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MAT208

MAT208. FALL 2009 Kolman Sec. 2.1-2.3. Row Echelon Form. All zero rows appear at the bottom of the matrix The first nonzero entry from the left of any nonzero row is a ONE. (Leading one) For each nonzero row, the leading ONE is to the right and below any leading ones in preceding rows.

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MAT208

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  1. MAT208 FALL 2009 Kolman Sec. 2.1-2.3

  2. Row Echelon Form • All zero rows appear at the bottom of the matrix • The first nonzero entry from the left of any nonzero row is a ONE. (Leading one) • For each nonzero row, the leading ONE is to the right and below any leading ones in preceding rows.

  3. Row Reduced Echelon Form(RREF) • All properties as in row echelon form and one more property: • If a column contains a leading ONE, then all other entries in that column are ZERO.

  4. Row (Column) Equivalent Matrices • A is equivalent to B if B can be obtained through a finite sequence of elementary row (column) operations on A. • Every matrix is equivalent to itself. • Two m x n matrices A and B are equivalent iff B=PAQ for some nonsingular matrices P and Q.

  5. Equivalent Matrices

  6. 1. REF--- Gaussian Elimination 2. RREF-- Gauss Jordan Elimination 3. Ax=b has a solution Iffb can be written as a linear combination of the columns of A. Solving Systems

  7. Two-Part Solutions If Ax=b is a consistent nonhomogeneous linear system, every solution may be written as:

  8. Square matrices (currently) Not all matrices have an inverse. Finding an inverse: A matrix A is invertible if there exists a matrix, A-1, such that: A A-1= A-1A=In Finding Inverses

  9. Equivalent Statements • A is invertible; • Ax = 0 has only the trivial solution; • The reduced row echelon form of A is In; • A can be expressed as a product of elementary matrices; • Ax = b is consistent for every n x 1 matrix b; • The solution to Ax = b is ; and • Ax = b has exactly one solution for every n x 1 matrix b. More to come.

  10. To prove: • A-1 is unique. • If A is invertible, then the unique solution to Ax=b is x= A-1b. • If A and B are invertible , then so is AB and (AB)-1=B-1A-1.

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