Mathematics 200, Spring 2013
Midterm Exam 2 Review Sheet
Solutions
2 3 -1
1. Find the determinant of the matrix A = 1 1 0 by hand, and then
3 4 7
-1
find A by using the adjoint matrix formula for the inverse.
Expanding the determinant along the first row, we find detHAL = 2 H7L - 3 H7L - 1 H1L = -8.
The formula is that A-1 = H1  A × adjHALL, where adjHAL is the transpose of the cofactor matrix cof HAL. The
cofactor matrix is the matrix whose Hi jL-entry is the signed subdeterminant associated to the Hi jL-entry of A. In our
7 -7 1
-25
17 1 .
case, we have that cof HAL =
1 -1 -1
7 -25 1
7 -25 1
-1
Therefore, the adjoint of A is adjHAL = -7 17 -1 , and A = H-1  8L × -7 17 -1 .
1
1 -1
1
1 -1
Check:
2 3 -1
7 - 25 1
MatrixFormB 1 1 0 . - 7 17 - 1 F
3 4 7
1
1 -1
-8 0 0
0 -8 0
0 0 -8
4
2. Solve the system A × x = 2 using Cramer's Rule, where A is the
3
matrix in problem 1.
Here is the solution vector computed with Cramer’s Rule:
2
exam 2 review sheet solutions.nb
4
DetB 2
3
2
MatrixFormB DetB 1
3
2
DetB 1
3
19
8
3
8
3
8
3
1
4
4
2
3
3
1
4
-1
0 F “ -8
7
-1
0 F “ -8 F
7
4
2 F “ -8
3
Check :
2 3 -1
MatrixFormB 1 1 0 .819  8, - 3  8, - 3  8<F
3 4 7
4
2
3
3. Find the value(s) of h for which the following set of vectors is
linearly dependent.
1
1
-5
-1 , 7 , 1
8
-3
h
The vectors are linearly dependent if there is a nontrivial linear combination of them that equals the 0-vector. In
1 -5 1 0
other words, they are linearly dependent if the homogeneous system with augmented matrix -1 7 1 0 has
-3 8 h 0
nontrivial
1 -5
-1 7
-3 8
solutions, or equivalently
1 0
1 -5
1
0
1 0 ® 0 2
2
0 ®
h 0
0 -7 3 + h 0
a free variable.
We row-reduce the matrix:
1 -5
1
0
0 1
1
0 . We see that x3 will be a free variable if and only
0 0 10 + h 0
if 10 + h = 0, so h = -10 is the unique value of h for which the vectors are linearly dependent.
4. Show that the following set of vectors H in R3 is a subspace, and
3 a + 2 b + 10 c
find a basis of this subspace: H = : a - 4 b - 6 c >
2 a + 3 b + 10 c
4. Show that the following set of vectors H in R3 is a subspace, and
3 a + 2 b + 10 c
exam 2 review sheet solutions.nb
find a basis of this subspace: H = : a - 4 b - 6 c >
2 a + 3 b + 10 c
3
2
10
We have that H = Span : 1 , -4 , -6 >; since the linear span of any set of vectors
2
3
10
3
2
3
subspace of that space, we see that H is a subspace of R . Note that 2 1 + 2 -4 =
2
3
3
in a vector space is a
10
-6 . Since the third
10
vector is a linear combination of the first two vectors in the spanning set, the third vector can be removed without
changing the span of the vectors. Furthermore, since the first two vectors are not scalar multiples of one another,
they are linearly independent. It follows that the first two vectors form a basis of H . A similar argument can be
used to show that any two of the vectors form a basis of H .
5. Find the dimension of Nul A and Col A, and a basis for each
subspace, if A is the following matrix:
1 0 -3 1 2
0 1 -4 -3 1
-3 2 1 -8 -6
2 -3 6 7 9
The null space is the set of solutions of the homogeneous system A.x = 0. We solve the system by row reduction,
and write the general solution in parametric vector form:
1 0 -3 1 2
0 1 -4 -3 1
MatrixFormBRowReduceB
-3 2 1 -8 -6
2 -3 6 7 9
1
0
0
0
0
0
FF
0
0
0 -3 0 4 0
1 -4 0 -5 0
0 0 1 -2 0
0 0 0 0 0
The basic variables are x1 , x2 , and x4 ; the free variables are x3 and x5 . Already this tells us that dimHnul AL = 2 and
dimHcol AL = 3. The general solution of the system has the form x5 = t, x4 = 2 t, x3 = s, x2 = 4 s + 5 t, and
3
-4
3s-4t
x1
4
5
4s+5t
x2
x1 = 3 s - 4 t. In parametric vector form, we have x3 =
= s 1 + t 0 ; the two vectors in the linear
s
x4
2t
0
2
x5
t
0
1
combination are a basis of nul A. A basis of col A is given by the pivot columns of A, which are the first, second,
and fourth columns of the original matrix A (NOT the row-reduced version of A).
x1
2 x1 - 3 x2
6. Let T : R3 ® R3 be defined by T x2 =
. Prove that T is
4 x3
x3
x2 - 5 x1
linear.
x1
2 x1 - 3 x2
6. Let T : R ® R be defined by T x2 =
. Prove that T is
4 x3
4
exam 2 review sheet solutions.nb
x3
x2 - 5 x1
linear.
3
3
One way to do this is to demonstrate that THxL = A.x, since a theorem states that
2
-3
0
2
x1
0
0
4
x
linear. But this is easy: note that T
= x1
+ x2
+ x3
= 0
2
x3
-5
1
0
-5
every matrix transformation is
-3 0
0 4 × x . We can also show
1 0
directly that Tsatisfies the two basic properties of linearity: THx + yL = THxL + THyL and THc xL = cTHxL.
8
7. Find the coordinate vector @xDB where x = -9 and
6
1
-3
2
B = : -1 , 4 , -2 >.
-3
9
4
The coordinate vector is the list of coefficients in the unique linear combination of the vectors in B that is equal to
x. So, to find the coordinate vector, we solve the inhomogeneous system with augmented matrix
In[1]:=
1 -3 2 8
MatrixFormBRowReduceB - 1 4 - 2 - 9 FF
-3 9 4 6
Out[1]//MatrixForm=
1 0 0 -1
0 1 0 -1
0 0 1 3
8
-1
It follows that B -9 F = -1
B
6
3
8. a) If the null space of an 8 x 5 matrix A is 2-dimensional, what is the
dimension of the row space of A? b) Suppose that a 4 x 7 matrix B
has four pivot columns. What can you say about Col B and Nul B?
a) Since the dimension of the null space is 2, we know that the homogeneous system A.x = 0 has two free variables,
and therefore 3 basic variables (and three pivots). Therefore, the rank of the matrix is 3, which is also equal to the
dimension of the row space.
b) Since there are four pivot columns and seven columns overall, we know that the homogeneous system B.x = 0
has three free variables. Therefore, the null space has dimension 3 and the column space has dimension 4.
9. Suppose a nonhomogeneous system of nine linear equations in ten
unknowns has a solution for all possible constants on the right sides
of the equations. Is it possible to find two nonzero solutions of the
associated homogeneous system that are not multiples of each
other? Discuss.
exam 2 review sheet solutions.nb
5
9. Suppose a nonhomogeneous system of nine linear equations in ten
unknowns has a solution for all possible constants on the right sides
of the equations. Is it possible to find two nonzero solutions of the
associated homogeneous system that are not multiples of each
other? Discuss.
The answer is NO. The hypothesis implies that the columns of the coefficient matrix span R9 , which means that
every row must have a pivot, so we have 9 pivots overall. This in turn implies that the homogeneous system has
one free variable, and so the null space has dimension 1. Therefore, there cannot be two nonzero solutions of the
homogeneous system that are not multiples of one another (if there were, the two solutions would be linearly
independent, and therefore the dimension of the null space would be at least 2).
10. Find the rank of each of the following matrices.
5 0 0
-3 -7 0
8 5 -1
1 -5 -4
0 3 4
-3 6 0
The first matrix has a non-zero determinant equal to the product of the diagonal entries. This implies that it is rowequivalent to the 3 x 3 identity matrix, and so has rank 3. The second matrix has rank 2, since it has two pivots:
In[2]:=
MatrixFormBRowReduceB
Out[2]//MatrixForm=
1 0
0 1
8
3
4
3
0 0 0
1 -5 -4
0 3 4 FF
-3 6 0