03.11.

2017

M AT H − 241
Engineering M athematics I
P roblem Set − 7

Question 1. Show that  

 tr(A)
1 1 
det(A) = 
2 2

tr(A ) tr(A)
for every 2 × 2 matrix.
( Note : Trace of an n × n matrix A = [aij ]n×n is defined as the sum of the main diagonal
n
X
entries i.e. tr(A) = aii ).
i=1

Question 1-Ans. For a 2 × 2 matrix A = [aij ]2×2 ;

2
X
det(A2×2 ) = a11 a22 − a21 a12 and tr(A2×2 ) = aii = a11 + a22 .
i=1
If B = A2 then;    
 tr(A)
1 1  1
a11 + a22 1 
det(A) = 

2
= 2  b11 + b22 a11 + a22  .
 
tr(B) tr(A)

where b11 = a211 + a12 a21 and b22 = a222 + a12 a21 . Then
 
1  tr(A) 1  1
det(A) = = ((a11 + a22 )2 − (a211 + a12 a21 + a222 + a12 a21 ))
2 tr(A2 ) tr(A) 2
1
= ((a11 + a22 )2 − (a211 + 2a12 a21 + a222 ))
2
1 2
= ((a + 2a11 a22 + a222 ) − (a211 + 2a12 a21 + a222 ))
2 11
1
= (2a11 a22 − 2a12 a21 ) = a11 a22 − a12 a21 = det(A).
2
Question 2. What is the maximum number of zeros that a 3 × 3 matrix can have without
having a zero determinant? Explain your reasoning.

Question 2-Ans. - For a square matrix which has a column or row of zeros, then deter-
minant of the matrix is zero. Therefore, for a 3 × 3 matrix we need at least 3 nonzero
entries.
- For a triangular matrix, its determinant is the multiplication of its diagonal entries.
- A diagonal matrix is a triangular matrix. (Both upper and lower triangular.)
- For A = [aij ]3×3 diagonal matrix determinant
 can be written
 as:
a11 0 0

det(A3×3 ) =  0 a22 0  = a11 a22 a33 .
0 0 a33 

1
- Therefore, number of possible zeros for a non-zero 3 × 3 matrix determinant is 6.

- For a n × n matrix with non-zero determinant, number of possible zero entries is n2 − n.

Question 3. Prove that (x1 , y1 ), (x2 ,y2 ), (x3 , y3 ) are collinear points if and only if
 x1 y 1 1 
 
x2 y2 1 = 0.
 
x3 y3 1

Question 3-Ans. (⇒) If the points (x1 , y1 ), (x2 , y2 ), (x3 , y3 ) are collinear, they should be
lying on the same line. For a line equation ax + by + c = 0 all of the points should satisfy
this equation. Then we can obtain the following   linear
 system
 of equations that:
x1 y 1 1 a 0
x2 y2 1  b  = 0
x3 y 3 1 c 0
Augmented matrix of the linear system can be written as:
     
x1 y 1 1 0 x1 y1 1 0 R2 x1 y1 1 0
R2 −R1 x2 −x1 y −y
x2 y2 1 0 − −−−→ x2 − x1 y2 − y1 0 0 −− −→  1 x22 −x11 0 0
−y1
x3 y 3 1 0 − R3 −R1
−−−→ x3 − x1 y3 − y1 0 0 x3 −x1
R3
1 xy33 −x 1
0 0
−−−→
−y1
Then, xy22 −x 1
= xy33 −y
−x1
1
which is the gradient of the line. Then second and third rows are the
same resulting with zero determinant.

(⇐) If the determinant is zero, the linear system:

   
x1 y 1 1 a 0
x2 y2 1  b  = 0
x3 y 3 1 c 0
should have a non-trivial solution meaning that:
   
a 0
 b  6= 0 .
c 0
If the system has a non-trivial solution, it means that there exists a line passing through all
these points or these points are collinear.

Question 4. Prove that the equation of the line through the distinct points (a1 , b1 ) and
(a2 , b2 ) can be written as:  
x y 1

 a1 b1 1 = 0.

 a2 b2 1

Question 4-Ans. The equation of the line through the distinct points (a1 , b1 ) and (a2 , b2 )
can be written as;
x−a1
a2 −a1
= by−b 1
2 −b1
.
which can be also written as;
x−a1
a2 −a1
= by−b 1
2 −b1
⇒ (a1 − x)(b2 − b1 ) − (a2 − a1 )(b1 − y) = 0.
This line equation can be written as a 2 × 2 matrix determinant or a cofactor of a 3 × 3

2
matrix determinant defined as;  
 x y 1 
 
 a1 − x b1 − y 0 = 0.
 
a2 − a1 b2 − b1 0
By applying
 elementary row
 operations  we can obtain the following:  
 x y 1  x y 1  x y 1
     
 a1 − x b1 − y 0 = 0 R +R  a1 b 1 1 = 0 R +R a1 b1 1 = 0.
2 1 3 2

a2 − a1 b2 − b1 0
 −−−−→ a − a b − b 0 −−−−→ a b 1
2 1 2 1 2 2
which is the desired determinant of the line through the distinct points (a1 , b1 ) and (a2 , b2 ).

Question 5. In the below figure, the area of the triangle ABC can be expressed as:
area(ABC) = area(ADEC) + area(CEF B) − area(ADF B).

Use this and the fact that area of a trapezoid equals 12 the altitude times the sum of the
parallel sides to show that:  
x1 y1 1
 
area(ABC) = 21 x2 y2 1.
x3 y3 1
( Note : In the definition of this formula, the vertices are labeled such that the triangle
is traced counter-clockwise proceeding from (x1 , y1 ) to (x2 , y2 ) to (x3 , y3 ). For a clockwise
orientation, the determinant above yields the negative of the area. )

3
Question 5-Ans. Lets calculate the area(ABC) by using trapezoid area formulation:
area(ABC) = area(ADEC) + area(CEF B) − area(ADF B)
1 1 1
= (y1 + y3 )(x3 − x1 ) + (y3 + y2 )(x2 − x3 ) − (y1 + y2 )(x2 − x1 )
2 2 2
1
= (y1 x3 − y1 x1 + y3 x3 − y3 x1 + y3 x2 − y3 x3 + y2 x2 − y2 x3
2
− y 1 x2 + y 1 x1 − y 2 x 2 + y 2 x1 )
1
= (y1 x3 − y3 x1 + y3 x2 − y2 x3 − y1 x2 + y2 x1 )
2
1
= ((1)(x2 y3 − x3 y2 ) + (−1)(x1 y3 − x3 y1 ) + (1)(x1 y2 − x2 y1 ))
2 
x1 y1 1
1  
= x2 y2 1 .
2
x3 y3 1

Question 6. If A is invertible n × n matrix, prove that:

det(adj(A)) = det(An−1 ).

Question 6-Ans. If A is invertible n × n matrix, then A−1 exists and can be written as
A−1 = adj(A)
det(A)
⇒ adj(A) = det(A)A−1 .
if we multiply both sides with A, we obtain:
A.adj(A) = det(A).I.
if we take the determinant of both sides, we obtain:
det(A.adj(A)) = det(det(A).I) ⇒ det(A)det(adj(A)) = det(A)n det(I)
which will result with the following: n
det(adj(A)) = det(A)
det(A)
= det(A)n−1 = det(An−1 ).