Find Reduced Row Echelon Form [[2n],[n-1]]

Solution:

Step 1:

Scale $R_1$ by multiplying with $\frac{1}{2n}$ to transform the first element into a $1$.

Step 1.1:

Apply the scalar multiplication to $R_1$: ${[\frac{2n}{2n},\frac{(n-1)}{2n}]}^T$.

Step 1.2:

After simplification, $R_1$ becomes: ${[1,\frac{n-1}{2n}]}^T$.

Step 2:

To zero out the first element of $R_2$, execute $R_2=R_2-(n-1)\cdot R_1$.

Step 2.1:

Carry out the row operation on $R_2$: ${[1,n-1-(n-1)\cdot 1]}^T$.

Step 2.2:

Post-simplification, $R_2$ is: ${[1,0]}^T$.

Knowledge Notes:

The process described is for finding the Reduced Row Echelon Form (RREF) of a matrix. The RREF is a form of a matrix where each leading entry (the first non-zero number from the left in a row) is $1$, and is the only non-zero entry in its column. All rows consisting entirely of zeros are at the bottom of the matrix. The steps to achieve RREF include:

1. Row Scaling: Multiplying a row by a non-zero scalar. This is used to make the leading entry of a row $1$.

2. Row Addition/Subtraction: Adding or subtracting one row from another. This is used to create zeros in a column below or above a leading $1$.

3. Row Swapping: Interchanging two rows. This is not used in this particular problem but is often necessary in other cases to achieve RREF.

In LaTeX, matrices are often written using the $pmatrix$or $bmatrix$environments, and row operations are typically denoted by arrow symbols or by explicitly stating the operation, such as $R_2=R_2-(n-1)R_1$.

In this problem, the matrix is a $2\times 1$ matrix, which simplifies the process since there is no need for row swapping or creating zeros above leading $1$s. The goal is to scale the first row to make the first element $1$ and then to subtract an appropriate multiple of the first row from the second row to make the second element $0$.