# Elementary Row (Column) Transformations

## Elementary Row (Column) Transformations

The following three operations applied on the rows (columns) of a matrix are called elementary row (column) transformations.

(i) Interchange of any two rows (columns): If ith row (column) of a matrix is interchanged with the jth row (column), it will be denoted by Ri ↔ Rj (Ci ↔ Cj).

For Example: $$A=\left[ \begin{matrix}1 & 2 & 3 \\ 5 & 6 & 1 \\ 3 & 5 & 6 \\\end{matrix} \right]$$, then by applying R₁ → R₂ we get $$B=\left[ \begin{matrix} 5 & 6 & 1 \\1 & 2 & 3 \\3 & 5 & 6 \\\end{matrix} \right]$$

(ii) Multiplying all elements of a row (column) of a matrix by a non-zero scalar. If the elements of ith row (column) are multiplied by non-zero scalar k, it will be denoted by Rᵢ → Rᵢk [Cᵢ →Cᵢ (k)] or Rᵢ → kRᵢ [Cᵢ → kCᵢ].

For Example: If $$A=\left[ \begin{matrix}3 & 2 & 1 \\0 & 6 & 2 \\-1 & 2 & -2 \\\end{matrix} \right]$$, then by applying R₂ → 3R₂, we obtain $$B=\left[ \begin{matrix}3 & 2 & 1 \\0 & 18 & 6 \\-1 & 2 & -2 \\\end{matrix} \right]$$.

(iii) Adding to the elements of a row (column), the corresponding elements of any other row (column) multiplied by any scalar k.

If k times the elements of jth row (column) are added to the corresponding elements of the ith row (column), it will be denoted by Rᵢ →Rᵢ + k Rᵢ (Cᵢ →Cᵢ + kCᵢ).

For Example: If $$A=\left[ \begin{matrix} 0 & 1 & 1 \\ -1 & 2 & 4 \\3 & 2 & 5 \\\end{matrix} \right]$$, then application of elementary operation R₃→R₃ + 2R₁ lead $$B=\left[ \begin{matrix} 0 & 1 & 1 \\-1 & 2 & 4 \\3 & 4 & 7 \\\end{matrix} \right]$$

Example: Using elementary row transformations find the rank of matrix  $$A=\left[ \begin{matrix}3 & -1 & -2 \\2 & 0 & -1 \\3 & -5 & 0 \\\end{matrix} \right]$$

Solution:

$$\sim \left[ \begin{matrix}3 & -1 & -2 \\2 & 0 & -1 \\3 & -5 & 0 \\\end{matrix} \right]$$

R₁ → 5R₁ –  R₃

$$\sim \left[ \begin{matrix}12 & 0 & -10 \\2 & 0 & -1 \\3 & -5 & 0 \\\end{matrix} \right]$$

R₂ ↔ R₃

$$\sim \left[ \begin{matrix}12 & 0 & -10 \\3 & -5 & 0 \\2 & 0 & -1 \\\end{matrix} \right]$$

R₂ →2 R₂ -3 R₃

$$\sim \left[ \begin{matrix}12 & 0 & -10 \\ 0 & -10 & 3 \\ 2 & 0 & -1 \\\end{matrix} \right]$$

R₁ → R₁ -10 R₃

$$\sim \left[ \begin{matrix}-8 & 0 & 0 \\0 & -10 & 3 \\2 & 0 & -1 \\\end{matrix} \right]$$

R₃ → R₁ +4 R₃

$$\sim \left[ \begin{matrix}-8 & 0 & 0 \\0 & -10 & 3 \\0 & 0 & -4 \\\end{matrix} \right]$$

R₂ → 4R₂ +3 R₃

$$\sim \left[ \begin{matrix}-8 & 0 & 0 \\0 & -40 & 0 \\0 & 0 & -4 \\\end{matrix} \right]$$

R₂ → R₂/-8, R₂ → R₂/-40, R₂ → R₂/-4

$$\sim \left[ \begin{matrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \\\end{matrix} \right]$$

Therefore rank of the matric is 3 (since non zero rows are 3).