math125-lecture 2.3

# math125-lecture 2.3 - A system of linear equations is...

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Unformatted text preview: A system of linear equations is consistent if it has one or more solutions A linear system that does not have any solutions is called inconsistent Sec 2.3 Tuesday, February 02, 2010 7:45 PM Section2dot3 Page 1 A system of linear equations is consistent if it has one or more solutions A linear system that does not have any solutions is called inconsistent 1 3 -1 1 1 0 1 3 -3 2 0 0 0 0 2 R' 3 = R 3- R 2 2.3.1 Tuesday, February 02, 2010 7:45 PM Section2dot3 Page 2 A system of linear equations is consistent if it has one or more solutions A linear system that does not have any solutions is called inconsistent R' 3 = R 3- R 2 1 3 -1 1 1 0 1 3 -3 2 0 0 0 0 2 x 1 x 2 x 3 x 4 The last row corresponds to the equation x 1 + 0 x 2 + 0 x 3 + 0 x 4 = 2 i.e. 0 = 2 a contradiction We conclude that there are no values of the x i 's that can satisfy the equations in the system. If there were, then 0 = 2 would be a true statement 2.3.2 Tuesday, February 02, 2010 7:45 PM Section2dot3 Page 3 x + y = 2 x- y = 6 1 1 2 rref 1 0 4 1 -1 6 0 1 -2 2.3.3 Tuesday, February 02, 2010 8:02 PM Section2dot3 Page 4 x + y = 2 x- y = 6 1 1 2 rref 1 0 4 1 -1 6 0 1 -2 2.3.4 Tuesday, February 02, 2010 8:02 PM Section2dot3 Page 5 x + y = 2 x- y = 6 1 1 2 rref 1 0 4 1 -1 6 0 1 -2 x + y = 2 x + y = 1 1 1 2 rref 1 1 2 1 1 1 0 0 -1 2.3.5 Tuesday, February 02, 2010 8:02 PM Section2dot3 Page 6 x + y = 2 x- y = 6 1 1 2 rref 1 0 4 1 -1 6 0 1 -2 x + y = 2 x + y = 1 1 1 2 rref 1 1 2 1 1 1 0 0 -1 2.3.6 Tuesday, February 02, 2010 8:02 PM Section2dot3 Page 7 x + y = 2 x- y = 6 1 1 2 rref 1 0 4 1 -1 6 0 1 -2 x + y = 2 x + y = 1 1 1 2 rref 1 1 2 1 1 1 0 0 -1 x + 3 y = 4 2 x + 6 y = 8 1 3 4 rref 1 3 4 2 6 8 0 0 divide by 2 to get x + 3 y = 4, the same as the first equation 2.3.7 Tuesday, February 02, 2010 8:02 PM Section2dot3 Page 8 x + y = 2 x- y = 6 1 1 2 rref 1 0 4 1 -1 6 0 1 -2 x + y = 2 x + y = 1 1 1 2 rref 1 1 2 1 1 1 0 0 -1 x + 3 y = 4 2 x + 6 y = 8 1 3 4 rref 1 3 4 2 6 8 0 0 divide by 2 to get x + 3 y = 4, the same as the first equation Every point on the line x + 3 y = 4 satisfies the system of equations If we solve this equation for y while treating x as a free/independent variable, we get the solution { ( x , -1/3 x + 4/3) | x ϵ R } OR equivalently { ( x , y ) | y = -(1/3) x + 4/3 and x ϵ R } 2.3.8 Tuesday, February 02, 2010 8:02 PM Section2dot3 Page 9 The row rank of a matrix is the number of non-zero rows the matrix has AFTER it has been put in row echelon form R' 3 = - R 3 ref 1 1 4 0 1 1 0 0 1 So, row rank = 3 2.3.9 Wednesday, February 03, 2010 10:31 AM Section2dot3 Page 10 The row rank of a matrix is the number of non-zero rows the matrix has AFTER it has been put in row echelon form R' 3 = - R 3 ref 1 1 4 0 1 1 0 0 1 So, row rank = 3 Already in ref so, row rank = 2 2.3.10 Wednesday, February 03, 2010 10:31 AM Section2dot3 Page 11 The row rank of a matrix is the number of non-zero rows the matrix has AFTER it has been put in row echelon form R' 3 = - R 3 ref 1 1 4 0 1 1 0 0 1 So, row rank = 3...
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math125-lecture 2.3 - A system of linear equations is...

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