213 - A11 - Row Reduction Function -- Sage

def RREF(A): reduced_echelon_form=True M = A.nrows() N = A.ncols() # if A.det()==0: # print html("<font size=+1 color=red>Matrix is Singular. Please pick another!</font>") print html("<font size=+2 color=blue>Using the matrix</font>") show(A) global Astart Astart = copy(A) # Now, let's start the step by step Gauss-Jordan routine ROWS = range(M) COLS = range(N) PIVOTS = range(min(M,N)) for i in PIVOTS: # Gauss-Jordan on column i # First, make the pivot element equal to 1 k = i # Need to fix this when a column is skipped when finding nonzero pivots # That is, if the remainder of a column is zero, go on to the next column till you reach the end. if A[i][i]==0: # swap for non-zero pivot # start looking in rows below the i-th while (k<M and A[k][i]==0): k += 1 if k==M: print 'All of the remaining rows are zero. Move to next columns or finish.' else: A.swap_rows(i,k) # So, now we should have a non-zero in the pivot position or k=M which means we are done. # Should consider using the function "nonzero_positions_in_column(i)" and use the min position > i. # Start zeroing out the rest of column i. if A[i][i]<>0: A[i]=A[i]/A[i][i] if reduced_echelon_form: for k in ROWS: if i<>k: A[k] = A[k]-A[i]*A[k][i] else: for k in ROWS: if i<k: A[k] = A[k]-A[i]*A[k][i] print '\n After',i+1,'step(s) the matrix is equivalent to:' show(A)

B = matrix(QQ,[[1,2,0],[-6,-6,2],[3,8,-1]]) RREF(B)

Using the matrix

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 1 & 2 & 0 \\ -6 & -6 & 2 \\ 3 & 8 & -1 \end{array}\right)$


 After 1 step(s) the matrix is equivalent to:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 1 & 2 & 0 \\ 0 & 6 & 2 \\ 0 & 2 & -1 \end{array}\right)$


 After 2 step(s) the matrix is equivalent to:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 1 & 0 & -\frac{2}{3} \\ 0 & 1 & \frac{1}{3} \\ 0 & 0 & -\frac{5}{3} \end{array}\right)$


 After 3 step(s) the matrix is equivalent to:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)$

Using the matrix


 After 1 step(s) the matrix is equivalent to:


 After 2 step(s) the matrix is equivalent to:


 After 3 step(s) the matrix is equivalent to:

B = matrix(QQ,[[1, -2, 0, 4, -1],[0, -1, -5, 5, 0],[0, 0, -3, 12, 3],[2, -1, 3, 20, 4]]) RREF(B)

Using the matrix

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr} 1 & -2 & 0 & 4 & -1 \\ 0 & -1 & -5 & 5 & 0 \\ 0 & 0 & -3 & 12 & 3 \\ 2 & -1 & 3 & 20 & 4 \end{array}\right)$


 After 1 step(s) the matrix is equivalent to:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr} 1 & -2 & 0 & 4 & -1 \\ 0 & -1 & -5 & 5 & 0 \\ 0 & 0 & -3 & 12 & 3 \\ 0 & 3 & 3 & 12 & 6 \end{array}\right)$


 After 2 step(s) the matrix is equivalent to:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr} 1 & 0 & 10 & -6 & -1 \\ 0 & 1 & 5 & -5 & 0 \\ 0 & 0 & -3 & 12 & 3 \\ 0 & 0 & -12 & 27 & 6 \end{array}\right)$


 After 3 step(s) the matrix is equivalent to:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr} 1 & 0 & 0 & 34 & 9 \\ 0 & 1 & 0 & 15 & 5 \\ 0 & 0 & 1 & -4 & -1 \\ 0 & 0 & 0 & -21 & -6 \end{array}\right)$


 After 4 step(s) the matrix is equivalent to:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrr} 1 & 0 & 0 & 0 & -\frac{5}{7} \\ 0 & 1 & 0 & 0 & \frac{5}{7} \\ 0 & 0 & 1 & 0 & \frac{1}{7} \\ 0 & 0 & 0 & 1 & \frac{2}{7} \end{array}\right)$

Using the matrix


 After 1 step(s) the matrix is equivalent to:


 After 2 step(s) the matrix is equivalent to:


 After 3 step(s) the matrix is equivalent to:


 After 4 step(s) the matrix is equivalent to:

C = matrix(QQ,[[1, 8, 9],[-3, -6, 9],[4, 3, -22]]) RREF(C)

Using the matrix

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 1 & 8 & 9 \\ -3 & -6 & 9 \\ 4 & 3 & -22 \end{array}\right)$


 After 1 step(s) the matrix is equivalent to:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 1 & 8 & 9 \\ 0 & 18 & 36 \\ 0 & -29 & -58 \end{array}\right)$


 After 2 step(s) the matrix is equivalent to:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 1 & 0 & -7 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right)$

All of the remaining rows are zero. Move to next columns or finish.

 After 3 step(s) the matrix is equivalent to:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 1 & 0 & -7 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right)$

Using the matrix


 After 1 step(s) the matrix is equivalent to:


 After 2 step(s) the matrix is equivalent to:

All of the remaining rows are zero. Move to next columns or finish.

 After 3 step(s) the matrix is equivalent to:

213 - A11 - Row Reduction Function

29 days ago by Professor213