3.5-16 Fall 2007 ======= Form the matrix C=aug(A,I) as frame one of a frame sequence of combo, swap and mult operations ending in last frame equal to rref(C). If the left half of rref(C) is the identity I, then the right half B is the inverse of A. A maple answer check: with(linalg): A:=matrix([[1,-3,-3],[-2,1,2],[2,-3,-3]]); inverse(A); The problem can be solved entirely in maple by using addrow, swaprow, mulrow to print the frames of the sequence. Then a worksheet print may be submitted. Please include the answer check above, somewhere in your print. 3.5-26 Fall 2007 ======= Form the matrix C=aug(A,B) as frame one of a frame sequence of combo, swap and mult operations ending in last frame equal to rref(C). If the left half of rref(C) is the identity I, then the right half D is the inverse of A multiplied against B, that is, D = inverse(A) B. A maple answer check: with(linalg): A:=matrix([[1,5,1],[2,1,-2],[1,7,2]]); B:=matrix([[2,0,1],[0,3,0],[1,0,2]]); evalm(inverse(A) &* B); The problem can be solved entirely in maple by using addrow, swaprow, mulrow to print the frames of the sequence. Then a worksheet print may be submitted. Please include the answer check above, somewhere in your print. 3.5-34 Fall 2005 ======= The result for dim(A)=3 is A=diag(a,b,c) and A invertible if and only if abc is not zero. The result for dim(A)= is A=diag(a1,a2,...,an) and A invertible if and only if a1*a2*...*an is not zero. You must present a proof for dim(A)=n. A dim(A)=3 proof earns 70% or less. Resource: Theorem page 190, which says A is invertible if and only if rref(A)=I. For dim(A)=3, the argument goes like this: I. If abc=0, then A has a row of zeros, hence rref(A) has a row of zeros and then rref(A) is not the identity I. By page 190, A fails to be invertible. II. If abc is not zero then three multiply rules can be applied to reduce matrix A to I, hence rref(A)=I. Page 190 implies A is invertible. III. In case II, the inverse of A can be found by using the same three mult rules on the augmented matrix C=aug(A,I) to give rref(C)=aug(I,B) where B=diag(1/a,1/b,1/c). 3.5-34 Spring 2006 ====== Other proofs are possible using just the definition of invertible matrix. The problem can be distilled to proving these two statements: (a) Given A=diag(a1,a2,...,an) and B=diag(1/a1,1/a2,...,1/an) and a1,...,an are not zero, then AB=I and therefore B is the inverse of A. (b) Given A=diag(a1,a2,...,an) and B is inverse to A, that is , AB=BA=I, then a1,...,an are not zero. Statement (a) is proved by direct verification. This means that A and B are written with "..." notation next to each other and multiplcation is carried out to get the diagonal matrix I. Statement (b) is proved by writing out the equation AB=I using a general matrix B with entries b_{ij}. The equations a_{11}b_{11}=1,...,a_{nn}b_nn}=1 are obtained. These equations imply that a_{11} is not zero, ..., a_{nn} is not zero. This problem may be submitted in place of 44. 3.5-44 Spring 2006 ====== The proof requires the use of (15), page 190, plus the fact that an inverse of an elementary matrix is also an elementary matrix. Expected details repeat parts of the proof of Theorem 6 [page 190], obtaining (15) for the present setting. Update Fall 2010: Extra credit problem Xc3.5-44a replaces 3.5-44. The logic here is that problem 44 is a proof of three pages, whereas the Xc problem is a (a) a little bit of theory and (b) a computation.