Math 3260
Practice Exam #1 – Answer key
1. a) Use Gauss-Jordan elimination:
1R
R R R
R
3  2 2 1 0 1
3 3
3 3
2
3 1 0 1
1 0 1
0 2  2  4 
~ 0 1  1  2
~ 0 1  1  2
~


0 1  2 5 
0 1  2 5 
0 0  1 7 
R  R R
2 3
2
3  R1 R3R1
1 0 1
1 0 0 10 
0 1  1  2 
~ 0 1 0  9 . Therefore the solution is


0 0 1  7
0 0 1  7
b) The reduced row echelon form of the augmented matrix is
1 0 0 3
3
0 1 0 0 and therefore the solution is 0  .


 
0 0 1 2
 2 
 10 
 9 .
 
  7 
2. The system has only the trivial solution if and only if the determinant of the
coefficient matrix is not zero, that is ad  bc  0. The system will have infinitely
many solutions if the determinant is zero. Since it is a homogeneous system, it can
never be inconsistent.
3. a) If the first system has a unique solution and A is n  n it means that A is invertible,
and therefore any other system with the same coefficient matrix will have a unique
solution, which can be written as x  A 1b .
If the coefficient matrix is not square, then it is possible to have a unique solution for
one system and no solution for another. Here is an example:
x  y  0
x  y  1


and 2 x  y  4
2 x  y  1
4 x  2 y  2
4 x  2 y  5


Notice that the first system has a unique solution (the third equation is just a multiple
of the second one), whereas the second system has no solution (the second and third
equations represent parallel lines).
b) The answer is Yes, and we will justify it by an example. Consider the following
two systems:
x  y  0
x  y  1
and 

2 x  2 y  0
2 x  2 y  4
Obviously, the first system has infinitely many solutions, ( y  t, x  t ), while the
second one is inconsistent.
4. A( A  B)  B( A  B)  ( A  B)( A  B) . Notice that it is not equal to A 2  B 2 .
5. We need to have f (1)  2, f (2)  2, and f (3)  4 . This leads to the following
systems in the unknowns a, b and c:
 a b c  2
a  1


4a  2b  c  2 whose solution is b  3
9a  3b  c  4
c  4


6. ABC is 3 3 , CA is 1 2 , and BA is not defined.
a b 
7. We want to find matrices B  
 such that AB  BA . Notice that B has to be a
c d 
2  2 matrix in order for both products to exist. We get:
a
b   a  b 0

 a  b  c  d 0 and therefore a  s, b  0, c  t , d  s  t , with s, t being arbitrary real numbers.

 

1 3 3
8. a) We have A  0 0 1  A , therefore A is not idempotent.
0 0 0
1  bc  1, bc  0
2
b  bd  b, bd  0
1 b 
1  bc b  bd 

b) 
, so 


2
c d 
c  cd bc  d 
c  cd  c, cd  0
bc  d 2  d so d 2  d since bc  0
2
From the above we have the following possibilities:
 b  c  0, d  1
 b  d  0, c arbitrary real number.
 c  d  0, b arbitrary real number.
c) ( AB) 2  ABAB  AABB  A2 B 2  AB proving that AB is idempotent. Notice that in
the second equality we used the fact that AB  BA .
 4 5
T
4  2 7 
9. Too easy to bother, but here it is: 
  2 3

5 3 0 
 7 0
1
 3

2 x x   2  1
 4 2


10. We have 


 so x  2 .
 5 3  5 2 
5 3
 2

11. If A is not invertible, it means that det( A)  0 and since det( AB)  det( A) det( B)  0 ,
it follows that AB cannot be invertible.
12. Using the cofactor expansion along the first row we get:
x[( x  1)( x  1)  4( x  1)]  2[2 x( x  1)  x( x  1)]  0  x( x  1)( x  3  6)  0, so x  0, x  1, x  3
13. The value of the determinant does not depend on a or b since a cofactor expansion
along the third row shows the value to be 2(3  45)  84.
1
14. If | A | 4, then | 3 A | 9  4  36, | A | 2  16, | 5 AT A 1 | 25  4   25 .
4
15. Since the matrices are triangular, the determinant is equal to the product of the entries
on the main diagonal, therefore the answers are: in a) –60, in b) –30.
16. Since the second row in the determinant has already two zeros, we will add the first
column to the second column and use the cofactor expansion along the second row:
1 1 0 2
1 0 0 2
0 0 2
1 1 0 0
1 0 0 0
0 0

 0 0 1 2
0
2 2 0 1
2 0 0 1
4 5
4 5 1
3
1 5 1
3 4 5 1
17. Since det( AB)  det( A) det( B)  det( I n )  1, it follows that det( A)  0 and det( B)  0 .
18. The systems will have a nontrivial solution if and only if the determinant of the
coefficient matrix is zero in both cases.
1 
6
a)
 2  3  28  0  (  7)(  4)  0, so   7,   4
5
2
b)
5
4
2 R2  2 R3  R2 5  
4
2
4
2
5
2
2
2

1 
2
0
2
2  2  (5   )[(1   )( 2   )  4(  1)]
2
 2[4(2  2)  2(1   )]  (  1)(5   )(  6)  20(  1)  (  1)( 2  11  10)
The above determinant is zero for   1 and   10 .
19. Denote , x, y, and z the determinants of the coefficient matrix, and the ones
obtained from the coefficient matrix when the column of coefficients of x, y, z are
replaced by the values on the right hand side of the equations, respectively.
2 1 3
7 1 3
2 7 3
We have   1
3
4
2
2 1
7
and z  1
3
4
2
2  35, x  10
1
0
4
2
x
10  140. Then x 
 2,

0
2  70, y  1 10 2  35 ,
1
3 0 1
y  1, z  4 .