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 3
A = 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 regards A:
(i) A is 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, A
has 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, A
is invertible, with inverse obtained, say, by applying the row-reduction steps
themselves to the 3 × 3 identity matrix, which yields
0 0 1
1 0 0 as the inverse.
0 1 0
As regards B:
(i) B is 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,
B has NO determinant. (v) Having rank only 3 while being 3 × 4,
B cannot be invertible.
As regards C:
(i & ii)) C IS in frre form already, and (iii) has rank zero.
(iv) Not being square, C has NO determinant (and no, saying det(C) = 0 does NOT mean the same thing, and is positively FALSE!); and since every
product of C with any matrix you can multiply it by becomes a matrix
full of zeros, C has no hope whatever of being invertible.
As regards D:
(i) D is both far and not far from being in frre form: there are no
zeros anywhere in D, 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 to I2×2 , 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 to I2×2 , as shown below:
1 0 -----\ 1 0 -----\ 7 -3
0 1 -----/ -2 1 -----/ -2 1 .
———
2. For which values, if any, of the real number a will the following
system of three equations (in the unknowns x , y , and z ) have
(i) no solutions?
(ii) exactly one solution?
(iii) more solutions than just one, but not infinitely many?
(iv) infinitely many solutions?
x + 2y – 3z = 4 ,
3x – y + 5z = 2 ,
4x + y + (a2 – 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):
1 2 -3 4 1 2 -3 4
3 -1 5 2 =====> 3 -1 5 2 .
4 1 a2-14 a+2 0 0 a2-16 a-4
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., when
a2 = 16 , yet a ≠ 4 , i.e.,
when a = –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 of four zeros, i.e.,
when a = 4 , the original system, while consistent, has rank only 2
and therefore has infinitely many solutions, which answers part (iv).
For all values of a other than ± 4 , the rank in question is 3,
same as the number of unknowns,
and so for those values of a there will be exactly one solution, which answers part (ii).
As for part (iii), the answer is: never (for no value of a can there ever
be more solutions than one, yet not infinitely many).
———
3. Can the 2 × 2 identity matrix I2×2 be expressed as the difference
I2×2 = AB – BA
between the two possible products AB and BA of two suitable 2 × 2 matrices
A and B ?
If so, give an explicit example; if not, give a coherent explanation of why not.
[Suggestion: Can the traces tr(I2×2) 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(I2×2) is easy enough to figure out:
tr(I2×2) = 1 + 1 = 2 .
As for the other, recall having seen that tr(AB) = tr(BA), and that
tr(P ± Q) = tr(P) ± tr(Q) .
But then
tr(AB – BA) = tr(AB) – tr(BA) =
tr(AB) – tr(AB) = 0 .
Consequently, were there, hypothetically, two 2 × 2 matrices
A and B with
I2×2 = AB – BA ,
we’d be forced to conclude that
2 = tr(I2×2) = tr(AB – BA) = 0 ,
a contradiction showing
just how untenable that hypothesis must be.
———
4. Given an n × n matrix A , with known determinant det(A) ,
calculate the determinant of the matrix product A•adj(A)
(of A with the adjoint matrix adj(A)).
Proposed Solution:
In any event, as regards the matrix product A•adj(A) , we know
A•adj(A) = det(A) In×n .
But the RHS above is what you get from the n × n
identity matrix In×n by multiplying each of its n rows by the factor det(A) .
Thus the
determinant of that RHS is the product of those n factors
det(A) along with the one additional factor
det(In×n) = 1 .
Summing up, then, we have:
det(A•adj(A)) =
det(det(A) In×n) =
(det(A))n (det(In×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 any n > 0 , we have tr(In×n) = n > 0 .
And, just as before, we have tr(AB) = tr(BA) and
tr(P ± Q) = tr(P) ± tr(Q)
(for n × n matrices A , B ,
P , and Q ).
So, just as before, the assumption that we could express
In×n as a difference
AB – BA would, upon calculating traces,
lead to the contradiction 0 < n = 0 .
———
Posted 4 Mar 2006