MAT 200, Midterm Exam Solution
1. (10 points total)
a. (5 points) Compute the determinant of the matrix
A =
2 2 0 −1
0 3 1 −2
0 3 −2 1
1 0 1 −1
Answer: det A =−3. The most efficient way is to develop the determinant along the first column, since there are two zeros in it.
b. (3 points) What is the dimension of the null-space of A? Explain.
Answer: Since det A 6= 0 the matrix is invertible, therefore Ax = 0 has only one solution: x = A−10 = 0. This means null(A) = {0}, so its dimension is 0.
Alternatively, you could have argued that since det A6= 0 the rank is 4, therefore by the fundamental theorem,
dim null(A) = #columns− rank = 4 − 4 = 0
c. (2 points) Find a basis for the row space of A.
Answer: Since det A 6= 0 the rank is 4 (=number of rows,columns). Therefore in RREF, each row of A contains a pivot, and it is the identity matrix; so you could choose the standard basis e!. . . e4 of R4. Alternatively any other basis on R4 would be acceptable as well.
2. (10 points total)
a. (5 points) For which values of k does the following system have zero / one / infinitely many solutions?
x + 2y − z = 1
kx + (2k− 1)y + z = 2
−x + (k − 3)y + z = −1
Answer: The row echelon form of the augmented matrix is
1 2 −1
0 1 1− k
0 0 (k + 1)(k− 1)
1 k− 2 k− 1
this case there is a unique solution. In the cases k = 1, the last row contains only zeros, so there are are solutions and also free variables, hence infinitely many solutions. If k =−1 the last row is (0 0 0 | − 2), corresponding to the equation 0 =−2, so there are no solutions.
b. (2 points) Let A be the coefficient matrix of this system. What is the dimension of the column space in each of the cases you found in (a)?
Answer: If k6= 1, −1 the column space has dimension 3. If k = 1, −1 there are two pivots in the RREF of the coefficient matrix, so the columns space has dimension 2.
c. (3 points) For k = 1, describe the row space of A using a system of (one or more) linear equations.
Answer: The RREF of A in this case is
1 0 3 0 1 −2 0 0 0
So n = (a, b, c) = (−3, 2, 1) is a solution to the equation An = 0. This is a normal vector to the space spanned by the rows, so the space is the solution space of the equation −3x + 2y + z = 0.
3. (10 points total)
a. (5 points) Compute the inverse of
A =
1 −1 1
1 0 3
2 −2 3
Answer:
A−1 =
6 1 −3 3 1 −2
−2 0 1
b. (3 points) Let B be the matrix obtained from A by swapping the second and third rows. Write B using A and an elementary matrix, and use this to compute the inverse of B.
Answer:
B =
1 0 0 0 0 1 0 1 0
A
so
B−1= A−1
1 0 0 0 0 1 0 1 0
−1
= A−1
1 0 0 0 0 1 0 1 0
=
6 −3 1 3 −2 1
−2 1 0
(the inverse of the elementary matrix is in this case the matrix itself).
Those of you who solved this in the same way as A got partial credit.
c. (2 points ) If C is another 3× 3 matrix and AC = CA, is it necessarily true that CA−1= A−1C?
Answer: yes, because we can multiply AC = CA by A−1 from the left and get C = IC = A−1AC = A−1CA
now multiply the equality C = A−1CA which we got by A−1 from the right, and get
CA−1= A−1CAA−1 = A−1CI = A−1C
4. (10 points total)
u =
1
−2 1 3
v =
2
−2
−1 1
w =
0
−2 3
−5
z =
−1
−4 8 12
Find a basis for the subspaceL(u, v, w, z) spanned by these vectors, and determine its dimension.
Answer: Placing these vectors columns as the columns of a matrix we find that the matrix has rank 3 and in its RREF form the pivots are in the first three columns, therefore u, v, w are a basis for the column space, which is the same as the space in questions.
b. (4 points) Does
b =
5
−2 1 2
belong toL(u, v, w, z)?
Answer: No, the equation su + tv + rw = b (in variables s, t) has no solution.
c. (2 points) What is the rank of the 4× 6 matrix whose columns are u, v, b, v, w, z?
Explain your answer.
Answer: The rank is 4. We know that the rank is the dimension of the column space. We also know that the vectors v, w depend on u, v, w, so the column space is spanned by u, v, w, b. We also know that b is not a LC of the other two, so these vectors are LI and form a basis for the column space; the space they span has dimension 3, so the column space has dimension 3, and this is the rank.
5. (10 points total)
a. (5 points) Find a parametric description of the plane through the origin which is orthogonal to n = (2,−1, 1).
Answer: This is the same as the null space of the matrix (2 − 1 1), which is
s
1 2
1 0
+ t
−12 0 1
: s, t ∈ R
b. (3 points) Does (2, 2,−3) belong to the plane which is orthogonal to n and passes through p = (1, 2, 3)?
Answer: No, because n· (p − (2, 2, −3)) = (2, −1, 1) · (−1, 0, 6) = −2 − −0 − 6 6= 0.
c. (2 points) Suppose that u, v are vectors, that kuk = 32kvk, and that kv − uk = 2kuk. Is the angle between u and v acute, right or obtuse? (Hint: express ku − vk2 using kuk , kvk and u · v).
Answer: Using the fact thatkwk2= w· w, have we ku − vk2 = (u− v) · (u − v)
= u· u − 2u · v + v · v
= kuk2− 2u · v + kvk2
using the information we have about the relation ofkuk , kvk , ku − vk we get
u· v = 1
2(−4 kuk2+kuk2+9
4kuk2) = (−4 + 13
4 )· kuk2< 0 so the angle is obtuse (assuming u6= 0).
For each of the following statements, indicate whether it is TRUE or FALSE by circling the correct answer. You do NOT have to justify your answer.
a. (2 points) Any three vectors in the plane with acute, non-zero angles between each pair, constitute a basis forR2.
FALSE: A basis ofR2 always contains 2 vectors, not 3.
b. (2 points) If A is a 3× 3 matrix and A2= I then A = I or A =−I.
FALSE: For example, A = (
1 0 0 −1
) .
c. (2 points) If a 4×3 matrix A has rank 3, then for any vector b the equation Ax = b has infinitely many solutions.
FALSE: The RREF of the augmented matrix of the equation could have a row of the form (0 0 0| 1).
d. (2 points) For a square matrix A, the column space of A has the same dimension as the column space of AT.
TRUE: The column space of AT is precisely the row space of A, which has di- mension equal to rank of A, which is also the dimension of the column space of A.
e. (2 points) If u, v, w ∈ R3 is a basis for R3 and A is a 4× 3 matrix such that Au = Av = Aw = 0, then all entries of A are 0.
TRUE: This means the null space of A has dimension 3, so by the fundamental theorem the row space has dimension 0. This can happen only when all the rows are the 0-vector, so all entries of A are 0.
