213 - A21 - Finding exact inverses -- Sage

%hide %auto def RREF(A): reduced_echelon_form=True M = A.nrows() N = A.ncols() # if A.det()==0: # pretty_print(html("<font size=+1 color=red>Matrix is Singular. Please pick another!</font>")) pretty_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)

num = 4 A = random_matrix(QQ, num, num) show(A) Ainv = A.inverse() show(Ainv) show(Ainv*A) show(A*Ainv)

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

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} -\frac{57}{2} & -1 & 5 & \frac{29}{2} \\ \frac{41}{2} & 1 & -3 & -\frac{21}{2} \\ -48 & -2 & 8 & 24 \\ 4 & 0 & 0 & -2 \end{array}\right)$

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

show(A.echelon_form())

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

# Here you can enter your own matrix or use the random one from above. # Comment this out if using the one from above A = matrix(QQ,[[1,0,-1,0],[0,-1,5,5],[6,-4,4,0],[0,0,1,1]]) show(A)

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

Aug = A.augment(identity_matrix(A.nrows())) show(Aug)

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

RREF(Aug)

Using the matrix

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


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

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrrrrr} 1 & 0 & -1 & 0 & 1 & 0 & 0 & 0 \\ 0 & -1 & 5 & 5 & 0 & 1 & 0 & 0 \\ 0 & -4 & 10 & 0 & -6 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 & 0 & 1 \end{array}\right)$


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

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrrrrr} 1 & 0 & -1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & -5 & -5 & 0 & -1 & 0 & 0 \\ 0 & 0 & -10 & -20 & -6 & -4 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 & 0 & 1 \end{array}\right)$


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

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrrrrr} 1 & 0 & 0 & 2 & \frac{8}{5} & \frac{2}{5} & -\frac{1}{10} & 0 \\ 0 & 1 & 0 & 5 & 3 & 1 & -\frac{1}{2} & 0 \\ 0 & 0 & 1 & 2 & \frac{3}{5} & \frac{2}{5} & -\frac{1}{10} & 0 \\ 0 & 0 & 0 & -1 & -\frac{3}{5} & -\frac{2}{5} & \frac{1}{10} & 1 \end{array}\right)$


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

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrrrrr} 1 & 0 & 0 & 0 & \frac{2}{5} & -\frac{2}{5} & \frac{1}{10} & 2 \\ 0 & 1 & 0 & 0 & 0 & -1 & 0 & 5 \\ 0 & 0 & 1 & 0 & -\frac{3}{5} & -\frac{2}{5} & \frac{1}{10} & 2 \\ 0 & 0 & 0 & 1 & \frac{3}{5} & \frac{2}{5} & -\frac{1}{10} & -1 \end{array}\right)$

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:


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

show(A.inverse()*A)

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

A = matrix(QQ,[[1, 0, -1, 0 ],[ 0, -1, 5 ,5],[ 6, -4, 4 ,0],[ 0, 0, 1, 1]]) Aug = A.augment(identity_matrix(A.nrows())) show(A)

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

RREF(Aug)

Using the matrix

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


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

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrrrrr} 1 & 0 & -1 & 0 & 1 & 0 & 0 & 0 \\ 0 & -1 & 5 & 5 & 0 & 1 & 0 & 0 \\ 0 & -4 & 10 & 0 & -6 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 & 0 & 1 \end{array}\right)$


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

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrrrrr} 1 & 0 & -1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & -5 & -5 & 0 & -1 & 0 & 0 \\ 0 & 0 & -10 & -20 & -6 & -4 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 & 0 & 1 \end{array}\right)$


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

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrrrrr} 1 & 0 & 0 & 2 & \frac{8}{5} & \frac{2}{5} & -\frac{1}{10} & 0 \\ 0 & 1 & 0 & 5 & 3 & 1 & -\frac{1}{2} & 0 \\ 0 & 0 & 1 & 2 & \frac{3}{5} & \frac{2}{5} & -\frac{1}{10} & 0 \\ 0 & 0 & 0 & -1 & -\frac{3}{5} & -\frac{2}{5} & \frac{1}{10} & 1 \end{array}\right)$


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

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrrrrr} 1 & 0 & 0 & 0 & \frac{2}{5} & -\frac{2}{5} & \frac{1}{10} & 2 \\ 0 & 1 & 0 & 0 & 0 & -1 & 0 & 5 \\ 0 & 0 & 1 & 0 & -\frac{3}{5} & -\frac{2}{5} & \frac{1}{10} & 2 \\ 0 & 0 & 0 & 1 & \frac{3}{5} & \frac{2}{5} & -\frac{1}{10} & -1 \end{array}\right)$

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:


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

B = block_matrix(QQ,[[A,identity_matrix(4)]]) RREF(B)

Using the matrix

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|rrrr} 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & -2 & 8 & 0 & 0 & 1 & 0 & 0 \\ 5 & 3 & 11 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 2 & -2 & 0 & 0 & 0 & 1 \end{array}\right)$


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

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|rrrr} 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & -2 & 8 & 0 & 0 & 1 & 0 & 0 \\ 0 & -2 & 6 & 1 & -5 & 0 & 1 & 0 \\ 0 & 0 & 2 & -2 & 0 & 0 & 0 & 1 \end{array}\right)$


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

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|rrrr} 1 & 0 & 5 & 0 & 1 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & -4 & 0 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 0 & -2 & 1 & -5 & -1 & 1 & 0 \\ 0 & 0 & 2 & -2 & 0 & 0 & 0 & 1 \end{array}\right)$


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

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|rrrr} 1 & 0 & 0 & \frac{5}{2} & -\frac{23}{2} & -2 & \frac{5}{2} & 0 \\ 0 & 1 & 0 & -2 & 10 & \frac{3}{2} & -2 & 0 \\ 0 & 0 & 1 & -\frac{1}{2} & \frac{5}{2} & \frac{1}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & -1 & -5 & -1 & 1 & 1 \end{array}\right)$


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

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr|rrrr} 1 & 0 & 0 & 0 & -24 & -\frac{9}{2} & 5 & \frac{5}{2} \\ 0 & 1 & 0 & 0 & 20 & \frac{7}{2} & -4 & -2 \\ 0 & 0 & 1 & 0 & 5 & 1 & -1 & -\frac{1}{2} \\ 0 & 0 & 0 & 1 & 5 & 1 & -1 & -1 \end{array}\right)$

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:


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

A = matrix(QQ,[[1, 1, 1, 0 ],[ 0, -2, 8, 0],[ 5, 3, 11 ,1],[ 0, 0, 2, -2]])

213 - A21 - Finding exact inverses

66 days ago by Professor213