Question

Use Normal form to find the rank of matrix A, where A = \left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 2&4&6 \\\ 4&8&{12} \end{array}} \right]

Answer 1

Hint : The normal form of a matrix is obtained from its original matrix by undergoing transformations on the rows and columns. The transformations include multiplying a row with a certain integer and subtracting the values of the row from another row and placing the result in its previous place. Convert the given matrix into its normal form and find its rank.

Complete step-by-step answer :
We are given a matrix A where A = \left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 2&4&6 \\\ 4&8&{12} \end{array}} \right] and we have to find its rank using its normal form.
The matrix A has 3 rows and 3 columns.
${R_1},{R_2},{R_3}$ are the 3 rows and ${C_1},{C_2},{C_3}$ are the 3 columns.
Convert the matrix A into its normal form by transformations.
First transformation is ${R_3} = {R_3} - 4\left( {{R_1}} \right)$ , which means transforming the 3rd row by subtracting 4 times of 1st row from 3rd row

$${R_3} - 4{R_1} = \left( {4,8,12} \right) - 4\left( {1,2,3} \right) \\\ = \left( {4,8,12} \right) - \left( {4,8,12} \right) \\\ {R_3} - 4{R_1} = \left( {0,0,0} \right) \\\$$

Replace the result in the place of 3rd row.

1&2&3 \\\ 2&4&6 \\\ 0&0&0 \end{array}} \right]$$ Second transformation is $${R_2} = {R_2} - 2\left( {{R_1}} \right)$$ , which means transforming the 2nd row by subtracting 2 times of 1st row from 2nd row

{R_2} - 2{R_1} = \left( {2,4,6} \right) - 2\left( {1,2,3} \right) \\
= \left( {2,4,6} \right) - \left( {2,4,6} \right) \\
{R_2} - 2{R_1} = \left( {0,0,0} \right) \\

Replace the result in the place of the 2nd row. $$A = \left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 0&0&0 \\\ 0&0&0 \end{array}} \right]$$ Here, we cannot undergo further transformation as the rows except 1st row are all zeros. Rank of a matrix can be told as the number of non-zero rows in its normal form. Here, there is only one no zero row. Therefore, Rank of the matrix $$A = \left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 2&4&6 \\\ 4&8&{12} \end{array}} \right]$$ is 1. **Note** : In the normal form of a matrix, every row can have a maximum of a single one and rest are all zeroes. There can also be rows with all zeros. Rank of the matrix can be found from its normal form by counting the no. of non-zero rows. Normal form is also known as canonical form or standard from.