E k E k 1 E 2 E 1 A = B

Transcription

1 Theorem.5. suggests that reducing a matrix A to (reduced) row echelon form is tha same as multiplying A from left by the appropriate elementary matrices. Hence if B is a matrix obtained from a matrix A by performing a finite sequence of elementary row operations, say k many, then there exists a sequence of elementary matrices E, E,..., E k such that E k E k E E A = B Definition: Matrices A and B are called row equivalent if either (hence each) can be obtained from the other by a sequence of elementary row operations. [ ] In our example following Theorem.5. we have E E A = = B In fact we can apply one more elementary row operation to B and get [ ] [ ] R +R [ ] [ ] R E : +R [ ] E E E A = = I Hence E E E is a candidate for A. In fact we will see that A = E E E. Theorem.5.: Every elementary matrix is invertible, and the inverse is also an elementary matrix. Observation: cr i c R i I E I inverse of E is the one obtained by applying c R i to I R i R j R j R i I E I inverse of E is the one obtained by switching back R i and R j cr i +R j R j cr i I E I inverse of E is the one obtained by substracting -c mupltiple of R i from R j Examples: [ ] [ ] [ ] R. R [ ] = [ ]

2 .. [ ] [ ] [ ] R R R R [ ] [ ] [ ] R +R R +R [ ] = [ ] = [ ] [ ] Theorem.5.: If A is an n n matrix, then the following statements are equivalent, that is, all true or all false.. A is invertible.. Ax = has only the trivial solution.. The reduced row echelon form of A is I n.. A is expressible as a product of elementary matrices. Proof: (a) (b) A invertible A (Ax) = A Ix = x = x x (b) (c) let x =, Ax = has only nontrivial solution means every unknown is a. x n leading variable. This means the reduced row echelon form of the augmented matrix will be.... Hence A... = I... (c) (d) If reduced row echelon form of A is I, it means there exists elementary matrices E, E,..., E k such that E k E k... E A = I. By Theorem.5. each E i is invertible, by mutiplying E k E k... E A = I from left by the corresponding inverses we get A = E E... E k. (d) (a) A is expressible as product of elementary matrices and each elementary matrix is invertible and by Theorem..6 their product is also invertible. Importance: If A is invertible matrix then we have elementary matrices such that E k E k... E A = I. By multiplying from right with A we get E k E k... E AA = IA A = E k E k... E. This idea leads to an algorithm to find inverse of an invertible square matrix.

3 The Inverse Algorithm: To find the inverse of an invertible matrix A, find a sequence of elementary row operations that reduces A to I and then perform the same sequence of operations on I to obtain A. Example: Find inverse of the matrix A =. R R R +R R +R R +R R +R A = R Theorem..: If R is the reduced row echelon form of an n n matrix A, then either R has a row of zeros or R is the identity matrix I. Idea: Last row of R can be all zero or not. If not then it means R do not have a roe of zeros then each row must have a leading and the rest of the entries should be zero which means R = I. Section.6 More on Linear Systems and Invertible Matrices Theorem.6.: If A is an invertible n n matrix, then for each n matrix b, the system of equations Ax = b has exactly one solution, namely, x = A b Proof: We have Ax = b and A invertible. Then A Ax = A b x = A b is the only solution. Example: x + x + x = x + x + x = x + x = R

4 x Coefficient matrix is A =, x = x, b =. Hence we can express x this system as Ax = b. If A is invertible then this system will have a unique solution x = A b. For that consider [A I] and apply elementary row operations toa to get I. Infact one finds that A is row equivalent to I and A =. Hence 9 x = = Solving Multiple linear Systems with a common coefficient matrix Let Ax = b, Ax = b,..., Ax = b k be a sequence of linear systems with the same coefficient matrix A. If A is invertible, then each system will have a unique solution as x = A b, x = A b,... x k = A b k If A is not invertible, then one can use the augmented matrix method, i.e. one can form the augmented matrix [A b b... b k ] row reduced to find solutions Example: Solve the following two linear systems. x + x + x = x + x + x = x + x = x + x + x = x + x + x = x + x = We have A = and A =. So the solution of the first x 9 x system is x = A = the soultion of the second system is x = x x A = If we solve the system by forming the augmented matrix then:

5 R +R R +R R +R R +R R +R Theorem.6.: Let A be a square matrix. If B is a square matrix satisfying BA = I, then B = A.. If B is a square matrix satisfying AB = I, then B = A. Proof: () Consider the system Ax = then B(Ax) = B = Ix = x = so system Ax = has only trivial solution and by Theorem.5. A is invertible so BI = BAA = IA = A. () Considet the system Bx =, then ABx = A = x = so system Bx = has only trivial solution and by Theorem.5. B is invertible and by part (a) A = B and B = (B ) = A by Theorem..7. Importance: To conclude B = A we need to show AB = I and BA = I in general but this theorem says for square matrices one of them is enough to conclude B = A. This theorem gives the following result Theorem.6.5: Let A and B be square matrices of the same size. If AB is invertible then A and B must also be invertible. Theorem.6.: If A is an n n matrix, then the following are equivalent. A is invertible.. Ax = has only the trivial solution.. The reduced row echelon form of A is I. A is expressible as a product of elementary matrices. 5. Ax = b is consistent for every n matrix b. 6. Ax = b has exactly one solution for every n matrix b. Example: Determine the conditions on b, b, b, if any, in order to guarantee that the linear system is consistent. x x + 5x = b x 5x + 8x = b x + x x = b 5

6 5 b From here we have augmented matrix 5 8 b By doing appropriate elementary b 5 b row operations one can get row echelon form b b. b + b + b Hence for this system to be consistent b = b + b. Section.7 Diagonal, Triangular, and Symmetric Matrices Diagonal Matrices: A square matrix in which all the entries of the main diagonal are zero is called a diagonal matrix. Powers and Inverses of Diagonal Matrices: 5 5 Example: Consider A = Then A = ( ) and A = 5 d... d k... d... In general if D =..., then d k Dk... = and if all d n... d k n d... d, d,..., d n are non zero then D is invertible and D d... = d n Triangular Matrices: A square matrix in which all the entries above the main diagonal are zero are called lower triangular, and a square matrix in which all the entries below the main diagonal are zeros is called upper triangular. A matrix that is either upper triangular or lower triangular is called triangular. Example: [ ]. is both diagonal, upper triangular and lower triangular. is upper triangular 6 is lower triangular. 5 6