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Linear algebra

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.

Note by Muhmd Sani
2 years, 4 months ago

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@muhmd sani You just need to realize a very basic fact that if A is a square matrix, that is 3x3 in your case, then AX=B would have a strictly unique solution if A is invertible. Every other result follows from this fact.

Now if a 3x3 A is invertible, surely its rank must have to be 3. Rank 3 implies that all three columns are pivot columns and consequently all the columns are independent thus providing a basis for R^3.

Also invertibility of A establishes the invertibility of A_inverse that in turn makes it necessary for the rank of A_inverse to be 3 also.

So, uniqueness of the solution of AX=B is 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 than 3 and 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 · 1 year, 10 months ago

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