Dr.

Ahlem Nemer 1

Course : Algebra 3 Year : 2023/2024

Chapter 1 : Determinants of matrices Department of Computer Science

Solutions

Solution 0.1 .
1)
1. A is a symmetric matrix if a = 5 and b = 6.
2. A is an upper triangular matrix if a = b = 0.
2) C is a 
diagonal matrix
 if a, e and i are not equal to 0 such that b = c = f = d = g = h = 0.
3 0 4
3) B T =  4 2 1 .
1 3 0
2 3 0 3 0 2
4)|B| = 3 −4 +1 = 31 6= 0 yes, it is possible to determine B −1
1 0 4 0 4 1

Solution 0.2 .
3 4 2 4 2 3
|A| = 5 −6 +7 = −21 6= 0. We deduce that A−1 exists.
5 2 1 2 1 5

Solution 0.3 .

2 α 0 α 0 2
|A| = α −1 +2
1 1 (α − 2) 1 (α − 2) 1
= −4(α − 2)

This leads to |A| = 0 =⇒ α = 2

Solution 0.4 .
1)     
2 1 3 x 3
 4 5 7  y  =  0 
0 1 8 z 2
.
|A| = 46.
3 1 3 2 3 3 2 1 3
0 5 7 4 0 7 4 5 0
2 1 8 83 0 2 8 −100 −50 0 1 2 24
x= = ,y = = = ,z = = .
46 46 46 46 23 46 46
Dr. Ahlem Nemer 2

2)     
3 2 9 x 4
 8 0 1  y  =  2 
7 5 4 z 1
.
|A| = 295.
4 2 9 3 4 9 3 2 4
2 0 1 8 2 1 8 0 2
1 5 4 56 7 1 4 −133 7 5 1 142
x= = ,y = = ,z = = .
295 295 295 295 295 295
Solution 0.5 .
1)     
5 1 x 5
= .
3 4 y 6
|A| = 17.  
1 4 −1
A−1 = .
17 −3 5
 
   14
1 4 −1 5
X = A−1 B = =  17
15  .
 
17 −3 5 6
17
2)     
1 2 4 x 1
 7 5 3  y  =  3 .
9 7 1 z 4
|A| = 40.  
−16 26 −14
1
A−1 =  20 −35 25  .
40
4 11 −9
6
 
 40 
−1
 15 
X=A B=  40  .

 1 
40
Solution 0.6 .
1) We have A2 = A and B = A − I. This leads to

B2 = (A − I)2
= A2 − 2A + I
= A − 2A + I
= −A + I
= −(A − I)
= −B
Dr. Ahlem Nemer 3

2) We have A2 = I and B = 3(A + I). This leads to

B2 = (3(A + I))2
= 9(A2 + 2A + I)
= 9(2I + 2A)
= 6(3(I + A))
= 6B

Solution 0.7 .
1)Take that

(A−1 )T AT = (AA−1 )T
= (I)T
= I

and that

AT (A−1 )T = (A−1 A)T

= (I)T
= I

Thus, we find
(A−1 )T AT = I, and AT (A−1 )T = I.

In the sequel, we deduce

(AT )−1 = (A−1 )T .
2).
a) From the property of the determinant of the product of two matrices, we have

|AB| = |A||B|

. We know that |A| =

6 0 and |B| =
6 0. This leads to

|AB| =
6 0

.
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b)We have |AB| =

6 0, then AB is invertible. Here, we take that

(B −1 A−1 )(AB) = B −1 (A−1 A)B

= B −1 IB
= B −1 B
= I

and that

(AB)(B −1 A−1 ) = A(BB −1 )A−1

= A(I)A−1
= AA−1
= I

Thus, we obtain

(B −1 A−1 )(AB) = I, and (AB)(B −1 A−1 ) = I

Then, we deduce that
(AB)−1 = B −1 A−1 .