1. For each of the following matrices (to which, alas, like those later, you must add your own matrix brackets), determine

(i) whether it is already in fully reduced row echelon (frre) form;

(ii) (if it is not already in frre form) what its frre form is;

(iii) what its rank is;

(iv) whether it has a determinant, and, if so, what that is; and

(v) whether it must be invertible, and, if so, what its inverse is:0 1 0 1 0 0 5 0 0 1 3A= 0 0 1 ,B= 0 0 1 3 ,C= 0 0 ,D= . 2 7 1 0 0 0 1 0 4 0 0

Proposed Solution:

As regardsA:

(i)Ais NOT in frre form; however, (ii) its frre form is the 3 × 3 identity matrix, as one sees by first interchanging rows 2 and 3 and then row 1 and (the new) row 2. (iii) Because frref(A) has none of its three rows all-zero,Ahas rank 3. (iv) Chasing through the multipliers (each a -1 )associated with the two steps in the row-reduction, one sees det(A) = (-1)(-1) = 1 . And (v), yes,Ais invertible, with inverse obtained, say, by applying the row-reduction steps themselves to the 3 × 3 identity matrix, which yieldsAs regards0 0 1 1 0 0 as the inverse. 0 1 0B:

(i)Bis NOT in frre form; (ii) but its frre form can be obtained by just exchanging rows 2 and 3. (iii) Its rank is again 3. (iv) Not being square,Bhas NO determinant. (v) Having rank only 3 while being 3 × 4,Bcannot be invertible.

As regardsC:

(i & ii))CIS in frre form already, and (iii) has rank zero. (iv) Not being square,Chas NO determinant (and no, saying det(C) = 0 does NOT mean the same thing, and is positively FALSE!); and since every product ofCwith any matrix you can multiply it by becomes a matrix full of zeros,Chas no hope whatever of being invertible.

As regardsD:

(i)Dis both far and not far from being in frre form: there are no zeros anywhere inD, but (ii) in just two steps (from row 2 subtract twice row 1, then from row 1 subtract thrice (the new) row 2) it reduces immediately toI_{2×2}, which is its frre form. (iii) Its rank is therefore 2; (iv) its determinant is therefore 1; and (v) it is invertible, with inverse the end result of carrying out the same row operation steps toI_{2×2}, as shown below:———1 0 -----\ 1 0 -----\ 7 -3 0 1 -----/ -2 1 -----/ -2 1 .

2. For which values, if any, of the real numberawill the following system of three equations (in the unknownsx,y, andz) have

(i) no solutions?

(ii) exactly one solution?

(iii) more solutions than just one, butnotinfinitely many?

(iv) infinitely many solutions?x+ 2y– 3z= 4 ,

3x–y+ 5z= 2 ,

4x+y+ (a^{2}– 14)z=a+ 2 .

Proposed Solution:

We begin by recording the augmented matrix associated with this system, and showing the partially row-reduced result after two steps in the process of row-reducing it (we only subtract from row 3 first row 1 and then row 2):We have not fully row-reduced our augmented matrix, but we have gone far enough to see that there's an embarrassing moment that comes when the bottom row in the frre form starts with three zeros, yet the fourth item is ≠ 0 , i.e., when1 2 -3 4 1 2 -3 4 3 -1 5 2 =====> 3 -1 5 2 . 4 1 a^{2}-14 a+2 0 0 a^{2}-16 a-4a^{2}= 16 , yeta≠ 4 , i.e., whena= –4 : it is in this case that the system has NO solution(s), which answers part (i).

When the bottom row in the frre form consists offourzeros, i.e., whena= 4 , the original system, while consistent, has rank only 2 and therefore has infinitely many solutions, which answers part (iv).

For all values ofaother than ± 4 , the rank in question is 3, same as the number of unknowns, and so for those values ofathere will be exactly one solution, which answers part (ii).

As for part (iii), the answer is: never (for no value ofacan there ever be more solutions than one, yet not infinitely many).

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3. Can the 2 × 2 identity matrixI_{2×2}be expressed as the differencebetween the two possible productsI_{2×2}=AB–BAABandBAof two suitable 2 × 2 matricesAandB?

If so, give an explicit example; if not, give a coherent explanation of why not.

[Suggestion: Can the traces tr(I_{2×2}) and tr(AB–BA) be of any use here?]

Proposed Solution:

Let’s follow the suggestion, and see whether those traces are of any use.

Anyway, tr(I_{2×2}) is easy enough to figure out:As for the other, recall having seen that tr(tr(I_{2×2}) = 1 + 1 = 2 .AB) = tr(BA), and that tr(P±Q) = tr(P) ± tr(Q) . But thenConsequently, were there, hypothetically, two 2 × 2 matricestr(AB–BA) = tr(AB) – tr(BA) = tr(AB) – tr(AB) = 0 .AandBwithI_{2×2}=AB–BA, we’d be forced to conclude thata contradiction showing just how untenable that hypothesis must be.2 = tr(I_{2×2}) = tr(AB–BA) = 0 ,

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4. Given ann×nmatrixA, with known determinant det(A) , calculate the determinant of the matrix productA•adj(A) (ofAwith the adjoint matrix adj(A)).

Proposed Solution:

In any event, as regards the matrix productA•adj(A) , we knowBut the RHS above is what you get from theA•adj(A) = det(A)I_{n×n}.n×nidentity matrixI_{n×n}by multiplyingeachof itsnrows by the factor det(A) .

Thus the determinant of that RHS is the product of thosenfactors det(A) along with the one additional factor det(I_{n×n}) = 1 .

Summing up, then, we have:———det(A•adj(A)) = det(det(A)I_{n×n}) = (det(A))^{n}(det(I_{n×n})) = (det(A))^{n}(1) = (det(A))^{n}.

5. Extend your Problem 3 findings to matrices of more general size(s) than just 2 × 2 .

Proposed Solution:

Well, for anyn> 0 , we have tr(I_{n×n}) =n> 0 .

And, just as before, we have tr(AB) = tr(BA) and tr(P±Q) = tr(P) ± tr(Q) (forn×nmatricesA,B,P, andQ)._{ }

So, just as before, the assumption that we could expressI_{n×n}as a differenceAB–BAwould, upon calculating traces, lead to the contradiction 0 <n= 0 .

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Posted 4 Mar 2006