Matrices & Determinants

This document contains every topic of Matrices and Determinants which is helpful for both college and school students:
Matrices
Types of Matrices
Operations of Matrices
Determinants
Minor of Matrix
Co-factor
Ad joint
Transpose
Inverse of matrix
Linear Equation Matrix Solution
Cramer's Rule
Gauss Jordan Elimination Method
Row Elementary Method

1. ISHANT JAIN, MALHAR SHAH,MEET DOSHI

2. 23−5
124
650
MEANING: -
m x n real numbers arranged in m rows and ncolumns and enclosed by a pair of
brackets is called m x n matrix.

3. TYPES OF MATRIX
Row Matrix
Column Matrix
Null Matrix/ Zero Matrix
Square Matrix
Diagonal Matrix
Scalar Matrix
Identity Matrix or Unit Matrix
Equal Matrix
Negative Matrix
Upper Triangular Matrix
Lower Triangular Matrix
Symmetric Matrix
Skew Symmetric Matrix

4. Sr.
No.
Typeof Matrix Meaning Example
1. Row Matrix A matrix consisting of asingle row.
Also called as row vector.
123
2. Column Matrix A matrix consisting of a single column.
Also called as column vector.
1
2
3
3. Null Matrix All the elements are zero.
Also known as zero matrix.
Denotedby “O”.
00
00
4. Square Matrix Matrix having same number of rowsand columns.
It can also be written as A
n.
23
45
213
162
524
5. DiagonalMatrix All elements are zero except main or principal diagonal.
20
05
300
040
001
6. Scalar Matrix All the diagonal elements are same.
30
03
200
020
002
7. UnitMatrix Ascalar matrix in which each diagonal element is 1.
Also called “Identity matrix”.
Denoted by I
n.
Every unit matrix is a diagonal matrix and also a scalar matrix.
10
01
100
010
001

5. Sr.
No.
Typeof Matrix Meaning Example
8. Equal Matrix Two matriceshaving the same order.
Each element of A= Corresponding to the element of B
A=
23
10
B=

4
2 9
2−10
9. Negative Matrix Replacing all the elements with its additive inverse.
A=
3−1
−24
5−3
B=
−31
2−4
−53
10. Upper Triangular
Matrix
Asquare matrix in which all elements below the principal
diagonal are zero.
137
028
007
11. LowerTriangular
Matrix
Asquare matrix in which all elements above the principal
diagonal are zero.
100
360
257
12. Symmetric Matrix
A square matrixhaving a
ij=a
ji
213
162
324
13. Skew Symmetric Matrix
A square matrixhaving a
ij= -a
ji
213
−162
−3−24
a
12=a
21
a
13=a
31
a
23=a
32
a
12=-a
21
a
13=-a
31
a
23=-a
32

6. 1.Matrix Addition
2.Matrix Subtraction
3.Scalar Multiplication
4.Matrix Multiplication

7. 46
02
32
85
78
87

8. 97
58
53
02
44
56

9. 43
69
1209
1827

10. 25
71
75
-30
-110
4635

12. Determinant is a scalar value that is calculated from a matrix.
It can only be calculated for a square matrix.
It is denoted by ∆(Delta).
Calculation For 2X2 Matrix
13
56
(1 x 6) –
(3 x 5) -9

13. 030502
040001
−20610
3{(0 X 10) –(1 X 6)} –
5{(4 X 10) –(-2 X 1)} +
2{(4 X 6) –(-2 X 0)}
-180

14. MEANING: -
The minor of a element in a matrix is the determinant obtained by deleting the row
and the column in which that element appears.
Minor of a particular element in a matrix
824
607
359
Step1:-Ignoretherowandcolumninwhichthe
elementa
11i.e.8is
Step 2: -Write the remaining elements in determinant
form
07
59
07
59
Minor of particular element is
(-35)

15. 824
607
359
07
59
67
39
60
35
24
59
84
39
82
35
24
07
84
67
82
60
Minor of a Matrix
−353330
−26034
1432−12
Minor
Matrix

16. In some cases, minor and
co-factor remains same but
it may be different or there
may be difference of minus.
C
ij=(-1)
i+j
M
ij
C
ij= co-factor of a
ij
M
ij= minor of a
ij
a
11a
12a
13
b
21b
22b
23
c
31c
32c
33
C
11= (-1)
1+1
M
11
C
11 =(-1)
2
M
11
C
11= M
11
a
11a
12a
13
b
21b
22b
23
c
31c
32c
33
C
12= (-1)
1+2
M
12
C
12 =(-1)
3
M
12
C
12= -M
12

17. MEANING: -
The Adjointof A is the transposed matrix of cofactors of A.
Adjointof a Square Matrix of order 2 x 2
Step1:-Changethepositionofprincipalelementi.e.
2&3.
Step2:-Changethesignofremainingtwoelements
i.e.-1&-5.
25
13
3
2
3−5
−12

18. Adjointof a Square Matrix of order
3 x 3
−124
102
31−1
Step1:-Findouttheminormatrixofgiven
matrix. 02
1−1
12
3−1
10
31
24
1−1
−14
3−1
−12
31
24
02
−14
12
−12
10
Step2:-FindouttheCofactormatrixof
Obtainedmatrixi.e.
+−+
−+−
+−+
−2−71
−6−11−7
4−6−2
−27 1
6−117
4 6−2
Step3:-FindouttheTransposeofObtained
matrix.
−��??????
�−���
��−�

19. 






135
742
3
2AA 2x3










17
34
52
3
2
T
T
AA T
jiijaa
For all iand j
Interchange rows and columns
Then transpose of A, denoted A
T
.
The dimensions of A
T
are the reverse
of the dimensions of A

20. MEANING: -
Consider a scalar k. The inverse is the reciprocal or division of 1 by the scalar.
Example: k=7the inverse of k or k
-1
= 1/k = 1/7

21. A
-1
=
??????????????????(??????)
??????
A =
312
2−3−1
121
= 8 ≠�
??????= 3 (-3 x 1 –2 x -1)
1 (2 x 1 –1 x -1)
2 (2 x2 –1 x -3)
??????
−13 7
−31 5
5−7−11
MINOR
Adj (A)
−13 5
−31 7
7−5−11
COFACTOR & TRANSPOSE
A
-1
=
1
8
−13 5
−31 7
7−5−11

23. MEANING: -
It is Solution by the method of inversion of the coefficient matrix.
3x + y + 2z = 3
2x +3y –z = -3
x + 2y + z = 4
Step1:-LetA=CoefficientMatrix. A =
312
2−3−1
121
Step2:-LetX=UnknownMatrix. X =
�
�
�
Step3:-LetB=ConstantMatrix. B =
3
−3
4
X = A
-1
B Required Solution

24. A
-1
=
??????????????????(??????)
??????
−13 7
−31 5
5−7−11
MINOR
A =
312
2−3−1
121
= 8 ≠�
Adj (A)
??????= 3 (-3 x 1 –2 x -1)
1 (2 x 1 –1 x -1)
2 (2 x2 –1 x -3)
??????
−13 5
−31 7
7−5−11
COFACTOR & TRANSPOSE
A =
312
2−3−1
121A
-1
=
1
8
−13 5
−31 7
7−5−11
�
�
�
=
1
8
−13 5
−31 7
7−5−11
X
3
−3
4
As, X = A
-1
B
�
�
�
=
1
2
−1
X = 1
Y = 2 Ans.
Z = -1

25. x -3y = 5
2x + y = 4
=
∆
∆
x
1
=
∆
∆
y
2
SOLUTIONS ARE: -
∆=
01−3
0201
= 7
∆
1=
05−3
0401
= 17
∆
2=
0105
0204
= -6
x=
17
7
y=
−6
7

26. x+2y+3z =1
x+3y+5z =2
2x+5y+9z =3
Step1:-LetA=CoefficientMatrix. A =
123
135
259
Step2:-LetX=UnknownMatrix. X =
�
�
�
Step3:-LetB=ConstantMatrix. B =
1
2
3
Step4:-Convertcoefficientmatrixintouppertriangularmatrix.

27. A =
123
135
259
�
�
�
=
1
2
3
~
123
012
013
X=
1
1
1
R
2–R
1
R
3-2R
1
R
3–R
2 ~
123
012
001
X =
1
1
0
Step5:-Convertmatrixformintoequationagain.
1x+2y+3z=1 …….(1)
0x+1y+2z=1 …….(2)
0x+0y+1z=0 …….(3)
From 3
rd
Eq
n
From 2
nd
Eq
n
y+2z=1
y=1
z=0
x+2y+3z=1
x+2+0=1
x= -1
From 1
st
Eq
n
X = -1
Y = 1 Ans.
Z = 0

28. 1.
•If |A|≠0���−�??????����??????��??????��??????��ℎ��??????
−1
��??????���
2.
•Let A = I
n∙??????;�ℎ���I
n=??????����??????���??????��??????��������′�′
3.
•Convert A=I
n∙A into I
n=BA
4.
•Convert B=??????
−1

29. A =
10−1
2−13
1−10
|A|=1
−13
−10
-0
23
10
+(-1)
2−1
1−1
= 1(3)-0(3)+(-1)(-1)
|A| = 4 ≠0
??????
10−1
2−13
1−10
=
100
010
001
A
A=I
n∙A
~
10−1
0−15
0−11
=
100
−210
−101
A
??????
2−2??????
1
??????
3−??????
1
~
10−1
0−15
00−4
=
100
−210
1−11
A??????
3−??????
2

30. 10−1
0−15
001
=
1 0 0
−21 0

−1
4
1
4
−1
4
A
−??????
3
4
10−1
01−5
001
=
1 0 0
2−10

−1
4
1
4
−1
4
A
−R
2
??????
−1
=

3
4
1
4
−1
4

3
4
1
4

−5
4

−1
4
1
4
−1
4
100
010
001
=

3
4
1
4
−1
4

3
4
1
4

−5
4

−1
4
1
4
−1
4
A
R
1+R
3
R2 + 5R3
I
n=BA
B=??????
−1