Row Reduction of Linear Systems

Question 1

elementary row operations

Answer 1

definition: there are 3 elementary row operations (ERO) that can be performed on matrices
1) multiply a row by a non-zero scalar kR_i
2) interchange 2 rows R_i (interchange i^th row and j^th row)
3) replace a row by the sum of itself and a multiple of another row

R_i + kR_j -> replace i^th row by sum of i^th row times j^th row

Question 2

row equvialent

Answer 2

• > if a matrix is obtained from another matrix by a finite number or ERO, we call these 2 matrices row equivalent
• > when performing elementary row operations (ERO) it must be done on A and B [A | B]

Question 3

row-reduced echelon form

Answer 3

definition: a matrix is in row-reduced echelon form (RREF) if
1) any zero rows (cosisting entirely zeros) is at the bottom
2) in each non-zero row, the first non-zero entry is 1 (leading 1)
3) in a column that contains a leading 1, all other entries are zero
4) in any two rows with some non-zero entries, the leading 1 of the higher row is father to the left (step-pattern)
- > augmented form [A | B] RREF only applies to (A) not (B)

Question 4

reducing to RREF (row-reduced echelon form)

Answer 4

1) choose it - choose an entry to be the next leading 1
2) move it - move its row up below all the rows which already have leading 1 (R₁ R₂)
3) turn it one - multiply its row by its inverse to make it 1 (cR₁)
4) reduce the column using the one (R₁ + kR₂)

Question 5

Gauss-Jordan Elimination Method

Answer 5

step 1) find the augmented matrix of the System of Linear Equations

step 2) use the Elementary Row Operations to transform the augmented matrix to Row Reduced Echelon Form

step 3) determine which one of the following cases happen and use it to find the solutions

case 1) no solution, if RREF has a row with zeros on the coefficient part and non-zero entry after the separating line [0 0 0 | (non-zero)]

case 2) unique solution: if every column of the coefficient part of the RREF has a leading 1 of some rows

to find the solution: turn back the augmented matrix to the equation form, that will give you the components of the unique solution

case 3) infinitely many solutions: some columns has no leading 1

assign parameters to the columns without a leading 1

of parameters = # of columns without leading 1

to each variable corresponding one of the columns without the leading 1, assign the corresponding parameter

write back the equation form and rewrite all the components in terms of the parameters

Question 6

augmented matrix with a variable entries example

Answer 6

if k² + k does not = 0 -> three leading 1 -> the SLE has a unique solution

k² + k = 0 if -> k = 0 and k = -1

for all values but k = 0,-1 the SLE has a unique solution

ex)

k = 0 -> [1 0 2 | 2]

0 1 0 | 1

0 0 0 | 0 -> zero so, there are infinitely many solutions

k = -1 -> [1 0 2 | 2]

0 1 0 | 1

0 0 0 | -1 -> non-zero so, there is no solution