EXAM
Exam #1
Math 2360
Summer II, 2000
Morning Class
Oct. 11, 2000
ANSWERS
i
50 pts.
Problem 1. In each part, you are given the augmented matrix of a linear
system. The coefficient matrix is already in Reduced Row Echelon Form. Determine if the system is consistent, and if it is consistent find all
solutions.
(a.)

1
 0

 0
0
0
1
0
0
0
0
1
0

5
−1 

3 
0
Answer : (x, y, z) = (5, −1, 3)
(b.)

1
 0

 0
0
−1 0
0 1
0 0
0 0
2
−2
0
0
0
0
1
0

2
3 

1 
5
Answer : The system is inconsistent, because of the last row.
(c.)

1
 0

 0
0
−1
0
0
0
0
1
0
0
2 0
−2 0
0 1
0 0

1
0 

−3 
0
Answer :
The system is consistent. Call the variables x1 through x5 . The variables
x1 ,x3 and x5 are leading variables and x2 and x4 are non-leading or free
variables. Say x2 = α and x4 = β. The third row of the matrix gives us
the equation x5 = −3. The second row of the matrix gives the equation
x3 − 2x4 = 0, so x3 = 2x4 = 2β. The first row of the matrix gives the
equation x1 − x2 + 2x4 = 1, and so x1 = 1 + x2 − 2x4 = 1 + α − 2β.
Putting all this together, we see the system has the two parameter family
of solutions
(x1 , x2 , x3 , x4 , x5 ) = (1 + α − 2β, α, 2β, β, −3)
1
45 pts.
Problem 2. Consider row operations on matrices with 3 rows.
Recall that for each row operation there is a corresponding elementary matrix
E so that EA is the same as the matrix obtained by applying the row operation
to A.
(a.) Consider the row operation R1 ↔ R3 .
i.) Find the corresponding elementary matrix E.
Answer :

0
E = 0
1

1
0 .
0
0
1
0
ii.) Find the inverse row operation.
Answer : R1 ↔ R3 (the same operation).
iii.) Find the elementary matrix that corresponds to the inverse row operation.
Answer : The same matrix:

0
0
1
0
1
0

1
0
0
(b.) Consider the row operation R3 ← 5R3 .
i.) Find the corresponding elementary matrix E.
Answer :

1
E = 0
0
0
1
0

0
0
5
ii.) Find the inverse row operation.
Answer : R3 ← 15 R3
iii.) Find the elementary matrix that corresponds to the inverse row operation.
2
Answer :

1 0
0 1
0 0

0
0 
1/5
(c.) Consider the row operation R2 ← R2 + 2R3
i.) Find the corresponding elementary matrix E.
Answer :

1
E = 0
0

0 0
1 2
0 1
ii.) Find the inverse row operation.
Answer : R2 ← R2 − 2R3 .
iii.) Find the elementary matrix that corresponds to the inverse row operation.
Answer :

1
0
0
40 pts.
0
1
0

0
−2
1
Problem 3. In each part, use row operations to determine if the matrix A
is invertible and, if so, to find the inverse. It is not necessary
to show the individual row operations (you can just use the rref key on the
calculator). Show the augmented matrix you start with and the augmented
matrix you finish with. Give the matrix entries in fractional form.
(a.)

1
A = 2
0
0
1
2

−2
−5
0
Answer :
To start, we form the augmented matrix

1 0 −2 1 0
 2 1 −5 0 1
0 2
0 0 0
3

0
0 
1
and perform row operations to put the part to the left of the bar in Reduced
Row Echelon Form. The result is


1 0 0 5 −2
1
 0 1 0 0
0 1/2 
0 0 1 2 −1 1/2
Since the reduced row echelon form of A
A is invertible. The inverse is the matrix

5 −2
0
A−1 =  0
2 −1
is the identity, we conclude that
to the right of the bar, i.e.,

1
1/2  .
1/2
(b.)

1
A = 0
2
1
1
1

1
−1
3
Answer :
We start by forming the augmented matrix


1 1
1 1 0 0
 0 1 −1 0 1 0  .
2 1
3 0 0 1
We perform row operations
The result is

1

 0

0
to put the part to the left of the bar in RREF.
0
2
1
−1
1
−1
0
1
0
0
−2
1
0


0 

1
Thus, the RREF of A is not the identity, so A is not invertible.
40 pts.
Problem 4. Consider the matrix


1 2
1 2
2 0
0
A = 1
1
4
(a.) Find the cofactors A22 and A32 .
Answer :
The cofactor Aij is obtained by crossing out the row i and column j, taking
the determinant, and adding the sign from the checkerboard (which can also
be described as (−1)i+j ). Thus,
2+2 0 2
A22 = (−1)
1 0 = (1)(0 − 2) = −2
and
3+2
A32 = (−1)
0
1
2
= (−1)(0 − 2) = 2
2
(b.) Show how to compute det(A) by using a cofactor expansion along some
row or column.
Answer :
Expand, for example, along
0 1
det(A) = 1 1
1 2
column 2, which gives
2
2
0
= 1A12 + 1A22 + 2A32
1 2
+ 1 + 0
= 1 − 1
1 0
0
2
+
2
− 0
1
= −(0 − 2) + (0 − 2) − 2(0 − 2)
=2−2+4= 4
40 pts.
Problem 5. Use row operations
0
1
2
to find the determinant.
2 6 2 1 6 10
5
2
2
Answer :
We proceed as follows:
0 2 6 1 2 1 2 6 10
1 2
= − 0 2
2 6
1 2
= − 0 2
0 2
1 2
= −2 0 1
0 2
1 2
= −2 0 1
0 0
apply R1 ↔ R2
1 6 10
1
6
8
1
3
8
1
3
2
R3 ← R3 − 2R1
R2 ←
1
R2
2
R3 ← R3 − 2R2
now upper triangular, multiply diagonal elements
= −2(1)(1)(2)
= −4
40 pts.
Problem 6. Use Cramer’s rule to solve the following system.
x
x
+
+
3
1
3y
y
=
=
2
5
Answer :
In matrix form, the system is
1
1
x
2
=
.
y
5
By Cramer’s rule,
and
2
5
x = 1
1
3
1
−13
=
= 13/2
−2
3
1
1
1
y = 1
1
2
5
3
=
= −3/2
−2
3
1
6
Thus, the solution is
x
13/2
=
.
y
−3/2
7