MATH 240, SOLUTIONS TO PRACTICE MIDTERM EXAM 1

Exercise 1. The 3 3 matrix A is given by:

A =

1 2 44 8 04 5 1

a) Find A ~b when ~b =

1024

.b) Find A + AT .

c) Is A + AT a symmetric matrix?

Solution:

a) We compute: 1 2 44 8 04 5 1

102

4

=10 + 4 + 1640 + 16 + 0

40 + 10 + 4

=3056

54

.b) The matrix AT equals

1 4 42 8 54 0 1

. Hence:A + AT =

2 6 86 16 58 5 2

.c) We note that:

(A + AT )T = A + AT =

2 6 86 16 58 5 2

.It follows that A + AT is a symmetric matrix.

Alternatively, we could have used the fact that:

(A + AT )T = AT + (AT )T = AT + A = A + AT .

Here, we recall the following fact about transposes: (AT )T = A. Exercise 2. Consider the following system of linear equations:

x1 2x2 + 3x3 = 22x1 + x2 + 5x3 = 7

7x1 5x2 8x3 = 3a) Use Gaussian elimination in order to determine the solution set to the system.

b) What is the rank of the matrix of coefficients A?

c) What is the rank of the augmented matrix A#?1

2 MATH 240, SOLUTIONS TO PRACTICE MIDTERM EXAM 1

d) How can we use only the results from c) and d) to determine whether the solution set is emptyor not?

Solution:

a) The augmented matrix of the system is:

A# =

1 2 3 22 1 5 77 5 8 3

.We first apply S12(2) and S13(7) in order to see that this matrix is:

1 2 3 20 5 1 3

0 9 29 17

.We then apply M2(

15 ) and see that this matrix is:

1 2 3 20 1 15 35

0 9 29 17

,which by application of S23(9) is:

1 2 3 20 1 15 35

0 0 1365 1125

.Finally, we apply M3( 5136 ) in order to deduce that this matrix is:

1 2 3 20 1 15 35

0 0 1 1417

.This matrix is in row-echelon form.

The obtained system is: x1 2x2 + 3x3 = 2x2 15x3 = 35x3 =

1417

By using backwards substitution, it follows that x3 =1417 , x2 =

1317 , x1 =

1817 . In other words, the

solution set is given by:

S ={(18

17,

13

17,

14

17

)}.

b) From part a), we know that:

A 1 2 30 1 15

0 0 1

.The matrix on the right-hand side is in row echelon form and has 3 leading ones. Hence, the rankof A equals 3.

c) Likewise, from part a), we know that:

A# 1 2 3 20 1 15 35

0 0 1 1417

.

MATH 240, SOLUTIONS TO PRACTICE MIDTERM EXAM 1 3

The matrix on the right-hand side is in row-echelon form and has 3 leading ones, hence the rank ofA# also equals 3.

d) By parts b) and c), it follows that rank (A) = rank (A#). Consequently, the solution set isnot empty. Exercise 3. a) Use elementary row operations in order to calculate the determinant of the matrixA given by:

A =

1 1 0 02 3 6 00 2 1 20 0 0 1

b) Suppose that B is a 4 4 matrix such that det (B) = 1. Find det (A BT ).Solution:

a) We first apply S12(2) in order to see that:

det (A) =

1 1 0 02 3 6 00 2 1 20 0 0 1

=1 1 0 00 5 6 00 2 1 20 0 0 1

By applying S23( 25 ), this determinant equals:

1 1 0 00 5 6 00 0 175 20 0 0 1

.Since the matrix is now upper-triangular, this determinant equals 1 5 ( 175 ) 1 = 17.

b) We know that: det (A BT ) = det (A) det (BT ) = det (A) det (B) = (17) 1 = 17. Here,we used the fact that det (BT ) = det (B). Exercise 4. Let S be the set of all 2 2 matrices A with real elements such that at least one entryof A equals zero. In other words:

S ={[0 b

c d

],

[a 0c d

],

[a b0 d

],

[a bc 0

]; a, b, c, d R

}a) Is the set S closed under addition?

b) Is the set S closed under multiplication by real scalars?

Solution:

a) The set S is not closed under addition. Let us note that

[1 00 0

]and

[0 11 1

]both belong to

S, whereas their sum

[1 00 0

]+

[0 11 1

]=

[1 11 1

], doesnt belong to S.

b) Suppose that A =

[a bc d

]belongs to S and suppose that r R. We want to check that r A

belongs to S. By definition of S, we know that at least one of a, b, c, d has to equal zero. Hence, at

least one of r a, r b, r c, r d has to equal zero. In particular, r A =[r a r br c r d

]belongs to S.

It follows that S is closed under multiplication by real scalars.