Matrices & Determinants

ISHANT JAIN, MALHAR SHAH,MEET DOSHI

2 3 −5
1 2 4
6 5 0
MEANING: -
m x n real numbers arranged in m rows and n columns and enclosed by a pair of
brackets is called m x n matrix.

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

Sr.
No.
Type of Matrix Meaning Example
1. Row Matrix  A matrix consisting of a single row.
 Also called as row vector.
1 2 3
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.
 Denoted by “O”.
0 0
0 0
4. Square Matrix  Matrix having same number of rows and columns.
 It can also be written as An.
2 3
4 5
2 1 3
1 6 2
5 2 4
5. Diagonal Matrix  All elements are zero except main or principal diagonal. 2 0
0 5
3 0 0
0 4 0
0 0 1
6. Scalar Matrix  All the diagonal elements are same. 3 0
0 3
2 0 0
0 2 0
0 0 2
7. Unit Matrix  A scalar matrix in which each diagonal element is 1.
 Also called “Identity matrix”.
 Denoted by In.
 Every unit matrix is a diagonal matrix and also a scalar matrix.
1 0
0 1
1 0 0
0 1 0
0 0 1

Sr.
No.
Type of Matrix Meaning Example
8. Equal Matrix  Two matrices having the same order.
 Each element of A= Corresponding to the element of B
A=
2 3
1 0
B=
4
2 9
2 − 1 0
9. Negative Matrix  Replacing all the elements with its additive inverse.
A=
3 −1
−2 4
5 −3
B=
−3 1
2 −4
−5 3
10. Upper Triangular
Matrix
 A square matrix in which all elements below the principal
diagonal are zero.
1 3 7
0 2 8
0 0 7
11. Lower Triangular
Matrix
 A square matrix in which all elements above the principal
diagonal are zero.
1 0 0
3 6 0
2 5 7
12. Symmetric Matrix
 A square matrix having aij=aji
2 1 3
1 6 2
3 2 4
13. Skew Symmetric Matrix
 A square matrix having aij= -aji
2 1 3
−1 6 2
−3 −2 4
a12 =a21
a13 =a31
a23 =a32
a12 =-a21
a13 =-a31
a23 =-a32

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

 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
1 3
5 6
(1 x 6) –
(3 x 5) - 9

03 05 02
04 00 01
−2 06 10
3{(0 X 10) – (1 X 6)} –
5{(4 X 10) – (-2 X 1)} +
2{(4 X 6) – (-2 X 0)}
- 180

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
8 2 4
6 0 7
3 5 9
Step 1:-Ignore the row and column in which the
element a11 i.e. 8 is
Step 2: -Write the remaining elements in determinant
form
0 7
5 9
0 7
5 9
Minor of particular element is
(-35)

8 2 4
6 0 7
3 5 9
0 7
5 9
6 7
3 9
6 0
3 5
2 4
5 9
8 4
3 9
8 2
3 5
2 4
0 7
8 4
6 7
8 2
6 0
 Minor of a Matrix
−35 33 30
−2 60 34
14 32 −12
Minor
Matrix

In some cases, minor and
co-factor remains same but
it may be different or there
may be difference of minus.
Cij = (-1)i+j Mij
Cij = co-factor of aij
Mij = minor of aij
a11 a12 a13
b21 b22 b23
c31 c32 c33
C11= (-1)1+1 M11
C11 =(-1)2 M11
C11= M11
a11 a12 a13
b21 b22 b23
c31 c32 c33
C12= (-1)1+2M12
C12 =(-1)3 M12
C12= -M12

MEANING: -
The Adjoint of A is the transposed matrix of cofactors of A.
Adjoint of a Square Matrix of order 2 x 2
Step 1:- Change the position of principal element i.e.
2 & 3.
Step 2:- Change the sign of remaining two elements
i.e. -1 & -5.
2 5
1 3
3
2
3 −5
−1 2

 Adjoint of a Square Matrix of order
3 x 3
−1 2 4
1 0 2
3 1 −1
Step 1:- Find out the minor matrix of given
matrix. 0 2
1 −1
1 2
3 −1
1 0
3 1
2 4
1 −1
−1 4
3 −1
−1 2
3 1
2 4
0 2
−1 4
1 2
−1 2
1 0
Step 2:- Find out the Cofactor matrix of
Obtained matrix i.e.
+ − +
− + −
+ − +
−2 −7 1
−6 −11 −7
4 −6 −2
−2 7 1
6 −11 7
4 6 −2
Step 3:- Find out the Transpose of Obtained
matrix.
−𝟐 𝟔 𝟒
𝟕 −𝟏𝟏 𝟔
𝟏 𝟕 −𝟐









1
3
5
7
4
2
3
2 A
A
2x3












1
7
3
4
5
2
3
2
T
T
A
A
T
ji
ij a
a  For all i and j
Interchange rows and columns
Then transpose of A, denoted AT
.
The dimensions of AT are the reverse
of the dimensions of A

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

A-1 =
𝐴𝑑𝑗 (𝐴)
𝐴
A =
3 1 2
2 −3 −1
1 2 1
= 8 ≠ 𝟎
𝐴 = 3 (-3 x 1 – 2 x -1)
1 (2 x 1 – 1 x -1)
2 (2 x2 – 1 x -3)
𝐴
−1 3 7
−3 1 5
5 −7 −11
MINOR
Adj (A)
−1 3 5
−3 1 7
7 −5 −11
COFACTOR & TRANSPOSE
A-1=
1
8
−1 3 5
−3 1 7
7 −5 −11

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
Step 1:- Let A = Coefficient Matrix. A =
3 1 2
2 −3 −1
1 2 1
Step 2:- Let X= Unknown Matrix. X =
𝑥
𝑦
𝑧
Step 3:- Let B= Constant Matrix. B =
3
−3
4
X = A-1B Required Solution

A-1 =
𝐴𝑑𝑗 (𝐴)
𝐴
−1 3 7
−3 1 5
5 −7 −11
MINOR
A =
3 1 2
2 −3 −1
1 2 1
= 8 ≠ 𝟎
Adj (A)
𝐴 = 3 (-3 x 1 – 2 x -1)
1 (2 x 1 – 1 x -1)
2 (2 x2 – 1 x -3)
𝐴
−1 3 5
−3 1 7
7 −5 −11
COFACTOR & TRANSPOSE
A =
3 1 2
2 −3 −1
1 2 1
A-1=
1
8
−1 3 5
−3 1 7
7 −5 −11
𝑥
𝑦
𝑧
=
1
8
−1 3 5
−3 1 7
7 −5 −11
X
3
−3
4
As, X = A-1B
𝑥
𝑦
𝑧
=
1
2
−1
X = 1
Y = 2 Ans.
Z = -1

x - 3y = 5
2x + y = 4
=
∆
∆
x
1
=
∆
∆
y
2
SOLUTIONS ARE: -
∆ =
01 −3
02 01
= 7
∆1 =
05 −3
04 01
= 17
∆2 =
01 05
02 04
= -6
x=
17
7
y=
−6
7

x+2y+3z =1
x+3y+5z =2
2x+5y+9z =3
Step 1:- Let A = Coefficient Matrix. A =
1 2 3
1 3 5
2 5 9
Step 2:- Let X= Unknown Matrix. X =
𝑥
𝑦
𝑧
Step 3:- Let B= Constant Matrix. B =
1
2
3
Step 4:- Convert coefficient matrix into upper triangular matrix.

A =
1 2 3
1 3 5
2 5 9
𝑥
𝑦
𝑧
=
1
2
3
~
1 2 3
0 1 2
0 1 3
X=
1
1
1
R2 – R1
R3 - 2R1
R3 – R2 ~
1 2 3
0 1 2
0 0 1
X =
1
1
0
Step 5:- Convert matrix form into equation again.
1x+2y+3z=1 …….(1)
0x+1y+2z=1 …….(2)
0x+0y+1z=0 …….(3)
From 3rd Eqn
From 2nd Eqn
y+2z=1
y=1
z=0
x+2y+3z=1
x+2+0=1
x= -1
From 1st Eqn
X = -1
Y = 1 Ans.
Z = 0

1.
• If |A|≠ 0 𝑛𝑜𝑛 − 𝑠𝑖𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑎𝑡𝑟𝑖𝑥 𝑡ℎ𝑒𝑛 𝐴−1
𝑒𝑥𝑖𝑠𝑡𝑠
2.
• Let A = In∙ 𝐴 ; 𝑤ℎ𝑒𝑟𝑒 In= 𝑖𝑑𝑒𝑛𝑡𝑖𝑡𝑦 𝑚𝑎𝑡𝑟𝑖𝑥 𝑜𝑓 𝑜𝑟𝑑𝑒𝑟 ′𝑛′
3.
• Convert A=In∙A into In=BA
4.
• Convert B=𝐴−1

A =
1 0 −1
2 −1 3
1 −1 0
|A|=1
−1 3
−1 0
-0
2 3
1 0
+(-1)
2 −1
1 −1
= 1(3)-0(3)+(-1)(-1)
 |A| = 4 ≠ 0
𝐴
1 0 −1
2 −1 3
1 −1 0
=
1 0 0
0 1 0
0 0 1
A
A=In∙A
~
1 0 −1
0 −1 5
0 −1 1
=
1 0 0
−2 1 0
−1 0 1
A
𝑅2 − 2𝑅1
𝑅3 − 𝑅1
~
1 0 −1
0 −1 5
0 0 −4
=
1 0 0
−2 1 0
1 −1 1
A 𝑅3 − 𝑅2

1 0 −1
0 −1 5
0 0 1
=
1 0 0
−2 1 0
−1
4
1
4
−1
4
A −𝑅3
4
1 0 −1
0 1 −5
0 0 1
=
1 0 0
2 −1 0
−1
4
1
4
−1
4
A −R2
𝐴−1 =
3
4
1
4
−1
4
3
4
1
4
−5
4
−1
4
1
4
−1
4
1 0 0
0 1 0
0 0 1
=
3
4
1
4
−1
4
3
4
1
4
−5
4
−1
4
1
4
−1
4
A
R1 + R3
R2 + 5R3 In=BA
B=𝐴−1

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