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Elementry Row Operations

Operations

TYPE ACTION EXAMPLE AND NOTATION Description
1 Interchange two rows \(R_1 \leftrightarrow R_2\) Interchanging \(R_1\) and \(R_2\)
2 Scalar multiplication of a row by constant \(t\) \(kR_1 \rightarrow R_1\) Multiplying Row 1 with constant \(k\)
3 Add a multiple of one row to another row \(R_1 (+/-) kR_2\) Adding or subtracting \(k \times R_2\) from \(R_1\)

What are these operations used for?

Row Reduction: Row Echelon Form

  • Row reduction is a sequence of elementry row operations that transforms a matrix into row echelon form.
  • \(A = \begin{bmatrix} 3 & 2 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 0 & 7 & 1 & 1 \\ \end{bmatrix}\)
  • To get the \(rref\) of the matrix \(A\) we need to reduce the first element to \(1\). This can be done by multiplying the first row by \(1/3\).
  • So the first operation is \(R_1/3\).
  • \(A = \begin{bmatrix} 1 & 2/3 & 1/3 & 1/3 \\ 1 & 1 & 0 & 0 \\ 0 & 7 & 1 & 1 \\ \end{bmatrix}\)
  • Now we need to make the second element of the second row \(0\). This can be done by subtracting \(1\) times the first row from the second row.
  • So the second operation is \(R_2 - R_1\).
  • \(A = \begin{bmatrix} 1 & 2/3 & 1/3 & 1/3 \\ 0 & 1/3 & -1/3 & -1/3 \\ 0 & 7 & 1 & 1 \\ \end{bmatrix}\)
  • Now we want to normalize the second row. This can be done by multiplying the second row by \(3\).
  • So the third operation is \(3R_2\).
  • \(A = \begin{bmatrix} 1 & 2/3 & 1/3 & 1/3 \\ 0 & 1 & -1 & -1 \\ 0 & 7 & 1 & 1 \\ \end{bmatrix}\)
  • Now we want to make the third element of the third row \(0\). This can be done by subtracting \(7\) times the second row from the third row.
  • So the fourth operation is \(R_3 - 7R_2\).
  • \(A = \begin{bmatrix} 1 & 2/3 & 1/3 & 1/3 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 8 & 8 \\ \end{bmatrix}\)
  • Now we want to normalize the third row. This can be done by multiplying the third row by \(1/8\).
  • So the fifth operation is \(R_3/8\).
  • \(A = \begin{bmatrix} 1 & 2/3 & 1/3 & 1/3 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 1 & 1 \\ \end{bmatrix}\)
  • Now we want to set all the values above the leading one to \(0\). This can be done by subtracting \(1\) times the third row from the first row.
  • So the sixth operation is \(R_2 - R_3 \text{ and } R_1 - \frac{1}{3}R_3\).
  • \(A = \begin{bmatrix} 1 & 2/3 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ \end{bmatrix}\)
  • Final Step is remove the leading zeros. This can be done by subtracting \(2/3\) times the second row from the first row.
  • So the seventh operation is \(R_1 - \frac{2}{3}R_2\).
  • \(A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ \end{bmatrix}\)
  • So we used the Row Operations to get the \(rref\) of the matrix \(A\).

Note

After all the row and column operations the determinant of the matrix always remains the same.