1326.sagews

EXAMPLE 1  Matrix Row Operations

a = [3, 18, -12, 21]
b = [1, 2, -3, 5]
c = [-2, -3, 4, -6]

R = matrix(QQ, [a, b, c])   

# note:  'QQ' designates the rationals.  Needed for Example 1b .
# If not specified, the matrix will only store integers.

R

Perform the following row operations:

1a.  $R_0 \leftrightarrow R_1$

A = copy(R)   # note:  copy of R made for Example 1a.  Same done in parts b and c.

A

A[0]

A[1]

A[0], A[1] = A[1], A[0]
A

1b.  $R_0 \leftarrow \frac{1}{3} R_0$

B = copy(R)

B

B[0] = 1/3*B[0]
B

1c.  $R_2 \leftarrow 2R_1 + R_2$

C = copy(R)
C

C[2] = 2*C[1] + C[2]
C

EXAMPLE 2  Gaussian Elimination with Back-Substitution

$3x + y + 2z = 31$
$x + y + 2z = 19$
$x + 3y + 2z = 25$

R = matrix([[3, 1, 2, 31], [1, 1, 2, 19], [1, 3, 2, 25]])
R

$R_0 \leftrightarrow R_1$

R[0], R[1] = R[1], R[0]
R

$R_1 \leftarrow -3R_0 + R_1$

R[1] = -3*R[0] + R[1]
R

$R_2 \leftarrow -R_0 + R_2$

R[2] = -R[0] + R[2]
R

$R_1 \leftarrow -\frac{1}{2}R_1$

R[1] = -1/2*R[1]
R

$R_2 \leftarrow -2R_1 + R_2$

R[2] = -2*R[1] + R[2]
R

$R_2 \leftarrow -\frac{1}{4}R_2$

R[2] = -1/4*R[2]
R

Back-substitution:

z = 5
y = 13 - 2*z
x = 19 - 2*z - y

x, y, z

Instead of back-substitution, one could also continue with a Gauss-Jordan elimination:

$R_1 \leftarrow -2R_2 + R_1$

R[1] = -2*R[2] + R[1]
R

$R_0 \leftarrow -2R_2 + R_0$

R[0] = -2*R[2] + R[0]
R

$R_0 \leftarrow -R_1 + R_0$

R[0] = -R[1] + R[0]
R

EXAMPLE 3  Gaussian Elimination

$2w + x + 3y - z = 6$
$w - x + 2y - 2z = -1$
$w - x - y + z = -4$
$-w + 2x - 2y - z = -7$

R = matrix(QQ, [[2, 1, 3, -1, 6], [1, -1, 2, -2, -1], [1, -1, -1, 1, -4], [-1, 2, -2, -1, -7]])
R

$R_0 \leftrightarrow R_1$

R.swap_rows(0, 1)
R

$R_1 \leftarrow -2R_0 + R_1$
$R_2 \leftarrow -R_0 + R_2$
$R_3 \leftarrow R_0 + R_3$

R[1] = -2*R[0] + R[1]
R[2] = -R[0] + R[2]
R[3] = R[0] + R[3]

R

$R_1 \leftarrow \frac{1}{3}R_1$

R[1] = 1/3*R[1]
R

$R_3 \leftarrow -R_1 + R_3$

R[3] = -R[1] + R[3]
R

$R_2 \leftarrow -\frac{1}{3} R_2$

R[2] = -1/3*R[2]
R

$R_3 \leftarrow -\frac{1}{3}R_2 + R_3$

R[3] = -1/3*R[2] + R[3]
R

$R_3 \leftarrow -\frac{3}{11} R_3$

R[3] = -3/11*R[3]
R

z = 3
y = 1 + z
x = 8/3 - z + 1/3*y
w = -1 + 2*z - 2*y + x

w, x, y, z

Or, continuing with Gauss-Jordan Elimination:

$R_2 \leftarrow R_3 + R_2$
$R_1 \leftarrow -R_3 + R_1$
$R_0 \leftarrow 2*R_3 + R_0$

R[2] = R[3] + R[2]
R[1] = -R[3] + R[1]
R[0] = 2*R[3] + R[0]

R

$R_1 \leftarrow \frac{1}{3}R_2 + R_1$
$R_0 \leftarrow -2R_2 + R_0$

R[1] = 1/3*R[2] + R[1]
R[0] = -2*R[2] + R[0]
R

$R_0 \leftarrow R_1 + R_0$

R[0] = R[1] + R[0]
R

And here's a secret!  Shhh ... don't tell anybody:

M = matrix(QQ, [[2, 1, 3, -1, 6], [1, -1, 2, -2, -1], [1, -1, -1, 1, -4], [-1, 2, -2, -1, -7]])
M

M.echelon_form()

EXAMPLE 4  Gauss-Jordan Elimination

Use Gauss-Jordan on the system in Example 2.

Here is a full Gauss-Jordan elimination again using a different set of operations:

R = matrix(QQ, [[3, 1, 2, 31], [1, 1, 2, 19], [1, 3, 2, 25]])
R

$R_0 \leftarrow \frac{1}{3}R_0$

R[0] = 1/3*R[0]
R

$R_1 \leftarrow -R_0 + R_1$
$R_2 \leftarrow -R_0 + R_2$

R[1] = -R[0] + R[1]
R[2] = -R[0] + R[2]
R

$R_1 = \frac{3}{2}R_1$

R[1] = 3/2*R[1]
R

$R_0 \leftarrow -\frac{1}{3}R_1 + R_0$
$R_2 \leftarrow -\frac{8}{3}R_1 + R_2$

R[0] = -1/3*R[1] + R[0]
R[2] = -8/3*R[1] + R[2]
R

$R_2 \leftarrow -\frac{1}{4}R_2$

R[2] = -1/4*R[2]
R

$R_1 \leftarrow -2R_2 + R_1$

R[1] = -2*R[2] + R[1]
R

R = matrix(QQ, [[3, 1, 2, 31], [1, 1, 2, 19], [1, 3, 2, 25]])
R

R.echelon_form()