Property of determinant

1) If a row or column of matrix A is zero, then det(A)=0.

2) For a square matrix A, det(A)=det(AT).
3) If two rows or columns of matrices are equal then the determinant is zero.
4) If two rows of matrix A are interchanged, the determinant changes sign.
5) If multiple of a row is added to another row, then the value of the determinant will not change.
6) If A is an nxn upper triangular, or lower triangular, or diagonal matrix, then det (A) is the product of
the elements in the main diagonal.
7) Let A be an nxn matrix and c be a scalar, then det (cA )=c n det ( A )
8) det ( A+ B)≠ det ( A)+det (B)
9) det ( AB)=det (A )det ( B)

Inverse of a Matrix and its Properties

Defn: A square matrix A is said to be invertible (non-singular) if there exists another matrix B such
that AB=BA=I .Here B is the inverse of A denoted by B= A−1.
If matrix A has no inverse we call it singular matrix .
Note that: the inverse of a matrix if it exists is unique.
Example:

[ ]
−1 −1
If A= [
−4 −2
5 5
−1
, then A =] 2
1
5
2
because

2 5

][ ] [ ] [ ][
−1 −1 −1 −1
−1
AA =
5 5 [
−4 −2 2
1
5
2
=
1 0
0 1
−1
, A A=
2
1
5 −4 −2
2 5 5
=
1 0
0 1 ][ ]
2 5 2 5

Example7: Determine the inverse of the matrix A= [−45 −25 ]

Solution:
We know that AI =A and A−1 I = A−1

¿>
[−45 −25 ][ 10 01 ] R → R + R [15 35][ 10 11] R → R −5 R
1 1 2 2 2 1
 [ ][ ] [ ]
−1 −1
1 1
[ 1 3
][
1 1
0 −10 −5 −4
R2→
−1
]R
1 3
10 2 0 1
1
2
2 R1 → R1−3 R 2
5
1 0
0 1 [ ] 2
1
5
2
2 5

[ ]
−1 −1
−1 2 5
¿> A =
1 2
2 5

Example8:

[ ]
3 1 0
Use the same method to determine the inverse of the matrix B= −1 2 2
5 0 −1
Exercise: 1. Show that the following matrix does not have an inverse (is singular matrix)

[ ]
3 3 6
C= 0 1 2
−2 0 0

[ ]
1 1
1

[ ]
3 3
1 1 −3
1 2 1
2) Show that the inverse of the matrix A= −1 2 0 is
6 3 2
1 −1 1
−1 1 1
6 3 2
Adjont method
−1 adj( A )
Given any non singular matrix A, its inverse is also given by A =
| A|
Note that:1) a matrix A is non-singular(invertible) if | A|≠ 0 , otherwise it has no inverse

2) for any 2x2 matrix A= [ ]

a b
c d
−1
then A =
1
ad−bc −c a [
d −b
]
Example9:

[ ]
1 2 3
Find the inverse of the matrix A= 1 7 1 using adjoint method.
5 3 1
Solution:

| |
1 2 3
| A|= 1 7 1 =1 ( 7−3 )−2 ( 1−5 ) +3 ( 3−35 )=4+8−96=−84
5 3 1
 [ ]
4 7 −19
And adjA = 4 −14 2
−32 7 5

[ ]
−1 −1 9

[ ]
21 12 84
adj( A ) −1 4 7 −19
−1 −1 1 −1
=> A = = 4 −14 2 =¿
| A| 84 21 6 42
−32 7 5
32 −1 −5
84 12 84
Exercise: show that the inverse of the matrix

[ ] [ ]
1 2 2 1 −3 6 −7
0 5 1 is −1 2 −1
4
−1 3 0 5 −6 5

Property of matrix inverse

If A and B are invertible matrices then
1) ( A¿¿−1)−1= A ¿ 2) ( A¿¿ T )−1=( A ¿¿−1)T ¿ ¿ 3) ( AB)−1=B−1 A−1

4) If C is invertible matrix , then AC=BC =¿ A=B

5) If A has an inverse matrix then it is unique

Solving system of linear equations

A general system of m linear equations with n-unknowns can be written as

a 11 x1 + a12 x 2+ a13 x 3 …+a 1n x n=b1

a 21 x1 + a22 x 2+ a23 x 3 …+a 2 n x n=b 2
⋮ ⋮⋮
a m 1 x 1+ am 2 x2 +a m 3 x 3 …+a mn x n=b m
Here x 1 , x 2 x 3 … , x n are unknowns a 11 , a12 ,a 13 , … , amn are the coefficients of the system and
b 1 , b2 , b3 , … ,b m are the constant terms.
The above system can be written in matrix equation form

[ ][ ] [ ]
a11 a12 … a1 n x1 b1
AX=b=¿ a21 a22 … a2 n x2 = b2
⋮⋮ ⋮ ⋮ ⋮
a m1 a m2 … a mn xn bm
 [ ] [] []
a11 a12 … a1 n x1 b1
a a … a2 n x2 b
Where A= 21 22 , X= and b= 2
⋮⋮ ⋮ ⋮ ⋮
a m1 a m2 … a mn xn bm

1. CRANER’S RULE FOR SOLVING LINEAR EQUATIONS

The unique solution to the system of equations

a 11 x1 + a12 x 2+ a13 x 3=b1
a 21 x1 + a22 x 2+ a23 x 3=b2
a 31 x1 + a32 x 2+ a33 x 3=b3

Δx Δx Δx
x 1= 1
x 2= 2
x 3= 3

Is Δ , Δ , Δ

| | | | | | | |
a11 a 12 a13 b1 a12 a 13 a11 b 1 a 13 a11 a12 b1
Where ∆= a a
21 22 23a ∆ =
, x b2 a22 a 23 ,
1
∆ x = a21 b 2 a 23 , ∆ x = a21 a22 b2
2 3

a31 a 32 a33 b3 a32 a 33 a31 b 3 a 33 a31 a32 b3

Note that: Cramer’s rule can’t apply if ∆=0

Example: use Cramer’s rule to solve the system

x 1−2 x2 + x 3=3
2 x1 + x 2−x 3=5
3 x 1−x 2+2 x 3=12

Solution: first check if ∆ ≠ 0

| |
1−21
∆= 2 1−1 =1 ( 2−1 ) +2 ( 4+ 3 ) +1 (−2−3 )=1+14−5=10
3−1 2

| |
3−2 1
∆ x = 5 1−1 =3 ( 2−1 ) +2 ( 10+12 ) +1 (−5−12 )=3+ 44−17=30
1

12−12

| |
131
∆ x = 25−1 =1 ( 10+12 )−3 ( 4 +3 ) +1 ( 24−15 )=22−21+ 9=10
2

3 12 2
 | |
1−23
∆ x = 2 1 5 =1 ( 12+ 5 ) +2 (24−15 )+ 3 (−2−3 )=17 +18−15=20
3

3−1 12

∆x 30 ∆ x 10 ∆ x 20
x 1= 1
= =3 , x 2= = =1 , x 3= = =2
2 3

∆ 10 ∆ 10 ∆ 10

Therefor the solution to the above system is x 1=3 , x2 =1, x 3=2

Note:

 If ∆ ≠ 0 and ∆ x =0 then the system has trivial solution

i

 If ∆ ≠ 0 and at least one ∆ x ≠ 0 then the system has unique solution.

i

 If ∆=0 and some ∆ x ≠ 0 then the system has no solution.

i

 If ∆=0 and ∆ x =0 then the system has many solution.

i

Exercise: use Cramer’s rule to solve the following system of equations

x−2 y+4 z=3 x−3 y−3 z=14

x−4 y+3z=−5 x−4 y+ z=2
a) x+3 y−2z=6 b) x+ y+2z=6

2. Solving system of linear equations by Gaussian’s method

Let us see the system
a 11 x1 + a12 x 2+ a13 x 3=b1
a 21 x1 + a22 x 2+ a23 x 3=b2
a 31 x1 + a32 x 2+ a33 x 3=b3

( |)
a11 a12 a13 b 1
Then the matrix a21 a 22 a23 b 2 is called augmented matrix of the above system.
a31 a32 a33 b3

This method is also called Gaussian elimination method.

To get the solution apply row operation on the augmented matrix to transform it in to upper
triangular matrix (roe echelon form).
Example: use Gaussian elimination method to solve the system
2x− y+ z=−3
x+ y−z=6
3 x− y−z=4
Solution: