ 3\ \ \ \3 \end{bmatrix} \$

Solution:

Given:

$A=\begin{bmatrix} 1\ \ \3\ \ \4\ -2\\ 2\ \ \6\ \ \8\ -4\\ 3\ \ \0\ \ \ 3\ \ \ \3 \end{bmatrix} \$

The order of matrix A is 3x4

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

The possible rank of matrix A is:

$rank\ (A_{3\times 4})\leq min(3,4)\leq 3$

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

$A_{1}=\begin{bmatrix} 1\ \ \3\ \ \4\\ 2\ \ \6\ \ \8\\ 3\ \ \0\ \ \ 3 \end{bmatrix} \ \ \ \ A_{2}=\begin{bmatrix} 3\ \ \4\ -2\\ 6\ \ \8\ -4\\ 0\ \ \ 3\ \ \ \3 \end{bmatrix} \$

$\\\left |A_{1} \right |=1\begin{vmatrix} 6\ \ \8\\ 0\ \ \3\\ \end{vmatrix}-3 \begin{vmatrix} 2\ \ \8\\ 3\ \ \3\\ \end{vmatrix}+4 \begin{vmatrix} 2\ \ \6\\ 3\ \ \0\\ \end{vmatrix}\\\\ \left |A_{1} \right |=1(18-0)-3(6-24)+4(0-18)\\ \left |A_{1} \right |=1(18)-3(-18)+4(-18)\\ \left |A_{1} \right |=18+54-72\\ \left |A_{1} \right |=72-72\\ \left |A_{1} \right |=0\\$

$\\\left |A_{2} \right |=3\begin{vmatrix} 8\ -4\\ 3\ \ \ \ \3\\ \end{vmatrix}-4 \begin{vmatrix} 6\ -4\\ 0\ \ \ \ \3\\ \end{vmatrix}+(-2) \begin{vmatrix} 6\ \ \8\\ 0\ \ \3\\ \end{vmatrix}\\\\ \left |A_{2} \right |=3(24-(-12))-4(18-0)-2(18-0)\\ \left |A_{2} \right |=3(24+12)-4(18)-2(18)\\ \left |A_{2} \right |=3(36)-72-36\\ \left |A_{2} \right |=108-108\\ \left |A_{2} \right |=0\\$

$\because$ Determinant of a 3x3 order is zero.

∴ The possible rank of matrix A is $rank\ (A)\leq 2$

The possible 2x2 order sub-matrices of the given matrix are:

Sub-matrices from 1st row and 2nd row by deleting the 3rd row:

$A_{1}=\begin{bmatrix} 1\ \ \3\\ 2\ \ \6\\ \end{bmatrix}\ \ \ A_{2}=\begin{bmatrix} 3\ \ \4\\ 6\ \ \8\\ \end{bmatrix}\ \ \ A_{3}=\begin{bmatrix} 4\ \ -2\\ 8\ \ -4\\ \end{bmatrix}$

$\\\left |A_{1} \right |=6-6=0\\ \left |A_{2} \right |=24-24=0\\ \left |A_{3} \right |=-16-(-16)=-16+16=0\\$

Sub-matrices from 2nd row and 3rd row by deleting the 1st row:

$A_{4}=\begin{bmatrix} 2\ \ \6\\ 3\ \ \0\\ \end{bmatrix}\ \ \ A_{5}=\begin{bmatrix} 6\ \ \8\\ 0\ \ \3\\ \end{bmatrix}\ \ \ A_{6}=\begin{bmatrix} 8\ -4\\ 3\ \ \ \3\\ \end{bmatrix}$