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Linear Algebra I 1016-
Final Solution
- (6p) Find the full solution of the system of equations:
x + 2y − z + u = 2 2 x + 4y − 3 z + 4u = 4
We compute a row echelon form for the augmented matrix of the system:
[ 1 2 − 1 1 2 2 4 − 3 4 4
]
R 2 ;− 2 R 1
[
]
We use the free variables y and u to find the pivot variables: z = 2u and x = 2 − u + z − 2 y = 2 + u − 2 y. So the complete solution is
x y z u
+^ y
+^ u
- (6p) Let
A =
(a) Find a basis for row (A).
We compute a row echelon form for A:
R 2 − 2 R 1 ,R 3 − 3 R 1 ,R4+2R 1 ;
R 3 − 5 R 2 ,R4+4R 2 ;
This gives the following basis for row (A):
[
]
[
]
(b) Find a basis for col (A).
Using the result from the previous part, this is
}^.
(c) Find the nullity of A.
We know that the sum of the rank and the nullity is equal to the number of columns. This means that the nullity of A is 1.
- (6p) Find the inverse of the matrix
A =
We can either use the Gauss-Jordan elimination or the classical adjoint. Let us use the second method:
AT^ =
det
[
]
−det
[
]
det
[
]
−det
[
]
det
[
]
−det
[
]
det
[
]
−det
[
]
det
[
]
T
Check:
AAT^ =
We can see immediately that when k = 1 , the rank is 1 ; when k = −1 , the rank
is 2 ; in all other cases, the rank is 3.
- (6p) Let A = [ a v 2 v 3 v 4 ] B = [ b v 2 v 3 v 4 ]
be 4 × 4 matrices, where a, b, v 2 , v 3 , and v 4 are column vectors in R^4. If det (A) = 4 and det (B) = 1, find det (A + B).
We use the multilinearity of the determinant:
det (A + B) = det ([ a v 2 v 3 v 4 ] + [ b v 2 v 3 v 4 ]) = det ([ a + b 2 v 2 2 v 3 2 v 4 ]) =
23 det ([ a + b v 2 v 3 v 4 ]) = 2^3 (det ([ a v 2 v 3 v 4 ]) + det ([ b v 2 v 3 v 4 ])) = = 2^3 (det (A) + det (B)) = 2^3 · 5 = 40.
- (6p) Let A ∈ Rn×n^ (an n × n matrix with real elements). True or false? Explain.
(a) If Ak^ = O for some positive integer k, then det (A) = 0.
True. We know that
(det (A))k^ = det (Ak) = det (O) = 0.
(b) If Ak^ = O for some positive integer k, then rank (A) = 0.
False. Consider
A =
[
]
Here A^2 = O but the rank of A is 1.
(c) If Ak^ = O for some positive integer k, then tr (A) = 0.
True. We know that if λ is an eigenvalue of A, then λk^ is an eigenvalue of Ak. Because Ak^ = O, all the eigenvalues of A have to be equal to 0. But then the trace of A is also 0, because the trace is the sum of the eigenvalues.
- (6p) Let A ∈ Rn×n^ (an n × n matrix with real elements). Show that if
A^2 + I = O,
then n must be even.
The assumption is A^2 = −I. Look at the determinant of the two sides:
(det (A))^2 = det (A^2 ) = det (−I) = (−1)ndet (I) = (−1)n.
If n is odd, the left hand side is nonnegative, the right hand side is negative. That is a contradiction, so n must be even.
- (6p) Let A be an n × n matrix, all of whose entries are either 1 or −1. Prove that det (A) is divisible by 2 n−^1.
Add or subtract the first column of the matrix to the other columns in such a manner that the elements in the first row turn to 0 (except of course a 11 = ±1). This does not change the value of the determinant. If we evaluate now the determinant by the first row, we simply get the cofactor obtained by deleting the first row and the first column. But at this point the elements in that minor are 2, −2 and 0. Using the multilinearity of the determinant we can factor out 2 from each column (so n − 1 times), and the remaining determinant now contains only 1s, −1s and 0s. Thus, the result is divisible by 2n−^1.
- (6p) Let A be an n × n invertible matrix. Show that
adj (adj (A)) = (det (A))n−^2 A.
We proved in class that for any invertible matrix B:
B adj (B) = det (B)I. (∗)
This of course just means that
(adj (B))−^1 =
B
det (B)
and det (B)B−^1 = adj (B) (∗∗)
Now plug in B = adj (A) into the second form in (∗∗), then use the first form in (∗∗) for B = A:
adj (adj (A)) = det (adj (A))(adj (A))−^1 = det (adj (A))
A
det (A)
det (adj (A)) det (A)
A.
