MA122 – Midterm Test Solutions Page 1 of 5

[10 marks] 1. In each question circle either True or False. No justification is needed.
(a) The solution set of the system of equations
3x1 + 4x2 − x3 = 0
2x1 + 3x2 + 2x3 = 0
3
is a subspace of R .
True False

(b) The rank of a matrix is the number of nonzero rows in any row echelon form (REF) of
the matrix.
True False
 
1
(c) If 1 is a solution of a homogeneous system of linear equations, then it is the unique

1
solution of the system.
True False

(d) Suppose →
−
u ,→
−
v ∈ R3 such that {→ −
u ,→
−v } is linearly independent. If →
−
w is a nonzero vector
in Span{→
−
u ,→
−
v } that is orthogonal to the vector → −
u , then {→
−
u ,→
−
v ,→
−w } is a basis for R3 .
True False

(e) A homogeneous system of 3 linear equations in 3 variables has ONLY the trivial solution.
True False

(f) Let          

 1 2 1 2 2 
         
2 ,   ,   ,   , 2
1 2 2

B= 2 2 1 2 2

 
2 1 2 1 2
 

and Span(B) = R4 . Then B is linearly dependent.

True False
 
1
(g) →
−
v = −8 is a solution of the following homogeneous system
−1
5x1 + x2 − 3x3 = 0
x1 − 2x2 + 3x3 = 0
True False

(h) Let →
−
u, →−
v and → −
w be nonzero vectors in R3 . If →
−
u ·→
−
v = 0, →
−
u ·→
−
w = 0 and →
−
v ·→
−
w = 0,
→
− →
− →
− 3
then { u , v , w } is a basis for R .
True False

(i) The set of vectors        


 1 1 0 0 
       
0 1 1 0

B=  , , ,
0 0 1 1

 
0 0 0 1
 

is a basis for R4 .
True False

(j) Let →
−
v,→
−
w be distinct nonzero vectors in R3 .
→
−
→ w ) × (perp−
[(proj− →
− →
−
→ w )] · w = 0.
v v

True False
 MA122 – Midterm Test Solutions Page 2 of 5



 x2 − x4 + x5 = 1
2. Given the system of linear equations x1 + x2 + x3 − x4 = 2

x1 + x3 + x5 = 3


[2 marks] (a) Write the augmented matrix of the above linear system.
Solution:  
0 1 0 −1 1 1
1 1 1 −1 0 2
1 0 1 0 1 3

[6 marks] (b) Use Gauss-Jordan elimination algorithm on the augmented matrix in part (a) to obtain
its row-equivalent matrix in reduced row echelon form (RREF), and circle the leading
ones.    
0 1 0 −1 1 1 1 1 1 −1 0 2
R ↔R2
Solution:  1 1 1 −1 0 2 −−1−−→  0 1 0 −1 1 1
1 0 1 0 1 3 1 0 1 0 1 3
   
1 1 1 −1 0 2 1 1 1 −1 0 2
−R +R3 →R3 R +R3 →R3
−−−1−−− −−→  0 1 0 −1 1 1 −−2−−− −−→  0 1 0 −1 1 1
0 −1 0 1 1 1 0 0 0 0 2 2
   
1
R →R3
1 1 1 −1 0 2 1 1 1 −1 0 2
2 3 −R3 +R2 →R2
−−−−−→ 0 1 0 −1 1
 1 −−−−−−−−→ 0 1 0 −1 0
  0
0 0 0 0 1 1 0 0 0 0 1 1
 
1 0 1 0 0 2
−R2 +R1 →R1
−−−−−−−−→ 0 1 0 −1 0
 0
0 0 0 0 1 1

[2 marks] (c) Use your answer in part (b) to determine which of x1 , x2 , x3 , x4 , x5 is (are) free vari-
able(s) (parameter(s)) of the system of linear equation.
Solution:
x3 = r and x4 = t, where r, t ∈ R.
[7 marks] (d) Using parts (b) and (c), solve the system of linear equations, and then write the general
solution of the linear system in the standard format of a linear combination of vectors
in R5 isolating the parameter(s) identified in part (c).
Solution:
x5 = 1;
x2 = x4 = t;
x1 = 2 − x3 = 2 − r.
         
x1 2−r 2 −1 0
x2   t  0 0 1
         
Thus,  x 3
 =  r  = 0 + r  1  + t 0, where r, t ∈ R.
         
x4   t  0 0 1
x5 1 1 0 0
[3 marks] (e) Verify that your answer in part (d) is the genuine solution of the system of linear
equations given at the top of the page.
Solution: the 1st equation: x2 − x4 + x5 = t − t + 1 = 1

the 2nd equation: x1 + x2 + x3 − x4 = (2 − r) + t + r − t = 2

the 3rd equation: x1 + x3 + x5 = (2 − r) + r + 1 = 3

[Reminder: You may need to correct possible errors in any part(s) above if one of the
original equations is not satisfied.]
 MA122 – Midterm Test Solutions Page 3 of 5

3. Let W be a plane in R3 that contains the points P (3, 0, 3), Q(0, 1, 0) and R(1, 0, −1).
−→ −→
[3 marks] (a) Find vectors P Q and P R and show that they are linearly independent.
Solution:
     
0 3 −3
−→ →
PQ = −
q −→−
p = 1 − 0 =  1 
0 3 −3
     
1 3 −2
−→ → − →
−
PR = r − p = 0 − 0 = 0 
    
−1 3 −4
They are not scalar multiple of each other, thus they are linearly independent.
[5 marks] (b) Use your answers in part (a) to write a vector equation of the plane.
Solution:
     
3 −3 −2
→
− →
− −→ −→  
x = p + t1 P Q + t2 P R = 0 + t1 1 + t2 0  , where t1 , t2 ∈ R
  
3 −3 −4

[5 marks] (c) Use your answers in part (a) to find a normal vector of the plane.
Solution:

       
−3 −2 1 × (−4) − 0 × (−3) −4
→
− −→ −→     
n = P Q × P R = 1 × 0 = −[(−3) × (−4) − (−3) × (−2)] = −6
 
−3 −4 (−3) × 0 − (−2) × 1 2
 
2
We may also take =  3  as a normal vector of the plane.
−1
[3 marks] (d) Use your answer of part (c) to find a scalar equation of the plane, and then simplify it
to its standard form.
Solution:
Using part (c) we get a scalar equation of the plane W :
→
− −−→
n · PX = 0

where X(x1 , x2 , x3 ) is an arbitrary point in the plane W .

It is
2(x1 − 3) + 3(x2 − 0) − (x3 − 3) = 0
or simlified as
2x1 + 3x2 − x3 = 3

[4 marks] (e) Find the area of the triangle 4P QR.

Solution:
The area of the triangle 4P QR is given by
 
−4
1 −→ −→ 1   1p 1√ √
PQ × PR = −6 = (−4)2 + (−6)2 + 22 = 56 = 14
2 2 2 2
2

Over
 MA122 – Midterm Test Solutions Page 4 of 5

4. Given a plane W in R3 that has a scalar equation x1 + 2x2 + 2x3 = 0.

[1 marks] (a) Find a normal vector →

−
n of the plane W .
 
1
→
−
Solution: The vector n = 2 is a normal vector of the plane.

2
[2 marks] (b) Is the point P (1, 1, 1) in the plane W ? Justify your answer.
Solution: The point P (1, 1, 1) is not in the plane because 1 + 2 × 1 + 2 × 1 = 5 6= 0.
−→
[4 marks] (c) Compute the orthogonal projection of OP for P (1, 1, 1) onto the normal vector →
−
n of the
plane found in part (a).
Solution:
−→ →
   
− 1 1
−→ OP · n → − 1×1+1×2+1×2  5 
proj− n OP =
→
2 n = 2 = 2
||→
−
n || 12 + 22 + 22
2
9
2

[2 marks] (d) Find the minimum distance from the point P in part (b) to the given plane W using the
projection of a vector onto the normal vector →
−
n found in part (a).
Solution: The distance from P to the plane is
−→ 5√ 2 5√ 5
n OP =
proj−
→ 1 + 22 + 2 2 = 9=
9 9 3

[3 marks] (e) Find the point R in the plane W that is closest to the point P (1, 1, 1) using the result
of part (c).
Solution:
4
 
   
1 1  9 
−→ −→ −→ −→   5    1
projW OP = perp−
→n OP = OP − proj− n OP = 1 −
→ 2 = − 
9  91 
 
1 2
−
9
 
4 1 1
Thus the point is R ,− ,− .
9 9 9
[3 marks] (f) Find an equation of the line, passing through the point R found in part (e), that is
perpendicular to the given plane W .
Solution: The normal vector of the plane W found in part (a) is a direction vector of
the line perpendicular to W . Thus the line has a vector equation
   
4 1
→
− 1 
x = −1 + t 2 , for t ∈ R

9
−1 2

OR    
1 1
→
−
x = 1 + s 2 , for s ∈ R
  
1 2
since the point P (1, 1, 1) is also on this line.

[6 marks] 5. Let →
−
u and →
−
w be vectors in Rn . Prove that

If k→
−
u −→
−
w k2 = k→
−
u k2 + k→
−
w k2 , then →
−
u and →
−
w are orthogonal (to each other).

Solution:
||→
−
u −→−
w || = (→−
u −→
−
w ) · (→
−
u −→
−
2
w)
=→ −
u ·→
−
u −→ −
u ·→
−
w −→
−
w ·→ −
u +→−
w ·→
−
w
= ||→
−
u || − 2(→
−
u ·→
−
w ) + ||→
−
2 2
u ||

If ||→
−
u −→
−
w || = ||→
−
u || + ||→
−
u || , it follows that →
−
u ·→
−
2 2 2
w = 0.
This implies that →
−u and → −
w are orthogonal.
 MA122 – Midterm Test Solutions Page 5 of 5

 
1 0 k
[5 marks] 6. Find all value(s) of the parameter k such that the matrix 2 0 4 has the rank 6= 3.
0 k 0
Solution:      
1 0 k 1 0 k 1 0 k
2 0 4 −→ 0 0 4 − k  −→ 0 k 0 =B
0 k 0 0 k 0 0 0 4 − 2k
The matrix B has a zero row if and only either k = 0 or k = 2. So rank(B) 6= 3, which implies
that rank(A) 6= 3 as A and B are row-equivalent.

7. Let V = Span(S) be the subspace of R4 spanned by the vectors in

           

 1 −1 2 −2 0 0 
           
1 −1 −1
 ,   ,   ,   ,   , 0 .
3 0

S=   2  −2  6  −2 1 7

 
−2 2 −4 4 0 0
 

[7 marks] (a) Find an ordered basis for V = Span(S) that consists of vectors from the set S.
Solution:
   
1 −1 2 −2 0 0 1 −1 2 −2 0 0
 1 −1 3 −1 0 0 −R1 +R2 →R2  0
  0 1 1 0 0

 2 −2 6 −−−−−−−−→  
−2 1 7 2 −2 6 −2 1 7
−2 2 −4 4 0 0 −2 2 −4 4 0 0
   
1 −1 2 −2 0 0 1 −1 2 −2 0 0
−2R1 +R3 →R3  0
 0 1 1 0 0 2R3 +R4 →R4 0
 0 1 1 0 0
−−−− −−−−→  − −−−−−−→  
0 0 2 2 1 7  0 0 2 2 1 7
−2 2 −4 4 0 0 0 0 0 0 0 0
 
1 −1 2 −2 0 0
−2R2 +R3 →R3  0 0 1 1 0 0
−−−− −−−−→  0

0 0 0 1 7
0 0 0 0 0 0
 
1 −1 0 −4 0 0
(−2)R2 +R1 →R1  0 0 1 1 0 0
May also proceed to −−−−−−−−−→  0

0 0 0 1 7
0 0 0 0 0 0
Thus the 1st, 3rd and 5th vectors in the set S form a basis
     

 1 2 0 
     
  ,   , 0
1 3

  2   6  1
 
−2 −4 0
 

[1 marks] (b) Find the dimension of V using your answer of part (a).
Solution: dim(V ) = 3

[6 marks] 8. Suppose that the set of vectors {→−v ,→

−
w } in Rn is linearly independent. Prove that the set of
vectors {→
−
v −→
−w,→−
v +→ −w } is also linearly independent.
Solution: Given the identity
→
−
a(→
−
v −→
−
w ) + b(→
−
v +→
−
w) = 0 .
Then
→
−
(a + b)→
−
v + (b − a)→
−
w = 0.

Because {→
−
v ,→
−
w } in Rn is linearly independent, we have
a + b = 0 and b − a = 0.

Thus a = b = 0.
Therefore {→
−
v −→ −
w,→
−
v +→
−
w } is also linearly independent by definition.

The END