York University
Department of Mathematics and Statistics
MATH:1025
Test - 1
17 Oct. 2014
Solution sketch and or some Hints
1. (14 points) Write TRUE of FALSE for each of the following statement. If a statement is false, then
you must explain (in the space on this page or give an example). You need not explain if true in all
cases.
a. A linear system of two equations in 3 unknowns has infinitely many solutions:
1
 x yz
FALSE, system may be inconsistent. Example 
2 x  2 y  2 z  3
b. If a linear system has same number of unknowns as number of equations, then the system
must have unique solution.
1
 x y
False: Example: The system 
has many sol.
2 x  2 y  2
c. If a matrix has more number of columns than number of rows, then columns are linearly
independent.
 1 0 1
 , column 3 is a linear comb. of first two.
False: In 
 0 1 3
d. A homogeneous system AX   , has a non-zero solution, then columns of the matrix are
linearly independent. _______
False, may have many sol. example or explanation
2
e. When A and B are square matrices of same size,  A  B   A 2  2 AB  B 2 .
False: For matrices AB not equal BA in general. example or explanation, i.e.
 A  B 2  A2  AB  BA  B 2  A2  2 AB  B 2 .
f. If a set u v  Rn is linearly independent, then so is u v w  Rn . ________
False: vector w  zero vector, then the set containing zero vector is dependent. or give
other explanation or example.
g. If a set u v w  Rn . is linearly dependent, then so is u v  Rn . ________
False, The set u,v,  is dependent where as subset u, v  e1 , e2  is independent. OR give
explanation
Math 1025
Test-1
Page 2 of 6
2. (15) Find the polynomial px   ax 3  bx 2  cx  d , whose graph passes through four points
A0, 1, B1, 0, C 1, 4 and D2, 4.

d 1
 a  b  c  1

 abc 3
 abcd  0



 abcd  4
8a  4b  2c  3
8a  4b  2c  d  4  using d  1.
 1 1 1  1  1 1

 
 1 1 1 3   0 2

Now reduce the Aug. matrix 
8 4 2 3  0  4

 

 

 
A student may reduce this further and read values for d  1,
1
5
Therefore the required polynomial is y  x 3  x 2  x  1
2
2
1  1  1 0 1  2 
 

0
2  0 2 0
2 

 6 11   0 0  6 15 
 

 

 

c  5 / 2, b  1 and a  c  2  1 / 2
Note: Chk my computation. Students are advised to state explicitly elementary row operations they do.
Deduct points for errors in operations and final answer.
Math 1025
Test-1
Page 3 of 6
 8x  9 y  2
3. (6) Find all values of the unknown k such that the system 
is consistent.
12 x  ky   1

9
2
 8  9 2   8


12
12 
Aug. matrix 12 k  1   0 k  9  1  2 This implies unique solutions when
8
8


 

 

27
k
 0  consissten t when k  27 / 2
2
Note: Chk my computations. This how I expect them to answer, but some might have done differently.
Give appropriate credit for the work.
4. (15)Find two non-zero vectors; u  v , of matching dimension such that they DO NOT belong to
 1   3 
   
span  4 ,   1 . Show work.
  5   1 
   
 x
 
 1  3
   
 y
A vector X     span if X  a 4   b  1 for real numbers a and b.
z
  5  1 
 
   
 
 



3 x  1 3
x  1 3
x
 1

 
 

Reduce the aug. matrix  4  1 y    0  13 y  4 x    0  13
y  4x

  5 1 z   0 16 z  5 x  
16
z  5 x   y  4 x 

 
 0 0
13


No solutions when 13z  5x  16 y  4x  0.  x  16 y  13z. Now choose two vectors e.g.
1
 0
 
 
X   2  and say X   0 
 3
1
 
 
Note: Chk my computation. Students are advised to state explicitly elementary row operations they do.
Deduct points for errors in operations and final answer.
Math 1025
Test-1
Page 4 of 6
5. (30) Use matrices given here to answer parts (a) through (g). You need minimal computation to
answer parts (b) to (g) if you know how to interpreter work and result of part (a). Notice that
matrices A, u, and v are columns of the matrix M.
1 0 0
1
 1 2
 1 2



M   3  2  1 1 0, A   3  2  1  C1 C2
 2
 2
0
4 0 1
0
4 
0
0


C3 , u  1, v  0
0
1
a. (15) Using elementary row operations, find Reduced Row Echelon matrix that is equivalent
to the matrix M.
b. (4) Explain why spanC1 C2 C3   R3 .
c. (2) State complete solution set for the homogeneous system AX   .
d. (2) Solve the system AX  u., X  R3 .
e. (2) Determine whether the linear transformation T  X   AX is or is not one-to-one.
Explain.
f. (2) Find X  R3 such that T  X   v
g. (3) Find the complete solution set for the homogenous system MX   ., X  R3 .
Answer key for # 5:
1 0 0 a m
1



(a) The RRE form of the matrix :  0 1 0 b n  i.e.  0
0 0 1 c p 
0



1 0 0
 1 2
  1 2 1 0 0   1 2 1


M   3  2  1 1 0,   0 4 2 1 0   0 4 2
0
4 0 1
 2
 0 4 6 0 1  0 0 4
R2  R2  3R1 , R3  R3  2 R1.
0 

1 0 3 / 8  1 / 8  ; for
0 1  1 / 4 1 / 4 
0 0
1/ 2
0   1 2 0 1 / 4  1 / 4
0   0 4 0 3 / 2  1 / 2
 1 1  0 0 4  1
1 
0
1
1
R3  R3  R2 . R2  R2  R3 , R1  R1  (1 / 4) R3 . R1  R1  1 / 2) R2 
2
0 
 1 0 0  1 / 2

  0 4 0 3 / 2  1 / 2
 0 0 4  1
1 
NOTE: All students must get same correct entries in columns 4 and 5.
(b) RRE form of the matrix M shows that columns of matrix A are linearly independent and hence they
span R3.
(c) Using part (b), or otherwise, deduce that homogeneous equation AX   has only trivial solution
and hence complete SS   , just one vector, NOT the empty set.
Math 1025
(d)
(e)
(f)
(g)
Test-1
Page 5 of 6
a
 4 
  1 
The system AX  u has unique solution X   b    3  as computed in part (a)
 c  8   2
 
 
Because the homo./ system AX   , in part (c) has only trivial sol. transformation is 1-1.
 m
0
  1 
As in part (d), the system T  X   AX  v, has unique solution X   n     1 as computed in
 p 8 2 
 
 
part (a).
Use RRE form of M to deduce that homo. System MX   has infinitely many solutions, The
complete SS is linear span of two vectors in R5, or all linear combinations of two free variables x4
and x5.
 x1 
 x1    ax4  mx5 
 a
  m   4  0 
 
  

 

    
 1 0 0 a m  x2 
 x2    bx4  nx5 
 b
  n    3  1 

 
  

 

    
 0 1 0 b n  x3     X   x3     cx4  px5   x4   c   x5   p   s 2   t   2 
 0 0 1 c p  x 
x4
 x4  

 1 
 0   8   0 

 4
x 
x  



 1   0   8 
x5
 0 

    
 5
 5 

where s and t are any real numbers. The complete
  4   0 
   
  3   1 


SS = span  2 ,   2   Span 4e1  3e2  2e3  8e3 , e2  2e3  8e5 
   
 8   0 
   
 0   8 
Math 1025
Test-1
Page 6 of 6