Let A be a 3by 3 matrix.Which of the following statements guarantee that AX=B has a unique solution.which guarantee that AX=B does not have a unique solution? (a)rank(A)=2 (b)rank(A inverse)=3 (c)dim{row space(A)}=dim{column space(A)} (d)two rows of A are equal. (e)the columns of A form a basis for R^3.

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TopNewest@muhmd sani You just need to realize a very basic fact that if

Ais a square matrix, that is3x3in your case, thenAX=Bwould have a strictlyuniquesolution ifAisinvertible. Every other result follows from this fact.Now if a

3x3Ais invertible, surely its rank must have to be3. Rank3implies that all three columns arepivotcolumns and consequently all the columns areindependentthus providing abasisforR^3.Also invertibility of

Aestablishes the invertibility ofA_inversethat in turn makes it necessary for the rank ofA_inverseto be3also.So, uniqueness of the solution of

AX=Bis guaranteed in options(b)and(e).For

(a), its clear that you'd get one free variable that would open the door for infinitely many solutions and thus uniqueness would be dead.For

(c), keep in mind that dimension of row space and that of column space of a matrix are always equal and equal to the rank of of that matrix, no matter what. We need them to be strictly equal to three in this case, so in essence option(c)won't suffice. We would have included it in our good books if option(c)had read as:dim{row space(A)}=dim{column space(A)}=3.In the case of

(d), independence of columns would be dead and thus rank would be less than3and further story won't allow a unique solution.To me, it is the elegance of linear systems that their reduced row echelon form describes each and every incredible detail about how would they behave. Two basic keys to understand the linear systems: Reduced Row Echelon Forms and Four fundamental subspaces. Hope this would help. :) – Saud Ahmad · 2 years ago

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