Chapter
stringclasses 18
values | sentence_range
stringlengths 3
9
| Text
stringlengths 7
7.34k
|
|---|---|---|
1 | 1518-1521 | Definition 1 Minor of an element aij of a determinant is the determinant obtained by
deleting its ith row and jth column in which element aij lies Minor of an element aij is
denoted by Mij Remark Minor of an element of a determinant of order n(n ≥ 2) is a determinant of
order n – 1 Example 8 Find the minor of element 6 in the determinant
1
2
3
4
5
6
7
8
9
∆ =
Solution Since 6 lies in the second row and third column, its minor M23 is given by
M23 =
1
2
7
8 = 8 – 14 = – 6 (obtained by deleting R2 and C3 in ∆) |
1 | 1519-1522 | Minor of an element aij is
denoted by Mij Remark Minor of an element of a determinant of order n(n ≥ 2) is a determinant of
order n – 1 Example 8 Find the minor of element 6 in the determinant
1
2
3
4
5
6
7
8
9
∆ =
Solution Since 6 lies in the second row and third column, its minor M23 is given by
M23 =
1
2
7
8 = 8 – 14 = – 6 (obtained by deleting R2 and C3 in ∆) Definition 2 Cofactor of an element aij, denoted by Aij is defined by
Aij = (–1)i + j Mij, where Mij is minor of aij |
1 | 1520-1523 | Remark Minor of an element of a determinant of order n(n ≥ 2) is a determinant of
order n – 1 Example 8 Find the minor of element 6 in the determinant
1
2
3
4
5
6
7
8
9
∆ =
Solution Since 6 lies in the second row and third column, its minor M23 is given by
M23 =
1
2
7
8 = 8 – 14 = – 6 (obtained by deleting R2 and C3 in ∆) Definition 2 Cofactor of an element aij, denoted by Aij is defined by
Aij = (–1)i + j Mij, where Mij is minor of aij Example 9 Find minors and cofactors of all the elements of the determinant
1
–2
4
3
Solution Minor of the element aij is Mij
Here a11 = 1 |
1 | 1521-1524 | Example 8 Find the minor of element 6 in the determinant
1
2
3
4
5
6
7
8
9
∆ =
Solution Since 6 lies in the second row and third column, its minor M23 is given by
M23 =
1
2
7
8 = 8 – 14 = – 6 (obtained by deleting R2 and C3 in ∆) Definition 2 Cofactor of an element aij, denoted by Aij is defined by
Aij = (–1)i + j Mij, where Mij is minor of aij Example 9 Find minors and cofactors of all the elements of the determinant
1
–2
4
3
Solution Minor of the element aij is Mij
Here a11 = 1 So M11 = Minor of a11= 3
M12
= Minor of the element a12 = 4
M21 = Minor of the element a21 = –2
M22 = Minor of the element a22 = 1
Now, cofactor of aij is Aij |
1 | 1522-1525 | Definition 2 Cofactor of an element aij, denoted by Aij is defined by
Aij = (–1)i + j Mij, where Mij is minor of aij Example 9 Find minors and cofactors of all the elements of the determinant
1
–2
4
3
Solution Minor of the element aij is Mij
Here a11 = 1 So M11 = Minor of a11= 3
M12
= Minor of the element a12 = 4
M21 = Minor of the element a21 = –2
M22 = Minor of the element a22 = 1
Now, cofactor of aij is Aij So
A11 = (–1)1 + 1 M11 = (–1)2 (3) = 3
A12 = (–1)1 + 2 M12 = (–1)3 (4) = – 4
A21 = (–1)2 + 1 M21 = (–1)3 (–2) = 2
A22 = (–1)2 + 2 M22 = (–1)4 (1) = 1
Rationalised 2023-24
DETERMINANTS 85
Example 10 Find minors and cofactors of the elements a11, a21 in the determinant
∆ =
11
12
13
21
22
23
31
32
33
a
a
a
a
a
a
a
a
a
Solution By definition of minors and cofactors, we have
Minor of a11 = M11 =
22
23
32
33
a
a
a
a
= a22 a33– a23 a32
Cofactor of a11 = A11 = (–1)1+1 M11 = a22 a33 – a23 a32
Minor of a21 = M21 =
12
13
32
33
a
a
a
a
= a12 a33 – a13 a32
Cofactor of a21 = A21 = (–1)2+1 M21 = (–1) (a12 a33 – a13 a32) = – a12 a33 + a13 a32
Remark Expanding the determinant ∆, in Example 21, along R1, we have
∆ = (–1)1+1 a11
22
23
32
33
a
a
a
a
+ (–1)1+2 a12
21
23
31
33
a
a
a
a
+ (–1)1+3 a13
21
22
31
32
a
a
a
a
= a11 A11 + a12 A12 + a13 A13, where Aij is cofactor of aij
= sum of product of elements of R1 with their corresponding cofactors
Similarly, ∆ can be calculated by other five ways of expansion that is along R2, R3,
C1, C2 and C3 |
1 | 1523-1526 | Example 9 Find minors and cofactors of all the elements of the determinant
1
–2
4
3
Solution Minor of the element aij is Mij
Here a11 = 1 So M11 = Minor of a11= 3
M12
= Minor of the element a12 = 4
M21 = Minor of the element a21 = –2
M22 = Minor of the element a22 = 1
Now, cofactor of aij is Aij So
A11 = (–1)1 + 1 M11 = (–1)2 (3) = 3
A12 = (–1)1 + 2 M12 = (–1)3 (4) = – 4
A21 = (–1)2 + 1 M21 = (–1)3 (–2) = 2
A22 = (–1)2 + 2 M22 = (–1)4 (1) = 1
Rationalised 2023-24
DETERMINANTS 85
Example 10 Find minors and cofactors of the elements a11, a21 in the determinant
∆ =
11
12
13
21
22
23
31
32
33
a
a
a
a
a
a
a
a
a
Solution By definition of minors and cofactors, we have
Minor of a11 = M11 =
22
23
32
33
a
a
a
a
= a22 a33– a23 a32
Cofactor of a11 = A11 = (–1)1+1 M11 = a22 a33 – a23 a32
Minor of a21 = M21 =
12
13
32
33
a
a
a
a
= a12 a33 – a13 a32
Cofactor of a21 = A21 = (–1)2+1 M21 = (–1) (a12 a33 – a13 a32) = – a12 a33 + a13 a32
Remark Expanding the determinant ∆, in Example 21, along R1, we have
∆ = (–1)1+1 a11
22
23
32
33
a
a
a
a
+ (–1)1+2 a12
21
23
31
33
a
a
a
a
+ (–1)1+3 a13
21
22
31
32
a
a
a
a
= a11 A11 + a12 A12 + a13 A13, where Aij is cofactor of aij
= sum of product of elements of R1 with their corresponding cofactors
Similarly, ∆ can be calculated by other five ways of expansion that is along R2, R3,
C1, C2 and C3 Hence ∆ = sum of the product of elements of any row (or column) with their
corresponding cofactors |
1 | 1524-1527 | So M11 = Minor of a11= 3
M12
= Minor of the element a12 = 4
M21 = Minor of the element a21 = –2
M22 = Minor of the element a22 = 1
Now, cofactor of aij is Aij So
A11 = (–1)1 + 1 M11 = (–1)2 (3) = 3
A12 = (–1)1 + 2 M12 = (–1)3 (4) = – 4
A21 = (–1)2 + 1 M21 = (–1)3 (–2) = 2
A22 = (–1)2 + 2 M22 = (–1)4 (1) = 1
Rationalised 2023-24
DETERMINANTS 85
Example 10 Find minors and cofactors of the elements a11, a21 in the determinant
∆ =
11
12
13
21
22
23
31
32
33
a
a
a
a
a
a
a
a
a
Solution By definition of minors and cofactors, we have
Minor of a11 = M11 =
22
23
32
33
a
a
a
a
= a22 a33– a23 a32
Cofactor of a11 = A11 = (–1)1+1 M11 = a22 a33 – a23 a32
Minor of a21 = M21 =
12
13
32
33
a
a
a
a
= a12 a33 – a13 a32
Cofactor of a21 = A21 = (–1)2+1 M21 = (–1) (a12 a33 – a13 a32) = – a12 a33 + a13 a32
Remark Expanding the determinant ∆, in Example 21, along R1, we have
∆ = (–1)1+1 a11
22
23
32
33
a
a
a
a
+ (–1)1+2 a12
21
23
31
33
a
a
a
a
+ (–1)1+3 a13
21
22
31
32
a
a
a
a
= a11 A11 + a12 A12 + a13 A13, where Aij is cofactor of aij
= sum of product of elements of R1 with their corresponding cofactors
Similarly, ∆ can be calculated by other five ways of expansion that is along R2, R3,
C1, C2 and C3 Hence ∆ = sum of the product of elements of any row (or column) with their
corresponding cofactors ANote If elements of a row (or column) are multiplied with cofactors of any
other row (or column), then their sum is zero |
1 | 1525-1528 | So
A11 = (–1)1 + 1 M11 = (–1)2 (3) = 3
A12 = (–1)1 + 2 M12 = (–1)3 (4) = – 4
A21 = (–1)2 + 1 M21 = (–1)3 (–2) = 2
A22 = (–1)2 + 2 M22 = (–1)4 (1) = 1
Rationalised 2023-24
DETERMINANTS 85
Example 10 Find minors and cofactors of the elements a11, a21 in the determinant
∆ =
11
12
13
21
22
23
31
32
33
a
a
a
a
a
a
a
a
a
Solution By definition of minors and cofactors, we have
Minor of a11 = M11 =
22
23
32
33
a
a
a
a
= a22 a33– a23 a32
Cofactor of a11 = A11 = (–1)1+1 M11 = a22 a33 – a23 a32
Minor of a21 = M21 =
12
13
32
33
a
a
a
a
= a12 a33 – a13 a32
Cofactor of a21 = A21 = (–1)2+1 M21 = (–1) (a12 a33 – a13 a32) = – a12 a33 + a13 a32
Remark Expanding the determinant ∆, in Example 21, along R1, we have
∆ = (–1)1+1 a11
22
23
32
33
a
a
a
a
+ (–1)1+2 a12
21
23
31
33
a
a
a
a
+ (–1)1+3 a13
21
22
31
32
a
a
a
a
= a11 A11 + a12 A12 + a13 A13, where Aij is cofactor of aij
= sum of product of elements of R1 with their corresponding cofactors
Similarly, ∆ can be calculated by other five ways of expansion that is along R2, R3,
C1, C2 and C3 Hence ∆ = sum of the product of elements of any row (or column) with their
corresponding cofactors ANote If elements of a row (or column) are multiplied with cofactors of any
other row (or column), then their sum is zero For example,
∆ = a11 A21 + a12 A22 + a13 A23
= a11 (–1)1+1
12
13
32
33
a
a
a
a
+ a12 (–1)1+2
11
13
31
33
a
a
a
a
+ a13 (–1)1+3
11
12
31
32
a
a
a
a
=
11
12
13
11
12
13
31
32
33
a
a
a
a
a
a
a
a
a
= 0 (since R1 and R2 are identical)
Similarly, we can try for other rows and columns |
1 | 1526-1529 | Hence ∆ = sum of the product of elements of any row (or column) with their
corresponding cofactors ANote If elements of a row (or column) are multiplied with cofactors of any
other row (or column), then their sum is zero For example,
∆ = a11 A21 + a12 A22 + a13 A23
= a11 (–1)1+1
12
13
32
33
a
a
a
a
+ a12 (–1)1+2
11
13
31
33
a
a
a
a
+ a13 (–1)1+3
11
12
31
32
a
a
a
a
=
11
12
13
11
12
13
31
32
33
a
a
a
a
a
a
a
a
a
= 0 (since R1 and R2 are identical)
Similarly, we can try for other rows and columns Rationalised 2023-24
86
MATHEMATICS
Example 11 Find minors and cofactors of the elements of the determinant
2
3
5
6
0
4
1
5
7
–
–
and verify that a11 A31 + a12 A32 + a13 A33= 0
Solution We have M11 =
0
4
5
–7
= 0 –20 = –20; A11 = (–1)1+1 (–20) = –20
M12 =
6
4
1
–7
= – 42 – 4 = – 46;
A12 = (–1)1+2 (– 46) = 46
M13 =
6
0
1
5 = 30 – 0 = 30;
A13 = (–1)1+3 (30) = 30
M21 =
3
5
5
7
–
– = 21 – 25 = – 4;
A21 = (–1)2+1 (– 4) = 4
M22 =
2
5
1
–7
= –14 – 5 = –19;
A22 = (–1)2+2 (–19) = –19
M23 =
2
3
1
5
–
= 10 + 3 = 13;
A23 = (–1)2+3 (13) = –13
M31 =
3
5
0
4
–
= –12 – 0 = –12;
A31 = (–1)3+1 (–12) = –12
M32 =
2
5
6
4 = 8 – 30 = –22;
A32 = (–1)3+2 (–22) = 22
and
M33 =
2
3
6
0
–
= 0 + 18 = 18;
A33 = (–1)3+3 (18) = 18
Now
a11 = 2, a12 = –3, a13 = 5; A31 = –12, A32 = 22, A33 = 18
So
a11 A31 + a12 A32 + a13 A33
= 2 (–12) + (–3) (22) + 5 (18) = –24 – 66 + 90 = 0
Rationalised 2023-24
DETERMINANTS 87
EXERCISE 4 |
1 | 1527-1530 | ANote If elements of a row (or column) are multiplied with cofactors of any
other row (or column), then their sum is zero For example,
∆ = a11 A21 + a12 A22 + a13 A23
= a11 (–1)1+1
12
13
32
33
a
a
a
a
+ a12 (–1)1+2
11
13
31
33
a
a
a
a
+ a13 (–1)1+3
11
12
31
32
a
a
a
a
=
11
12
13
11
12
13
31
32
33
a
a
a
a
a
a
a
a
a
= 0 (since R1 and R2 are identical)
Similarly, we can try for other rows and columns Rationalised 2023-24
86
MATHEMATICS
Example 11 Find minors and cofactors of the elements of the determinant
2
3
5
6
0
4
1
5
7
–
–
and verify that a11 A31 + a12 A32 + a13 A33= 0
Solution We have M11 =
0
4
5
–7
= 0 –20 = –20; A11 = (–1)1+1 (–20) = –20
M12 =
6
4
1
–7
= – 42 – 4 = – 46;
A12 = (–1)1+2 (– 46) = 46
M13 =
6
0
1
5 = 30 – 0 = 30;
A13 = (–1)1+3 (30) = 30
M21 =
3
5
5
7
–
– = 21 – 25 = – 4;
A21 = (–1)2+1 (– 4) = 4
M22 =
2
5
1
–7
= –14 – 5 = –19;
A22 = (–1)2+2 (–19) = –19
M23 =
2
3
1
5
–
= 10 + 3 = 13;
A23 = (–1)2+3 (13) = –13
M31 =
3
5
0
4
–
= –12 – 0 = –12;
A31 = (–1)3+1 (–12) = –12
M32 =
2
5
6
4 = 8 – 30 = –22;
A32 = (–1)3+2 (–22) = 22
and
M33 =
2
3
6
0
–
= 0 + 18 = 18;
A33 = (–1)3+3 (18) = 18
Now
a11 = 2, a12 = –3, a13 = 5; A31 = –12, A32 = 22, A33 = 18
So
a11 A31 + a12 A32 + a13 A33
= 2 (–12) + (–3) (22) + 5 (18) = –24 – 66 + 90 = 0
Rationalised 2023-24
DETERMINANTS 87
EXERCISE 4 3
Write Minors and Cofactors of the elements of following determinants:
1 |
1 | 1528-1531 | For example,
∆ = a11 A21 + a12 A22 + a13 A23
= a11 (–1)1+1
12
13
32
33
a
a
a
a
+ a12 (–1)1+2
11
13
31
33
a
a
a
a
+ a13 (–1)1+3
11
12
31
32
a
a
a
a
=
11
12
13
11
12
13
31
32
33
a
a
a
a
a
a
a
a
a
= 0 (since R1 and R2 are identical)
Similarly, we can try for other rows and columns Rationalised 2023-24
86
MATHEMATICS
Example 11 Find minors and cofactors of the elements of the determinant
2
3
5
6
0
4
1
5
7
–
–
and verify that a11 A31 + a12 A32 + a13 A33= 0
Solution We have M11 =
0
4
5
–7
= 0 –20 = –20; A11 = (–1)1+1 (–20) = –20
M12 =
6
4
1
–7
= – 42 – 4 = – 46;
A12 = (–1)1+2 (– 46) = 46
M13 =
6
0
1
5 = 30 – 0 = 30;
A13 = (–1)1+3 (30) = 30
M21 =
3
5
5
7
–
– = 21 – 25 = – 4;
A21 = (–1)2+1 (– 4) = 4
M22 =
2
5
1
–7
= –14 – 5 = –19;
A22 = (–1)2+2 (–19) = –19
M23 =
2
3
1
5
–
= 10 + 3 = 13;
A23 = (–1)2+3 (13) = –13
M31 =
3
5
0
4
–
= –12 – 0 = –12;
A31 = (–1)3+1 (–12) = –12
M32 =
2
5
6
4 = 8 – 30 = –22;
A32 = (–1)3+2 (–22) = 22
and
M33 =
2
3
6
0
–
= 0 + 18 = 18;
A33 = (–1)3+3 (18) = 18
Now
a11 = 2, a12 = –3, a13 = 5; A31 = –12, A32 = 22, A33 = 18
So
a11 A31 + a12 A32 + a13 A33
= 2 (–12) + (–3) (22) + 5 (18) = –24 – 66 + 90 = 0
Rationalised 2023-24
DETERMINANTS 87
EXERCISE 4 3
Write Minors and Cofactors of the elements of following determinants:
1 (i) 2
4
0
3
–
(ii)
a
c
b
d
2 |
1 | 1529-1532 | Rationalised 2023-24
86
MATHEMATICS
Example 11 Find minors and cofactors of the elements of the determinant
2
3
5
6
0
4
1
5
7
–
–
and verify that a11 A31 + a12 A32 + a13 A33= 0
Solution We have M11 =
0
4
5
–7
= 0 –20 = –20; A11 = (–1)1+1 (–20) = –20
M12 =
6
4
1
–7
= – 42 – 4 = – 46;
A12 = (–1)1+2 (– 46) = 46
M13 =
6
0
1
5 = 30 – 0 = 30;
A13 = (–1)1+3 (30) = 30
M21 =
3
5
5
7
–
– = 21 – 25 = – 4;
A21 = (–1)2+1 (– 4) = 4
M22 =
2
5
1
–7
= –14 – 5 = –19;
A22 = (–1)2+2 (–19) = –19
M23 =
2
3
1
5
–
= 10 + 3 = 13;
A23 = (–1)2+3 (13) = –13
M31 =
3
5
0
4
–
= –12 – 0 = –12;
A31 = (–1)3+1 (–12) = –12
M32 =
2
5
6
4 = 8 – 30 = –22;
A32 = (–1)3+2 (–22) = 22
and
M33 =
2
3
6
0
–
= 0 + 18 = 18;
A33 = (–1)3+3 (18) = 18
Now
a11 = 2, a12 = –3, a13 = 5; A31 = –12, A32 = 22, A33 = 18
So
a11 A31 + a12 A32 + a13 A33
= 2 (–12) + (–3) (22) + 5 (18) = –24 – 66 + 90 = 0
Rationalised 2023-24
DETERMINANTS 87
EXERCISE 4 3
Write Minors and Cofactors of the elements of following determinants:
1 (i) 2
4
0
3
–
(ii)
a
c
b
d
2 (i)
1
0
0
0
1
0
0
0
1
(ii)
1
0
4
3
5
1
0
1
2
–
3 |
1 | 1530-1533 | 3
Write Minors and Cofactors of the elements of following determinants:
1 (i) 2
4
0
3
–
(ii)
a
c
b
d
2 (i)
1
0
0
0
1
0
0
0
1
(ii)
1
0
4
3
5
1
0
1
2
–
3 Using Cofactors of elements of second row, evaluate ∆ =
5
3
8
2
0
1
1
2
3 |
1 | 1531-1534 | (i) 2
4
0
3
–
(ii)
a
c
b
d
2 (i)
1
0
0
0
1
0
0
0
1
(ii)
1
0
4
3
5
1
0
1
2
–
3 Using Cofactors of elements of second row, evaluate ∆ =
5
3
8
2
0
1
1
2
3 4 |
1 | 1532-1535 | (i)
1
0
0
0
1
0
0
0
1
(ii)
1
0
4
3
5
1
0
1
2
–
3 Using Cofactors of elements of second row, evaluate ∆ =
5
3
8
2
0
1
1
2
3 4 Using Cofactors of elements of third column, evaluate ∆ =
1
1
1
x
yz
y
zx
z
xy |
1 | 1533-1536 | Using Cofactors of elements of second row, evaluate ∆ =
5
3
8
2
0
1
1
2
3 4 Using Cofactors of elements of third column, evaluate ∆ =
1
1
1
x
yz
y
zx
z
xy 5 |
1 | 1534-1537 | 4 Using Cofactors of elements of third column, evaluate ∆ =
1
1
1
x
yz
y
zx
z
xy 5 If ∆ =
11
12
13
21
22
23
31
32
33
a
a
a
a
a
a
a
a
a
and Aij is Cofactors of aij, then value of ∆ is given by
(A)
a11 A31+ a12 A32 + a13 A33
(B)
a11 A11+ a12 A21 + a13 A31
(C)
a21 A11+ a22 A12 + a23 A13
(D)
a11 A11+ a21 A21 + a31 A31
4 |
1 | 1535-1538 | Using Cofactors of elements of third column, evaluate ∆ =
1
1
1
x
yz
y
zx
z
xy 5 If ∆ =
11
12
13
21
22
23
31
32
33
a
a
a
a
a
a
a
a
a
and Aij is Cofactors of aij, then value of ∆ is given by
(A)
a11 A31+ a12 A32 + a13 A33
(B)
a11 A11+ a12 A21 + a13 A31
(C)
a21 A11+ a22 A12 + a23 A13
(D)
a11 A11+ a21 A21 + a31 A31
4 5 Adjoint and Inverse of a Matrix
In the previous chapter, we have studied inverse of a matrix |
1 | 1536-1539 | 5 If ∆ =
11
12
13
21
22
23
31
32
33
a
a
a
a
a
a
a
a
a
and Aij is Cofactors of aij, then value of ∆ is given by
(A)
a11 A31+ a12 A32 + a13 A33
(B)
a11 A11+ a12 A21 + a13 A31
(C)
a21 A11+ a22 A12 + a23 A13
(D)
a11 A11+ a21 A21 + a31 A31
4 5 Adjoint and Inverse of a Matrix
In the previous chapter, we have studied inverse of a matrix In this section, we shall
discuss the condition for existence of inverse of a matrix |
1 | 1537-1540 | If ∆ =
11
12
13
21
22
23
31
32
33
a
a
a
a
a
a
a
a
a
and Aij is Cofactors of aij, then value of ∆ is given by
(A)
a11 A31+ a12 A32 + a13 A33
(B)
a11 A11+ a12 A21 + a13 A31
(C)
a21 A11+ a22 A12 + a23 A13
(D)
a11 A11+ a21 A21 + a31 A31
4 5 Adjoint and Inverse of a Matrix
In the previous chapter, we have studied inverse of a matrix In this section, we shall
discuss the condition for existence of inverse of a matrix To find inverse of a matrix A, i |
1 | 1538-1541 | 5 Adjoint and Inverse of a Matrix
In the previous chapter, we have studied inverse of a matrix In this section, we shall
discuss the condition for existence of inverse of a matrix To find inverse of a matrix A, i e |
1 | 1539-1542 | In this section, we shall
discuss the condition for existence of inverse of a matrix To find inverse of a matrix A, i e , A–1 we shall first define adjoint of a matrix |
1 | 1540-1543 | To find inverse of a matrix A, i e , A–1 we shall first define adjoint of a matrix 4 |
1 | 1541-1544 | e , A–1 we shall first define adjoint of a matrix 4 5 |
1 | 1542-1545 | , A–1 we shall first define adjoint of a matrix 4 5 1 Adjoint of a matrix
Definition 3 The adjoint of a square matrix A = [aij]n × n is defined as the transpose of
the matrix [Aij]n × n, where Aij is the cofactor of the element aij |
1 | 1543-1546 | 4 5 1 Adjoint of a matrix
Definition 3 The adjoint of a square matrix A = [aij]n × n is defined as the transpose of
the matrix [Aij]n × n, where Aij is the cofactor of the element aij Adjoint of the matrix A
is denoted by adj A |
1 | 1544-1547 | 5 1 Adjoint of a matrix
Definition 3 The adjoint of a square matrix A = [aij]n × n is defined as the transpose of
the matrix [Aij]n × n, where Aij is the cofactor of the element aij Adjoint of the matrix A
is denoted by adj A Let
11
12
13
21
22
23
31
32
33
A =
a
a
a
a
a
a
a
a
a
Rationalised 2023-24
88
MATHEMATICS
Then
11
12
13
21
22
23
31
32
33
A
A
A
A =Transposeof A
A
A
A
A
A
adj
11
21
31
12
22
32
13
23
33
A
A
A
= A
A
A
A
A
A
Example 12
2
3
Find
A for A = 1
4
adj
Solution We have A11 = 4, A12 = –1, A21 = –3, A22 = 2
Hence
adj A =
11
21
12
22
A
A
4
–3
=
A
A
–1
2
Remark For a square matrix of order 2, given by
A =
11
12
21
22
a
a
a
a
The adj A can also be obtained by interchanging a11 and a22 and by changing signs
of a12 and a21, i |
1 | 1545-1548 | 1 Adjoint of a matrix
Definition 3 The adjoint of a square matrix A = [aij]n × n is defined as the transpose of
the matrix [Aij]n × n, where Aij is the cofactor of the element aij Adjoint of the matrix A
is denoted by adj A Let
11
12
13
21
22
23
31
32
33
A =
a
a
a
a
a
a
a
a
a
Rationalised 2023-24
88
MATHEMATICS
Then
11
12
13
21
22
23
31
32
33
A
A
A
A =Transposeof A
A
A
A
A
A
adj
11
21
31
12
22
32
13
23
33
A
A
A
= A
A
A
A
A
A
Example 12
2
3
Find
A for A = 1
4
adj
Solution We have A11 = 4, A12 = –1, A21 = –3, A22 = 2
Hence
adj A =
11
21
12
22
A
A
4
–3
=
A
A
–1
2
Remark For a square matrix of order 2, given by
A =
11
12
21
22
a
a
a
a
The adj A can also be obtained by interchanging a11 and a22 and by changing signs
of a12 and a21, i e |
1 | 1546-1549 | Adjoint of the matrix A
is denoted by adj A Let
11
12
13
21
22
23
31
32
33
A =
a
a
a
a
a
a
a
a
a
Rationalised 2023-24
88
MATHEMATICS
Then
11
12
13
21
22
23
31
32
33
A
A
A
A =Transposeof A
A
A
A
A
A
adj
11
21
31
12
22
32
13
23
33
A
A
A
= A
A
A
A
A
A
Example 12
2
3
Find
A for A = 1
4
adj
Solution We have A11 = 4, A12 = –1, A21 = –3, A22 = 2
Hence
adj A =
11
21
12
22
A
A
4
–3
=
A
A
–1
2
Remark For a square matrix of order 2, given by
A =
11
12
21
22
a
a
a
a
The adj A can also be obtained by interchanging a11 and a22 and by changing signs
of a12 and a21, i e ,
We state the following theorem without proof |
1 | 1547-1550 | Let
11
12
13
21
22
23
31
32
33
A =
a
a
a
a
a
a
a
a
a
Rationalised 2023-24
88
MATHEMATICS
Then
11
12
13
21
22
23
31
32
33
A
A
A
A =Transposeof A
A
A
A
A
A
adj
11
21
31
12
22
32
13
23
33
A
A
A
= A
A
A
A
A
A
Example 12
2
3
Find
A for A = 1
4
adj
Solution We have A11 = 4, A12 = –1, A21 = –3, A22 = 2
Hence
adj A =
11
21
12
22
A
A
4
–3
=
A
A
–1
2
Remark For a square matrix of order 2, given by
A =
11
12
21
22
a
a
a
a
The adj A can also be obtained by interchanging a11 and a22 and by changing signs
of a12 and a21, i e ,
We state the following theorem without proof Theorem 1 If A be any given square matrix of order n, then
A(adj A) = (adj A) A = A I ,
where I is the identity matrix of order n
Verification
Let
A =
11
12
13
21
22
23
31
32
33
a
a
a
a
a
a
a
a
a
, then adj A =
11
21
31
12
22
32
13
23
33
A
A
A
A
A
A
A
A
A
Since sum of product of elements of a row (or a column) with corresponding
cofactors is equal to |A| and otherwise zero, we have
Rationalised 2023-24
DETERMINANTS 89
A (adj A) =
A
0
0
0
A
0
0
0
A
= A
1
0
0
0
1
0
0
0
1
= A I
Similarly, we can show (adj A) A = A I
Hence A (adj A) = (adj A) A = A I
Definition 4 A square matrix A is said to be singular if A = 0 |
1 | 1548-1551 | e ,
We state the following theorem without proof Theorem 1 If A be any given square matrix of order n, then
A(adj A) = (adj A) A = A I ,
where I is the identity matrix of order n
Verification
Let
A =
11
12
13
21
22
23
31
32
33
a
a
a
a
a
a
a
a
a
, then adj A =
11
21
31
12
22
32
13
23
33
A
A
A
A
A
A
A
A
A
Since sum of product of elements of a row (or a column) with corresponding
cofactors is equal to |A| and otherwise zero, we have
Rationalised 2023-24
DETERMINANTS 89
A (adj A) =
A
0
0
0
A
0
0
0
A
= A
1
0
0
0
1
0
0
0
1
= A I
Similarly, we can show (adj A) A = A I
Hence A (adj A) = (adj A) A = A I
Definition 4 A square matrix A is said to be singular if A = 0 For example, the determinant of matrix A =
1
2
4
8
is zero
Hence A is a singular matrix |
1 | 1549-1552 | ,
We state the following theorem without proof Theorem 1 If A be any given square matrix of order n, then
A(adj A) = (adj A) A = A I ,
where I is the identity matrix of order n
Verification
Let
A =
11
12
13
21
22
23
31
32
33
a
a
a
a
a
a
a
a
a
, then adj A =
11
21
31
12
22
32
13
23
33
A
A
A
A
A
A
A
A
A
Since sum of product of elements of a row (or a column) with corresponding
cofactors is equal to |A| and otherwise zero, we have
Rationalised 2023-24
DETERMINANTS 89
A (adj A) =
A
0
0
0
A
0
0
0
A
= A
1
0
0
0
1
0
0
0
1
= A I
Similarly, we can show (adj A) A = A I
Hence A (adj A) = (adj A) A = A I
Definition 4 A square matrix A is said to be singular if A = 0 For example, the determinant of matrix A =
1
2
4
8
is zero
Hence A is a singular matrix Definition 5 A square matrix A is said to be non-singular if A ≠ 0
Let
A =
1
2
3
4
|
1 | 1550-1553 | Theorem 1 If A be any given square matrix of order n, then
A(adj A) = (adj A) A = A I ,
where I is the identity matrix of order n
Verification
Let
A =
11
12
13
21
22
23
31
32
33
a
a
a
a
a
a
a
a
a
, then adj A =
11
21
31
12
22
32
13
23
33
A
A
A
A
A
A
A
A
A
Since sum of product of elements of a row (or a column) with corresponding
cofactors is equal to |A| and otherwise zero, we have
Rationalised 2023-24
DETERMINANTS 89
A (adj A) =
A
0
0
0
A
0
0
0
A
= A
1
0
0
0
1
0
0
0
1
= A I
Similarly, we can show (adj A) A = A I
Hence A (adj A) = (adj A) A = A I
Definition 4 A square matrix A is said to be singular if A = 0 For example, the determinant of matrix A =
1
2
4
8
is zero
Hence A is a singular matrix Definition 5 A square matrix A is said to be non-singular if A ≠ 0
Let
A =
1
2
3
4
Then A =
1
2
3
4 = 4 – 6 = – 2 ≠ 0 |
1 | 1551-1554 | For example, the determinant of matrix A =
1
2
4
8
is zero
Hence A is a singular matrix Definition 5 A square matrix A is said to be non-singular if A ≠ 0
Let
A =
1
2
3
4
Then A =
1
2
3
4 = 4 – 6 = – 2 ≠ 0 Hence A is a nonsingular matrix
We state the following theorems without proof |
1 | 1552-1555 | Definition 5 A square matrix A is said to be non-singular if A ≠ 0
Let
A =
1
2
3
4
Then A =
1
2
3
4 = 4 – 6 = – 2 ≠ 0 Hence A is a nonsingular matrix
We state the following theorems without proof Theorem 2 If A and B are nonsingular matrices of the same order, then AB and BA
are also nonsingular matrices of the same order |
1 | 1553-1556 | Then A =
1
2
3
4 = 4 – 6 = – 2 ≠ 0 Hence A is a nonsingular matrix
We state the following theorems without proof Theorem 2 If A and B are nonsingular matrices of the same order, then AB and BA
are also nonsingular matrices of the same order Theorem 3 The determinant of the product of matrices is equal to product of their
respective determinants, that is, AB = A B , where A and B are square matrices of
the same order
Remark We know that (adj A) A = A I =
A
A
A
A
0
0
0
0
0
0
0
≠
,
Writing determinants of matrices on both sides, we have
(
adjA)A
=
A
0
0
0
A
0
0
0
A
Rationalised 2023-24
90
MATHEMATICS
i |
1 | 1554-1557 | Hence A is a nonsingular matrix
We state the following theorems without proof Theorem 2 If A and B are nonsingular matrices of the same order, then AB and BA
are also nonsingular matrices of the same order Theorem 3 The determinant of the product of matrices is equal to product of their
respective determinants, that is, AB = A B , where A and B are square matrices of
the same order
Remark We know that (adj A) A = A I =
A
A
A
A
0
0
0
0
0
0
0
≠
,
Writing determinants of matrices on both sides, we have
(
adjA)A
=
A
0
0
0
A
0
0
0
A
Rationalised 2023-24
90
MATHEMATICS
i e |
1 | 1555-1558 | Theorem 2 If A and B are nonsingular matrices of the same order, then AB and BA
are also nonsingular matrices of the same order Theorem 3 The determinant of the product of matrices is equal to product of their
respective determinants, that is, AB = A B , where A and B are square matrices of
the same order
Remark We know that (adj A) A = A I =
A
A
A
A
0
0
0
0
0
0
0
≠
,
Writing determinants of matrices on both sides, we have
(
adjA)A
=
A
0
0
0
A
0
0
0
A
Rationalised 2023-24
90
MATHEMATICS
i e |(adj A)| |A| =
3
1
0
0
A
0
1
0
0
0
1
(Why |
1 | 1556-1559 | Theorem 3 The determinant of the product of matrices is equal to product of their
respective determinants, that is, AB = A B , where A and B are square matrices of
the same order
Remark We know that (adj A) A = A I =
A
A
A
A
0
0
0
0
0
0
0
≠
,
Writing determinants of matrices on both sides, we have
(
adjA)A
=
A
0
0
0
A
0
0
0
A
Rationalised 2023-24
90
MATHEMATICS
i e |(adj A)| |A| =
3
1
0
0
A
0
1
0
0
0
1
(Why )
i |
1 | 1557-1560 | e |(adj A)| |A| =
3
1
0
0
A
0
1
0
0
0
1
(Why )
i e |
1 | 1558-1561 | |(adj A)| |A| =
3
1
0
0
A
0
1
0
0
0
1
(Why )
i e |(adj A)| |A| = |A|3 (1)
i |
1 | 1559-1562 | )
i e |(adj A)| |A| = |A|3 (1)
i e |
1 | 1560-1563 | e |(adj A)| |A| = |A|3 (1)
i e |(adj A)| = | A |2
In general, if A is a square matrix of order n, then |adj(A)| = |A|n – 1 |
1 | 1561-1564 | |(adj A)| |A| = |A|3 (1)
i e |(adj A)| = | A |2
In general, if A is a square matrix of order n, then |adj(A)| = |A|n – 1 Theorem 4 A square matrix A is invertible if and only if A is nonsingular matrix |
1 | 1562-1565 | e |(adj A)| = | A |2
In general, if A is a square matrix of order n, then |adj(A)| = |A|n – 1 Theorem 4 A square matrix A is invertible if and only if A is nonsingular matrix Proof Let A be invertible matrix of order n and I be the identity matrix of order n |
1 | 1563-1566 | |(adj A)| = | A |2
In general, if A is a square matrix of order n, then |adj(A)| = |A|n – 1 Theorem 4 A square matrix A is invertible if and only if A is nonsingular matrix Proof Let A be invertible matrix of order n and I be the identity matrix of order n Then, there exists a square matrix B of order n such that AB = BA = I
Now
AB = I |
1 | 1564-1567 | Theorem 4 A square matrix A is invertible if and only if A is nonsingular matrix Proof Let A be invertible matrix of order n and I be the identity matrix of order n Then, there exists a square matrix B of order n such that AB = BA = I
Now
AB = I So AB = I or A B = 1 (since I
1, AB
A B )
=
=
This gives
A ≠ 0 |
1 | 1565-1568 | Proof Let A be invertible matrix of order n and I be the identity matrix of order n Then, there exists a square matrix B of order n such that AB = BA = I
Now
AB = I So AB = I or A B = 1 (since I
1, AB
A B )
=
=
This gives
A ≠ 0 Hence A is nonsingular |
1 | 1566-1569 | Then, there exists a square matrix B of order n such that AB = BA = I
Now
AB = I So AB = I or A B = 1 (since I
1, AB
A B )
=
=
This gives
A ≠ 0 Hence A is nonsingular Conversely, let A be nonsingular |
1 | 1567-1570 | So AB = I or A B = 1 (since I
1, AB
A B )
=
=
This gives
A ≠ 0 Hence A is nonsingular Conversely, let A be nonsingular Then A ≠ 0
Now
A (adj A) = (adj A) A = A I
(Theorem 1)
or
A
1
1
A
A A
I
| A |
| A |
adj
adj
=
=
or
AB = BA = I, where B =
1
A
| A | adj
Thus
A is invertible and A–1 =
1
A
| A | adj
Example 13 If A =
1
3
3
1
4
3
1
3
4
, then verify that A adj A = |A| I |
1 | 1568-1571 | Hence A is nonsingular Conversely, let A be nonsingular Then A ≠ 0
Now
A (adj A) = (adj A) A = A I
(Theorem 1)
or
A
1
1
A
A A
I
| A |
| A |
adj
adj
=
=
or
AB = BA = I, where B =
1
A
| A | adj
Thus
A is invertible and A–1 =
1
A
| A | adj
Example 13 If A =
1
3
3
1
4
3
1
3
4
, then verify that A adj A = |A| I Also find A–1 |
1 | 1569-1572 | Conversely, let A be nonsingular Then A ≠ 0
Now
A (adj A) = (adj A) A = A I
(Theorem 1)
or
A
1
1
A
A A
I
| A |
| A |
adj
adj
=
=
or
AB = BA = I, where B =
1
A
| A | adj
Thus
A is invertible and A–1 =
1
A
| A | adj
Example 13 If A =
1
3
3
1
4
3
1
3
4
, then verify that A adj A = |A| I Also find A–1 Solution We have A = 1 (16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1 ≠ 0
Now A11 = 7, A12 = –1, A13 = –1, A21 = –3, A22 = 1,A23 = 0, A31 = –3, A32 = 0,
A33 = 1
Therefore
adj A =
7
3
3
1
1
0
1
0
1
−
−
−
−
Rationalised 2023-24
DETERMINANTS 91
Now
A (adj A) =
1
3
3
7
3
3
1
4
3
1
1
0
1
3
4
1
0
1
−
−
−
−
=
7
3
3
3
3
0
3
0
3
7
4
3
3
4
0
3
0
3
7
3
4
3
3
0
3
0
4
−
−
− +
+
− +
+
−
−
− +
+
− +
+
−
−
− +
+
− +
+
=
1
0
0
0
1
0
0
0
1
= (1)
1
0
0
0
1
0
0
0
1
= A |
1 | 1570-1573 | Then A ≠ 0
Now
A (adj A) = (adj A) A = A I
(Theorem 1)
or
A
1
1
A
A A
I
| A |
| A |
adj
adj
=
=
or
AB = BA = I, where B =
1
A
| A | adj
Thus
A is invertible and A–1 =
1
A
| A | adj
Example 13 If A =
1
3
3
1
4
3
1
3
4
, then verify that A adj A = |A| I Also find A–1 Solution We have A = 1 (16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1 ≠ 0
Now A11 = 7, A12 = –1, A13 = –1, A21 = –3, A22 = 1,A23 = 0, A31 = –3, A32 = 0,
A33 = 1
Therefore
adj A =
7
3
3
1
1
0
1
0
1
−
−
−
−
Rationalised 2023-24
DETERMINANTS 91
Now
A (adj A) =
1
3
3
7
3
3
1
4
3
1
1
0
1
3
4
1
0
1
−
−
−
−
=
7
3
3
3
3
0
3
0
3
7
4
3
3
4
0
3
0
3
7
3
4
3
3
0
3
0
4
−
−
− +
+
− +
+
−
−
− +
+
− +
+
−
−
− +
+
− +
+
=
1
0
0
0
1
0
0
0
1
= (1)
1
0
0
0
1
0
0
0
1
= A I
Also
A–1
1
A
A
a d j
=
=
7
3
3
1
1
1
0
1
1
0
1
−
−
−
−
=
7
3
3
1
1
0
1
0
1
−
−
−
−
Example 14 If A =
2
3
1
2
and B
1
4
1
−3
=
−
−
, then verify that (AB)–1 = B–1A–1 |
1 | 1571-1574 | Also find A–1 Solution We have A = 1 (16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1 ≠ 0
Now A11 = 7, A12 = –1, A13 = –1, A21 = –3, A22 = 1,A23 = 0, A31 = –3, A32 = 0,
A33 = 1
Therefore
adj A =
7
3
3
1
1
0
1
0
1
−
−
−
−
Rationalised 2023-24
DETERMINANTS 91
Now
A (adj A) =
1
3
3
7
3
3
1
4
3
1
1
0
1
3
4
1
0
1
−
−
−
−
=
7
3
3
3
3
0
3
0
3
7
4
3
3
4
0
3
0
3
7
3
4
3
3
0
3
0
4
−
−
− +
+
− +
+
−
−
− +
+
− +
+
−
−
− +
+
− +
+
=
1
0
0
0
1
0
0
0
1
= (1)
1
0
0
0
1
0
0
0
1
= A I
Also
A–1
1
A
A
a d j
=
=
7
3
3
1
1
1
0
1
1
0
1
−
−
−
−
=
7
3
3
1
1
0
1
0
1
−
−
−
−
Example 14 If A =
2
3
1
2
and B
1
4
1
−3
=
−
−
, then verify that (AB)–1 = B–1A–1 Solution We have AB =
2
3
1
2
1
5
1
4
1
3
5
14
−
−
=
−
−
−
Since,
AB = –11 ≠ 0, (AB)–1 exists and is given by
(AB)–1 =
14
5
1
1
(AB)
5
1
AB
11
adj
−
−
=−
−
−
14
5
1
5
1
11
=
Further, A = –11 ≠ 0 and B = 1 ≠ 0 |
1 | 1572-1575 | Solution We have A = 1 (16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1 ≠ 0
Now A11 = 7, A12 = –1, A13 = –1, A21 = –3, A22 = 1,A23 = 0, A31 = –3, A32 = 0,
A33 = 1
Therefore
adj A =
7
3
3
1
1
0
1
0
1
−
−
−
−
Rationalised 2023-24
DETERMINANTS 91
Now
A (adj A) =
1
3
3
7
3
3
1
4
3
1
1
0
1
3
4
1
0
1
−
−
−
−
=
7
3
3
3
3
0
3
0
3
7
4
3
3
4
0
3
0
3
7
3
4
3
3
0
3
0
4
−
−
− +
+
− +
+
−
−
− +
+
− +
+
−
−
− +
+
− +
+
=
1
0
0
0
1
0
0
0
1
= (1)
1
0
0
0
1
0
0
0
1
= A I
Also
A–1
1
A
A
a d j
=
=
7
3
3
1
1
1
0
1
1
0
1
−
−
−
−
=
7
3
3
1
1
0
1
0
1
−
−
−
−
Example 14 If A =
2
3
1
2
and B
1
4
1
−3
=
−
−
, then verify that (AB)–1 = B–1A–1 Solution We have AB =
2
3
1
2
1
5
1
4
1
3
5
14
−
−
=
−
−
−
Since,
AB = –11 ≠ 0, (AB)–1 exists and is given by
(AB)–1 =
14
5
1
1
(AB)
5
1
AB
11
adj
−
−
=−
−
−
14
5
1
5
1
11
=
Further, A = –11 ≠ 0 and B = 1 ≠ 0 Therefore, A–1 and B–1 both exist and are given by
A–1 = −
−
−
−
=
−
1
11
4
3
1
2
3
2
1
1
1
,B
Therefore
B A
−
− = −
−
−
−
1
1
1
11
3
2
1
1
4
3
1
2
= −
−
−
−
−
1
11
14
5
5
1
14
5
1
5
1
11
=
Hence (AB)–1 = B–1 A–1
Rationalised 2023-24
92
MATHEMATICS
Example 15 Show that the matrix A =
2
3
1
2
satisfies the equation A2 – 4A + I = O,
where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix |
1 | 1573-1576 | I
Also
A–1
1
A
A
a d j
=
=
7
3
3
1
1
1
0
1
1
0
1
−
−
−
−
=
7
3
3
1
1
0
1
0
1
−
−
−
−
Example 14 If A =
2
3
1
2
and B
1
4
1
−3
=
−
−
, then verify that (AB)–1 = B–1A–1 Solution We have AB =
2
3
1
2
1
5
1
4
1
3
5
14
−
−
=
−
−
−
Since,
AB = –11 ≠ 0, (AB)–1 exists and is given by
(AB)–1 =
14
5
1
1
(AB)
5
1
AB
11
adj
−
−
=−
−
−
14
5
1
5
1
11
=
Further, A = –11 ≠ 0 and B = 1 ≠ 0 Therefore, A–1 and B–1 both exist and are given by
A–1 = −
−
−
−
=
−
1
11
4
3
1
2
3
2
1
1
1
,B
Therefore
B A
−
− = −
−
−
−
1
1
1
11
3
2
1
1
4
3
1
2
= −
−
−
−
−
1
11
14
5
5
1
14
5
1
5
1
11
=
Hence (AB)–1 = B–1 A–1
Rationalised 2023-24
92
MATHEMATICS
Example 15 Show that the matrix A =
2
3
1
2
satisfies the equation A2 – 4A + I = O,
where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix Using this equation, find A–1 |
1 | 1574-1577 | Solution We have AB =
2
3
1
2
1
5
1
4
1
3
5
14
−
−
=
−
−
−
Since,
AB = –11 ≠ 0, (AB)–1 exists and is given by
(AB)–1 =
14
5
1
1
(AB)
5
1
AB
11
adj
−
−
=−
−
−
14
5
1
5
1
11
=
Further, A = –11 ≠ 0 and B = 1 ≠ 0 Therefore, A–1 and B–1 both exist and are given by
A–1 = −
−
−
−
=
−
1
11
4
3
1
2
3
2
1
1
1
,B
Therefore
B A
−
− = −
−
−
−
1
1
1
11
3
2
1
1
4
3
1
2
= −
−
−
−
−
1
11
14
5
5
1
14
5
1
5
1
11
=
Hence (AB)–1 = B–1 A–1
Rationalised 2023-24
92
MATHEMATICS
Example 15 Show that the matrix A =
2
3
1
2
satisfies the equation A2 – 4A + I = O,
where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix Using this equation, find A–1 Solution We have
2
2
3
2
3
7
12
A
A |
1 | 1575-1578 | Therefore, A–1 and B–1 both exist and are given by
A–1 = −
−
−
−
=
−
1
11
4
3
1
2
3
2
1
1
1
,B
Therefore
B A
−
− = −
−
−
−
1
1
1
11
3
2
1
1
4
3
1
2
= −
−
−
−
−
1
11
14
5
5
1
14
5
1
5
1
11
=
Hence (AB)–1 = B–1 A–1
Rationalised 2023-24
92
MATHEMATICS
Example 15 Show that the matrix A =
2
3
1
2
satisfies the equation A2 – 4A + I = O,
where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix Using this equation, find A–1 Solution We have
2
2
3
2
3
7
12
A
A A
1
2
1
2
4
7
=
=
=
Hence
2
7
12
8
12
1
0
A
4A
I
4
7
4
8
0
1
−
+ =
−
+
0
0
O
0
0
=
=
Now
A2 – 4A + I = O
Therefore
A A – 4A = – I
or
A A (A–1) – 4 A A–1 = – I A–1 (Post multiplying by A–1 because |A| ≠ 0)
or
A (A A–1) – 4I = – A–1
or
AI – 4I = – A–1
or
A–1 = 4I – A =
4
0
2
3
2
3
0
4
1
2
1
2
−
−
=
−
Hence
1
2
3
A
1
2
−
−
=
−
EXERCISE 4 |
1 | 1576-1579 | Using this equation, find A–1 Solution We have
2
2
3
2
3
7
12
A
A A
1
2
1
2
4
7
=
=
=
Hence
2
7
12
8
12
1
0
A
4A
I
4
7
4
8
0
1
−
+ =
−
+
0
0
O
0
0
=
=
Now
A2 – 4A + I = O
Therefore
A A – 4A = – I
or
A A (A–1) – 4 A A–1 = – I A–1 (Post multiplying by A–1 because |A| ≠ 0)
or
A (A A–1) – 4I = – A–1
or
AI – 4I = – A–1
or
A–1 = 4I – A =
4
0
2
3
2
3
0
4
1
2
1
2
−
−
=
−
Hence
1
2
3
A
1
2
−
−
=
−
EXERCISE 4 4
Find adjoint of each of the matrices in Exercises 1 and 2 |
1 | 1577-1580 | Solution We have
2
2
3
2
3
7
12
A
A A
1
2
1
2
4
7
=
=
=
Hence
2
7
12
8
12
1
0
A
4A
I
4
7
4
8
0
1
−
+ =
−
+
0
0
O
0
0
=
=
Now
A2 – 4A + I = O
Therefore
A A – 4A = – I
or
A A (A–1) – 4 A A–1 = – I A–1 (Post multiplying by A–1 because |A| ≠ 0)
or
A (A A–1) – 4I = – A–1
or
AI – 4I = – A–1
or
A–1 = 4I – A =
4
0
2
3
2
3
0
4
1
2
1
2
−
−
=
−
Hence
1
2
3
A
1
2
−
−
=
−
EXERCISE 4 4
Find adjoint of each of the matrices in Exercises 1 and 2 1 |
1 | 1578-1581 | A
1
2
1
2
4
7
=
=
=
Hence
2
7
12
8
12
1
0
A
4A
I
4
7
4
8
0
1
−
+ =
−
+
0
0
O
0
0
=
=
Now
A2 – 4A + I = O
Therefore
A A – 4A = – I
or
A A (A–1) – 4 A A–1 = – I A–1 (Post multiplying by A–1 because |A| ≠ 0)
or
A (A A–1) – 4I = – A–1
or
AI – 4I = – A–1
or
A–1 = 4I – A =
4
0
2
3
2
3
0
4
1
2
1
2
−
−
=
−
Hence
1
2
3
A
1
2
−
−
=
−
EXERCISE 4 4
Find adjoint of each of the matrices in Exercises 1 and 2 1 1
2
3
4
2 |
1 | 1579-1582 | 4
Find adjoint of each of the matrices in Exercises 1 and 2 1 1
2
3
4
2 1
1
2
2
3
5
2
0
1
−
−
Verify A (adj A) = (adj A) A = |A| I in Exercises 3 and 4
3 |
1 | 1580-1583 | 1 1
2
3
4
2 1
1
2
2
3
5
2
0
1
−
−
Verify A (adj A) = (adj A) A = |A| I in Exercises 3 and 4
3 2
3
4
6
−
−
4 |
1 | 1581-1584 | 1
2
3
4
2 1
1
2
2
3
5
2
0
1
−
−
Verify A (adj A) = (adj A) A = |A| I in Exercises 3 and 4
3 2
3
4
6
−
−
4 1
1
2
3
0
2
1
0
3
−
−
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11 |
1 | 1582-1585 | 1
1
2
2
3
5
2
0
1
−
−
Verify A (adj A) = (adj A) A = |A| I in Exercises 3 and 4
3 2
3
4
6
−
−
4 1
1
2
3
0
2
1
0
3
−
−
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11 5 |
1 | 1583-1586 | 2
3
4
6
−
−
4 1
1
2
3
0
2
1
0
3
−
−
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11 5 2
2
4
−3
6 |
1 | 1584-1587 | 1
1
2
3
0
2
1
0
3
−
−
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11 5 2
2
4
−3
6 −−
1
5
3
2
7 |
1 | 1585-1588 | 5 2
2
4
−3
6 −−
1
5
3
2
7 1
2
3
0
2
4
0
0
5
Rationalised 2023-24
DETERMINANTS 93
8 |
1 | 1586-1589 | 2
2
4
−3
6 −−
1
5
3
2
7 1
2
3
0
2
4
0
0
5
Rationalised 2023-24
DETERMINANTS 93
8 1
0
0
3
3
0
5
2
1
−
9 |
1 | 1587-1590 | −−
1
5
3
2
7 1
2
3
0
2
4
0
0
5
Rationalised 2023-24
DETERMINANTS 93
8 1
0
0
3
3
0
5
2
1
−
9 2
1
3
4
1
0
7
2
1
−
−
10 |
1 | 1588-1591 | 1
2
3
0
2
4
0
0
5
Rationalised 2023-24
DETERMINANTS 93
8 1
0
0
3
3
0
5
2
1
−
9 2
1
3
4
1
0
7
2
1
−
−
10 1
1
2
0
2
3
3
2
4
−
−
−
11 |
1 | 1589-1592 | 1
0
0
3
3
0
5
2
1
−
9 2
1
3
4
1
0
7
2
1
−
−
10 1
1
2
0
2
3
3
2
4
−
−
−
11 1
0
0
0
cos
sin
0
sin
cos
α
α
α
−
α
12 |
1 | 1590-1593 | 2
1
3
4
1
0
7
2
1
−
−
10 1
1
2
0
2
3
3
2
4
−
−
−
11 1
0
0
0
cos
sin
0
sin
cos
α
α
α
−
α
12 Let A =
3
7
2
5
and B = 6
8
7
9
|
1 | 1591-1594 | 1
1
2
0
2
3
3
2
4
−
−
−
11 1
0
0
0
cos
sin
0
sin
cos
α
α
α
−
α
12 Let A =
3
7
2
5
and B = 6
8
7
9
Verify that (AB)–1 = B–1 A–1 |
1 | 1592-1595 | 1
0
0
0
cos
sin
0
sin
cos
α
α
α
−
α
12 Let A =
3
7
2
5
and B = 6
8
7
9
Verify that (AB)–1 = B–1 A–1 13 |
1 | 1593-1596 | Let A =
3
7
2
5
and B = 6
8
7
9
Verify that (AB)–1 = B–1 A–1 13 If A =
3
1
1
2
−
, show that A2 – 5A + 7I = O |
1 | 1594-1597 | Verify that (AB)–1 = B–1 A–1 13 If A =
3
1
1
2
−
, show that A2 – 5A + 7I = O Hence find A–1 |
1 | 1595-1598 | 13 If A =
3
1
1
2
−
, show that A2 – 5A + 7I = O Hence find A–1 14 |
1 | 1596-1599 | If A =
3
1
1
2
−
, show that A2 – 5A + 7I = O Hence find A–1 14 For the matrix A =
3
2
1
1
, find the numbers a and b such that A2 + aA + bI = O |
1 | 1597-1600 | Hence find A–1 14 For the matrix A =
3
2
1
1
, find the numbers a and b such that A2 + aA + bI = O 15 |
1 | 1598-1601 | 14 For the matrix A =
3
2
1
1
, find the numbers a and b such that A2 + aA + bI = O 15 For the matrix A =
1
1
1
1
2
3
2
1
−3
−
Show that A3– 6A2 + 5A + 11 I = O |
1 | 1599-1602 | For the matrix A =
3
2
1
1
, find the numbers a and b such that A2 + aA + bI = O 15 For the matrix A =
1
1
1
1
2
3
2
1
−3
−
Show that A3– 6A2 + 5A + 11 I = O Hence, find A–1 |
1 | 1600-1603 | 15 For the matrix A =
1
1
1
1
2
3
2
1
−3
−
Show that A3– 6A2 + 5A + 11 I = O Hence, find A–1 16 |
1 | 1601-1604 | For the matrix A =
1
1
1
1
2
3
2
1
−3
−
Show that A3– 6A2 + 5A + 11 I = O Hence, find A–1 16 If A =
2
1
1
1
2
1
1
1
2
−
−
−
−
Verify that A3 – 6A2 + 9A – 4I = O and hence find A–1
17 |
1 | 1602-1605 | Hence, find A–1 16 If A =
2
1
1
1
2
1
1
1
2
−
−
−
−
Verify that A3 – 6A2 + 9A – 4I = O and hence find A–1
17 Let A be a nonsingular square matrix of order 3 × 3 |
1 | 1603-1606 | 16 If A =
2
1
1
1
2
1
1
1
2
−
−
−
−
Verify that A3 – 6A2 + 9A – 4I = O and hence find A–1
17 Let A be a nonsingular square matrix of order 3 × 3 Then |adj A| is equal to
(A) | A |
(B) | A |2
(C) | A |3
(D) 3|A|
18 |
1 | 1604-1607 | If A =
2
1
1
1
2
1
1
1
2
−
−
−
−
Verify that A3 – 6A2 + 9A – 4I = O and hence find A–1
17 Let A be a nonsingular square matrix of order 3 × 3 Then |adj A| is equal to
(A) | A |
(B) | A |2
(C) | A |3
(D) 3|A|
18 If A is an invertible matrix of order 2, then det (A–1) is equal to
(A) det (A)
(B)
1
det (A)
(C) 1
(D) 0
4 |
1 | 1605-1608 | Let A be a nonsingular square matrix of order 3 × 3 Then |adj A| is equal to
(A) | A |
(B) | A |2
(C) | A |3
(D) 3|A|
18 If A is an invertible matrix of order 2, then det (A–1) is equal to
(A) det (A)
(B)
1
det (A)
(C) 1
(D) 0
4 6 Applications of Determinants and Matrices
In this section, we shall discuss application of determinants and matrices for solving the
system of linear equations in two or three variables and for checking the consistency of
the system of linear equations |
1 | 1606-1609 | Then |adj A| is equal to
(A) | A |
(B) | A |2
(C) | A |3
(D) 3|A|
18 If A is an invertible matrix of order 2, then det (A–1) is equal to
(A) det (A)
(B)
1
det (A)
(C) 1
(D) 0
4 6 Applications of Determinants and Matrices
In this section, we shall discuss application of determinants and matrices for solving the
system of linear equations in two or three variables and for checking the consistency of
the system of linear equations Rationalised 2023-24
94
MATHEMATICS
Consistent system A system of equations is said to be consistent if its solution (one
or more) exists |
1 | 1607-1610 | If A is an invertible matrix of order 2, then det (A–1) is equal to
(A) det (A)
(B)
1
det (A)
(C) 1
(D) 0
4 6 Applications of Determinants and Matrices
In this section, we shall discuss application of determinants and matrices for solving the
system of linear equations in two or three variables and for checking the consistency of
the system of linear equations Rationalised 2023-24
94
MATHEMATICS
Consistent system A system of equations is said to be consistent if its solution (one
or more) exists Inconsistent system A system of equations is said to be inconsistent if its solution
does not exist |
1 | 1608-1611 | 6 Applications of Determinants and Matrices
In this section, we shall discuss application of determinants and matrices for solving the
system of linear equations in two or three variables and for checking the consistency of
the system of linear equations Rationalised 2023-24
94
MATHEMATICS
Consistent system A system of equations is said to be consistent if its solution (one
or more) exists Inconsistent system A system of equations is said to be inconsistent if its solution
does not exist ANote In this chapter, we restrict ourselves to the system of linear equations
having unique solutions only |
1 | 1609-1612 | Rationalised 2023-24
94
MATHEMATICS
Consistent system A system of equations is said to be consistent if its solution (one
or more) exists Inconsistent system A system of equations is said to be inconsistent if its solution
does not exist ANote In this chapter, we restrict ourselves to the system of linear equations
having unique solutions only 4 |
1 | 1610-1613 | Inconsistent system A system of equations is said to be inconsistent if its solution
does not exist ANote In this chapter, we restrict ourselves to the system of linear equations
having unique solutions only 4 6 |
1 | 1611-1614 | ANote In this chapter, we restrict ourselves to the system of linear equations
having unique solutions only 4 6 1 Solution of system of linear equations using inverse of a matrix
Let us express the system of linear equations as matrix equations and solve them using
inverse of the coefficient matrix |
1 | 1612-1615 | 4 6 1 Solution of system of linear equations using inverse of a matrix
Let us express the system of linear equations as matrix equations and solve them using
inverse of the coefficient matrix Consider the system of equations
a1 x + b1 y + c1 z = d1
a2 x + b2 y + c2 z = d 2
a3 x + b3 y + c3 z = d 3
Let
A =
1
1
1
1
2
2
2
2
3
3
3
3
, X
and B
a
b
c
x
d
a
b
c
y
d
a
b
c
z
d
=
=
Then, the system of equations can be written as, AX = B, i |
1 | 1613-1616 | 6 1 Solution of system of linear equations using inverse of a matrix
Let us express the system of linear equations as matrix equations and solve them using
inverse of the coefficient matrix Consider the system of equations
a1 x + b1 y + c1 z = d1
a2 x + b2 y + c2 z = d 2
a3 x + b3 y + c3 z = d 3
Let
A =
1
1
1
1
2
2
2
2
3
3
3
3
, X
and B
a
b
c
x
d
a
b
c
y
d
a
b
c
z
d
=
=
Then, the system of equations can be written as, AX = B, i e |
1 | 1614-1617 | 1 Solution of system of linear equations using inverse of a matrix
Let us express the system of linear equations as matrix equations and solve them using
inverse of the coefficient matrix Consider the system of equations
a1 x + b1 y + c1 z = d1
a2 x + b2 y + c2 z = d 2
a3 x + b3 y + c3 z = d 3
Let
A =
1
1
1
1
2
2
2
2
3
3
3
3
, X
and B
a
b
c
x
d
a
b
c
y
d
a
b
c
z
d
=
=
Then, the system of equations can be written as, AX = B, i e ,
1
1
1
2
2
2
3
3
3
a
b
c
x
a
b
c
y
a
b
c
z
=
1
2
3
d
d
d
Case I If A is a nonsingular matrix, then its inverse exists |
1 | 1615-1618 | Consider the system of equations
a1 x + b1 y + c1 z = d1
a2 x + b2 y + c2 z = d 2
a3 x + b3 y + c3 z = d 3
Let
A =
1
1
1
1
2
2
2
2
3
3
3
3
, X
and B
a
b
c
x
d
a
b
c
y
d
a
b
c
z
d
=
=
Then, the system of equations can be written as, AX = B, i e ,
1
1
1
2
2
2
3
3
3
a
b
c
x
a
b
c
y
a
b
c
z
=
1
2
3
d
d
d
Case I If A is a nonsingular matrix, then its inverse exists Now
AX = B
or
A–1 (AX) = A–1 B
(premultiplying by A–1)
or
(A–1A) X = A–1 B
(by associative property)
or
I X = A–1 B
or
X = A–1 B
This matrix equation provides unique solution for the given system of equations as
inverse of a matrix is unique |
1 | 1616-1619 | e ,
1
1
1
2
2
2
3
3
3
a
b
c
x
a
b
c
y
a
b
c
z
=
1
2
3
d
d
d
Case I If A is a nonsingular matrix, then its inverse exists Now
AX = B
or
A–1 (AX) = A–1 B
(premultiplying by A–1)
or
(A–1A) X = A–1 B
(by associative property)
or
I X = A–1 B
or
X = A–1 B
This matrix equation provides unique solution for the given system of equations as
inverse of a matrix is unique This method of solving system of equations is known as
Matrix Method |
1 | 1617-1620 | ,
1
1
1
2
2
2
3
3
3
a
b
c
x
a
b
c
y
a
b
c
z
=
1
2
3
d
d
d
Case I If A is a nonsingular matrix, then its inverse exists Now
AX = B
or
A–1 (AX) = A–1 B
(premultiplying by A–1)
or
(A–1A) X = A–1 B
(by associative property)
or
I X = A–1 B
or
X = A–1 B
This matrix equation provides unique solution for the given system of equations as
inverse of a matrix is unique This method of solving system of equations is known as
Matrix Method Case II If A is a singular matrix, then |A| = 0 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.