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1 | 1618-1621 | 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 In this case, we calculate (adj A) B |
1 | 1619-1622 | This method of solving system of equations is known as
Matrix Method Case II If A is a singular matrix, then |A| = 0 In this case, we calculate (adj A) B If (adj A) B ≠ O, (O being zero matrix), then solution does not exist and the
system of equations is called inconsistent |
1 | 1620-1623 | Case II If A is a singular matrix, then |A| = 0 In this case, we calculate (adj A) B If (adj A) B ≠ O, (O being zero matrix), then solution does not exist and the
system of equations is called inconsistent Rationalised 2023-24
DETERMINANTS 95
If (adj A) B = O, then system may be either consistent or inconsistent according
as the system have either infinitely many solutions or no solution |
1 | 1621-1624 | In this case, we calculate (adj A) B If (adj A) B ≠ O, (O being zero matrix), then solution does not exist and the
system of equations is called inconsistent Rationalised 2023-24
DETERMINANTS 95
If (adj A) B = O, then system may be either consistent or inconsistent according
as the system have either infinitely many solutions or no solution Example 16 Solve the system of equations
2x + 5y = 1
3x + 2y = 7
Solution The system of equations can be written in the form AX = B, where
A =
2
5
1
,X
and B
3
2
7
x
y
=
=
Now, A = –11 ≠ 0, Hence, A is nonsingular matrix and so has a unique solution |
1 | 1622-1625 | If (adj A) B ≠ O, (O being zero matrix), then solution does not exist and the
system of equations is called inconsistent Rationalised 2023-24
DETERMINANTS 95
If (adj A) B = O, then system may be either consistent or inconsistent according
as the system have either infinitely many solutions or no solution Example 16 Solve the system of equations
2x + 5y = 1
3x + 2y = 7
Solution The system of equations can be written in the form AX = B, where
A =
2
5
1
,X
and B
3
2
7
x
y
=
=
Now, A = –11 ≠ 0, Hence, A is nonsingular matrix and so has a unique solution Note that
A–1 = −
−
−
1
11
2
5
3
2
Therefore
X = A–1B = – 1
11
2
5
3
2
71
−
−
i |
1 | 1623-1626 | Rationalised 2023-24
DETERMINANTS 95
If (adj A) B = O, then system may be either consistent or inconsistent according
as the system have either infinitely many solutions or no solution Example 16 Solve the system of equations
2x + 5y = 1
3x + 2y = 7
Solution The system of equations can be written in the form AX = B, where
A =
2
5
1
,X
and B
3
2
7
x
y
=
=
Now, A = –11 ≠ 0, Hence, A is nonsingular matrix and so has a unique solution Note that
A–1 = −
−
−
1
11
2
5
3
2
Therefore
X = A–1B = – 1
11
2
5
3
2
71
−
−
i e |
1 | 1624-1627 | Example 16 Solve the system of equations
2x + 5y = 1
3x + 2y = 7
Solution The system of equations can be written in the form AX = B, where
A =
2
5
1
,X
and B
3
2
7
x
y
=
=
Now, A = –11 ≠ 0, Hence, A is nonsingular matrix and so has a unique solution Note that
A–1 = −
−
−
1
11
2
5
3
2
Therefore
X = A–1B = – 1
11
2
5
3
2
71
−
−
i e x
y
= −
−
= −
1
11
33
11
3
1
Hence
x = 3, y = – 1
Example 17 Solve the following system of equations by matrix method |
1 | 1625-1628 | Note that
A–1 = −
−
−
1
11
2
5
3
2
Therefore
X = A–1B = – 1
11
2
5
3
2
71
−
−
i e x
y
= −
−
= −
1
11
33
11
3
1
Hence
x = 3, y = – 1
Example 17 Solve the following system of equations by matrix method 3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4
Solution The system of equations can be written in the form AX = B, where
3
2
3
8
A
2
1
1 , X
and B
1
4
3
2
4
x
y
z
−
=
−
=
=
−
We see that
A = 3 (2 – 3) + 2(4 + 4) + 3 (– 6 – 4) = – 17 ≠ 0
Hence, A is nonsingular and so its inverse exists |
1 | 1626-1629 | e x
y
= −
−
= −
1
11
33
11
3
1
Hence
x = 3, y = – 1
Example 17 Solve the following system of equations by matrix method 3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4
Solution The system of equations can be written in the form AX = B, where
3
2
3
8
A
2
1
1 , X
and B
1
4
3
2
4
x
y
z
−
=
−
=
=
−
We see that
A = 3 (2 – 3) + 2(4 + 4) + 3 (– 6 – 4) = – 17 ≠ 0
Hence, A is nonsingular and so its inverse exists Now
A11 = –1,
A12 = – 8,
A13 = –10
A21 = –5,
A22 = – 6,
A23 = 1
A31 = –1,
A32 = 9,
A33 = 7
Rationalised 2023-24
96
MATHEMATICS
Therefore
A–1 =
1
5
1
1
8
6
9
17
10
1
7
−
−
−
−
−
−
−
So
X =
–1
1
5
1
8
1
A B =
8
6
9
1
17
10
1
7
4
−
−
−
−
−
−
−
i |
1 | 1627-1630 | x
y
= −
−
= −
1
11
33
11
3
1
Hence
x = 3, y = – 1
Example 17 Solve the following system of equations by matrix method 3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4
Solution The system of equations can be written in the form AX = B, where
3
2
3
8
A
2
1
1 , X
and B
1
4
3
2
4
x
y
z
−
=
−
=
=
−
We see that
A = 3 (2 – 3) + 2(4 + 4) + 3 (– 6 – 4) = – 17 ≠ 0
Hence, A is nonsingular and so its inverse exists Now
A11 = –1,
A12 = – 8,
A13 = –10
A21 = –5,
A22 = – 6,
A23 = 1
A31 = –1,
A32 = 9,
A33 = 7
Rationalised 2023-24
96
MATHEMATICS
Therefore
A–1 =
1
5
1
1
8
6
9
17
10
1
7
−
−
−
−
−
−
−
So
X =
–1
1
5
1
8
1
A B =
8
6
9
1
17
10
1
7
4
−
−
−
−
−
−
−
i e |
1 | 1628-1631 | 3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4
Solution The system of equations can be written in the form AX = B, where
3
2
3
8
A
2
1
1 , X
and B
1
4
3
2
4
x
y
z
−
=
−
=
=
−
We see that
A = 3 (2 – 3) + 2(4 + 4) + 3 (– 6 – 4) = – 17 ≠ 0
Hence, A is nonsingular and so its inverse exists Now
A11 = –1,
A12 = – 8,
A13 = –10
A21 = –5,
A22 = – 6,
A23 = 1
A31 = –1,
A32 = 9,
A33 = 7
Rationalised 2023-24
96
MATHEMATICS
Therefore
A–1 =
1
5
1
1
8
6
9
17
10
1
7
−
−
−
−
−
−
−
So
X =
–1
1
5
1
8
1
A B =
8
6
9
1
17
10
1
7
4
−
−
−
−
−
−
−
i e x
y
z
=
17
1
1
34
2
17
51
3
−
−
−
=
−
Hence
x = 1, y = 2 and z = 3 |
1 | 1629-1632 | Now
A11 = –1,
A12 = – 8,
A13 = –10
A21 = –5,
A22 = – 6,
A23 = 1
A31 = –1,
A32 = 9,
A33 = 7
Rationalised 2023-24
96
MATHEMATICS
Therefore
A–1 =
1
5
1
1
8
6
9
17
10
1
7
−
−
−
−
−
−
−
So
X =
–1
1
5
1
8
1
A B =
8
6
9
1
17
10
1
7
4
−
−
−
−
−
−
−
i e x
y
z
=
17
1
1
34
2
17
51
3
−
−
−
=
−
Hence
x = 1, y = 2 and z = 3 Example 18 The sum of three numbers is 6 |
1 | 1630-1633 | e x
y
z
=
17
1
1
34
2
17
51
3
−
−
−
=
−
Hence
x = 1, y = 2 and z = 3 Example 18 The sum of three numbers is 6 If we multiply third number by 3 and add
second number to it, we get 11 |
1 | 1631-1634 | x
y
z
=
17
1
1
34
2
17
51
3
−
−
−
=
−
Hence
x = 1, y = 2 and z = 3 Example 18 The sum of three numbers is 6 If we multiply third number by 3 and add
second number to it, we get 11 By adding first and third numbers, we get double of the
second number |
1 | 1632-1635 | Example 18 The sum of three numbers is 6 If we multiply third number by 3 and add
second number to it, we get 11 By adding first and third numbers, we get double of the
second number Represent it algebraically and find the numbers using matrix method |
1 | 1633-1636 | If we multiply third number by 3 and add
second number to it, we get 11 By adding first and third numbers, we get double of the
second number Represent it algebraically and find the numbers using matrix method Solution Let first, second and third numbers be denoted by x, y and z, respectively |
1 | 1634-1637 | By adding first and third numbers, we get double of the
second number Represent it algebraically and find the numbers using matrix method Solution Let first, second and third numbers be denoted by x, y and z, respectively Then, according to given conditions, we have
x + y + z = 6
y + 3z = 11
x + z = 2y or x – 2y + z = 0
This system can be written as A X = B, where
A =
1
1
1
0
1
3
1
2
1
�
, X =
x
y
z
and B =
6
11
0
Here
(
)
(
)
A
1 1
6 – (0 – 3)
0 –1
9
0
=
+
+
=
≠ |
1 | 1635-1638 | Represent it algebraically and find the numbers using matrix method Solution Let first, second and third numbers be denoted by x, y and z, respectively Then, according to given conditions, we have
x + y + z = 6
y + 3z = 11
x + z = 2y or x – 2y + z = 0
This system can be written as A X = B, where
A =
1
1
1
0
1
3
1
2
1
�
, X =
x
y
z
and B =
6
11
0
Here
(
)
(
)
A
1 1
6 – (0 – 3)
0 –1
9
0
=
+
+
=
≠ Now we find adj A
A11 = 1 (1 + 6) = 7,
A12 = – (0 – 3) = 3,
A13 = – 1
A21 = – (1 + 2) = – 3,
A22 = 0,
A23 = – (– 2 – 1) = 3
A31 = (3 – 1) = 2,
A32 = – (3 – 0) = – 3,
A33 = (1 – 0) = 1
Hence
adj A =
7
–3
2
3
0
–3
–1
3
1
Rationalised 2023-24
DETERMINANTS 97
Thus
A –1 =
1
A adj (A) =
7
3
2
1
3
0
3
9
1
3
1
–
–
–
Since
X = A–1 B
X =
7
3
2
6
1
3
0
3
11
9
1
3
1
0
–
–
–
or
x
y
z
= 1
9
42
33
0
18
0
0
6
33
0
−
+
+
+
− +
+
= 1
9
9
18
27
=
1
2
3
Thus
x = 1, y = 2, z = 3
EXERCISE 4 |
1 | 1636-1639 | Solution Let first, second and third numbers be denoted by x, y and z, respectively Then, according to given conditions, we have
x + y + z = 6
y + 3z = 11
x + z = 2y or x – 2y + z = 0
This system can be written as A X = B, where
A =
1
1
1
0
1
3
1
2
1
�
, X =
x
y
z
and B =
6
11
0
Here
(
)
(
)
A
1 1
6 – (0 – 3)
0 –1
9
0
=
+
+
=
≠ Now we find adj A
A11 = 1 (1 + 6) = 7,
A12 = – (0 – 3) = 3,
A13 = – 1
A21 = – (1 + 2) = – 3,
A22 = 0,
A23 = – (– 2 – 1) = 3
A31 = (3 – 1) = 2,
A32 = – (3 – 0) = – 3,
A33 = (1 – 0) = 1
Hence
adj A =
7
–3
2
3
0
–3
–1
3
1
Rationalised 2023-24
DETERMINANTS 97
Thus
A –1 =
1
A adj (A) =
7
3
2
1
3
0
3
9
1
3
1
–
–
–
Since
X = A–1 B
X =
7
3
2
6
1
3
0
3
11
9
1
3
1
0
–
–
–
or
x
y
z
= 1
9
42
33
0
18
0
0
6
33
0
−
+
+
+
− +
+
= 1
9
9
18
27
=
1
2
3
Thus
x = 1, y = 2, z = 3
EXERCISE 4 5
Examine the consistency of the system of equations in Exercises 1 to 6 |
1 | 1637-1640 | Then, according to given conditions, we have
x + y + z = 6
y + 3z = 11
x + z = 2y or x – 2y + z = 0
This system can be written as A X = B, where
A =
1
1
1
0
1
3
1
2
1
�
, X =
x
y
z
and B =
6
11
0
Here
(
)
(
)
A
1 1
6 – (0 – 3)
0 –1
9
0
=
+
+
=
≠ Now we find adj A
A11 = 1 (1 + 6) = 7,
A12 = – (0 – 3) = 3,
A13 = – 1
A21 = – (1 + 2) = – 3,
A22 = 0,
A23 = – (– 2 – 1) = 3
A31 = (3 – 1) = 2,
A32 = – (3 – 0) = – 3,
A33 = (1 – 0) = 1
Hence
adj A =
7
–3
2
3
0
–3
–1
3
1
Rationalised 2023-24
DETERMINANTS 97
Thus
A –1 =
1
A adj (A) =
7
3
2
1
3
0
3
9
1
3
1
–
–
–
Since
X = A–1 B
X =
7
3
2
6
1
3
0
3
11
9
1
3
1
0
–
–
–
or
x
y
z
= 1
9
42
33
0
18
0
0
6
33
0
−
+
+
+
− +
+
= 1
9
9
18
27
=
1
2
3
Thus
x = 1, y = 2, z = 3
EXERCISE 4 5
Examine the consistency of the system of equations in Exercises 1 to 6 1 |
1 | 1638-1641 | Now we find adj A
A11 = 1 (1 + 6) = 7,
A12 = – (0 – 3) = 3,
A13 = – 1
A21 = – (1 + 2) = – 3,
A22 = 0,
A23 = – (– 2 – 1) = 3
A31 = (3 – 1) = 2,
A32 = – (3 – 0) = – 3,
A33 = (1 – 0) = 1
Hence
adj A =
7
–3
2
3
0
–3
–1
3
1
Rationalised 2023-24
DETERMINANTS 97
Thus
A –1 =
1
A adj (A) =
7
3
2
1
3
0
3
9
1
3
1
–
–
–
Since
X = A–1 B
X =
7
3
2
6
1
3
0
3
11
9
1
3
1
0
–
–
–
or
x
y
z
= 1
9
42
33
0
18
0
0
6
33
0
−
+
+
+
− +
+
= 1
9
9
18
27
=
1
2
3
Thus
x = 1, y = 2, z = 3
EXERCISE 4 5
Examine the consistency of the system of equations in Exercises 1 to 6 1 x + 2y = 2
2 |
1 | 1639-1642 | 5
Examine the consistency of the system of equations in Exercises 1 to 6 1 x + 2y = 2
2 2x – y = 5
3 |
1 | 1640-1643 | 1 x + 2y = 2
2 2x – y = 5
3 x + 3y = 5
2x + 3y = 3
x + y = 4
2x + 6y = 8
4 |
1 | 1641-1644 | x + 2y = 2
2 2x – y = 5
3 x + 3y = 5
2x + 3y = 3
x + y = 4
2x + 6y = 8
4 x + y + z = 1
5 |
1 | 1642-1645 | 2x – y = 5
3 x + 3y = 5
2x + 3y = 3
x + y = 4
2x + 6y = 8
4 x + y + z = 1
5 3x–y – 2z = 2
6 |
1 | 1643-1646 | x + 3y = 5
2x + 3y = 3
x + y = 4
2x + 6y = 8
4 x + y + z = 1
5 3x–y – 2z = 2
6 5x – y + 4z = 5
2x + 3y + 2z = 2
2y – z = –1
2x + 3y + 5z = 2
ax + ay + 2az = 4
3x – 5y = 3
5x – 2y + 6z = –1
Solve system of linear equations, using matrix method, in Exercises 7 to 14 |
1 | 1644-1647 | x + y + z = 1
5 3x–y – 2z = 2
6 5x – y + 4z = 5
2x + 3y + 2z = 2
2y – z = –1
2x + 3y + 5z = 2
ax + ay + 2az = 4
3x – 5y = 3
5x – 2y + 6z = –1
Solve system of linear equations, using matrix method, in Exercises 7 to 14 7 |
1 | 1645-1648 | 3x–y – 2z = 2
6 5x – y + 4z = 5
2x + 3y + 2z = 2
2y – z = –1
2x + 3y + 5z = 2
ax + ay + 2az = 4
3x – 5y = 3
5x – 2y + 6z = –1
Solve system of linear equations, using matrix method, in Exercises 7 to 14 7 5x + 2y = 4
8 |
1 | 1646-1649 | 5x – y + 4z = 5
2x + 3y + 2z = 2
2y – z = –1
2x + 3y + 5z = 2
ax + ay + 2az = 4
3x – 5y = 3
5x – 2y + 6z = –1
Solve system of linear equations, using matrix method, in Exercises 7 to 14 7 5x + 2y = 4
8 2x – y = –2
9 |
1 | 1647-1650 | 7 5x + 2y = 4
8 2x – y = –2
9 4x – 3y = 3
7x + 3y = 5
3x + 4y = 3
3x – 5y = 7
10 |
1 | 1648-1651 | 5x + 2y = 4
8 2x – y = –2
9 4x – 3y = 3
7x + 3y = 5
3x + 4y = 3
3x – 5y = 7
10 5x + 2y = 3
11 |
1 | 1649-1652 | 2x – y = –2
9 4x – 3y = 3
7x + 3y = 5
3x + 4y = 3
3x – 5y = 7
10 5x + 2y = 3
11 2x + y + z = 1
12 |
1 | 1650-1653 | 4x – 3y = 3
7x + 3y = 5
3x + 4y = 3
3x – 5y = 7
10 5x + 2y = 3
11 2x + y + z = 1
12 x – y + z = 4
3x + 2y = 5
x – 2y – z = 3
2
2x + y – 3z = 0
3y – 5z = 9
x + y + z = 2
13 |
1 | 1651-1654 | 5x + 2y = 3
11 2x + y + z = 1
12 x – y + z = 4
3x + 2y = 5
x – 2y – z = 3
2
2x + y – 3z = 0
3y – 5z = 9
x + y + z = 2
13 2x + 3y +3 z = 5
14 |
1 | 1652-1655 | 2x + y + z = 1
12 x – y + z = 4
3x + 2y = 5
x – 2y – z = 3
2
2x + y – 3z = 0
3y – 5z = 9
x + y + z = 2
13 2x + 3y +3 z = 5
14 x – y + 2z = 7
x – 2y + z = – 4
3x + 4y – 5z = – 5
3x – y – 2z = 3
2x – y + 3z = 12
Rationalised 2023-24
98
MATHEMATICS
15 |
1 | 1653-1656 | x – y + z = 4
3x + 2y = 5
x – 2y – z = 3
2
2x + y – 3z = 0
3y – 5z = 9
x + y + z = 2
13 2x + 3y +3 z = 5
14 x – y + 2z = 7
x – 2y + z = – 4
3x + 4y – 5z = – 5
3x – y – 2z = 3
2x – y + 3z = 12
Rationalised 2023-24
98
MATHEMATICS
15 If A =
2
–3
5
3
2
– 4
1
1
–2
, find A–1 |
1 | 1654-1657 | 2x + 3y +3 z = 5
14 x – y + 2z = 7
x – 2y + z = – 4
3x + 4y – 5z = – 5
3x – y – 2z = 3
2x – y + 3z = 12
Rationalised 2023-24
98
MATHEMATICS
15 If A =
2
–3
5
3
2
– 4
1
1
–2
, find A–1 Using A–1 solve the system of equations
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3
16 |
1 | 1655-1658 | x – y + 2z = 7
x – 2y + z = – 4
3x + 4y – 5z = – 5
3x – y – 2z = 3
2x – y + 3z = 12
Rationalised 2023-24
98
MATHEMATICS
15 If A =
2
–3
5
3
2
– 4
1
1
–2
, find A–1 Using A–1 solve the system of equations
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3
16 The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ` 60 |
1 | 1656-1659 | If A =
2
–3
5
3
2
– 4
1
1
–2
, find A–1 Using A–1 solve the system of equations
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3
16 The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ` 60 The cost of 2 kg onion,
4 kg wheat and 6 kg rice is ` 90 |
1 | 1657-1660 | Using A–1 solve the system of equations
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3
16 The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ` 60 The cost of 2 kg onion,
4 kg wheat and 6 kg rice is ` 90 The cost of 6 kg onion 2 kg wheat and 3 kg rice
is ` 70 |
1 | 1658-1661 | The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ` 60 The cost of 2 kg onion,
4 kg wheat and 6 kg rice is ` 90 The cost of 6 kg onion 2 kg wheat and 3 kg rice
is ` 70 Find cost of each item per kg by matrix method |
1 | 1659-1662 | The cost of 2 kg onion,
4 kg wheat and 6 kg rice is ` 90 The cost of 6 kg onion 2 kg wheat and 3 kg rice
is ` 70 Find cost of each item per kg by matrix method Miscellaneous Examples
Example 19 Use product
1
1
2
0
2
3
3
2
4
2
0
1
9
2
3
6
1
2
�
�
�
�
�
�
to solve the system of equations
x – y + 2z = 1
2y – 3z = 1
3x – 2y + 4z = 2
Solution Consider the product
1
1
2
2
0
1
0
2
3
9
2
3
3
2
4
6
1
2
–
–
–
–
–
–
=
2
9
12
0
2
2
1
3
4
0
18 18
0
4
3
0
6
6
6 18
24
0
4
4
3
6
8
− −
+
−
+
+
−
+
−
+
−
−
+
− −
+
−
+
+
−
=
1
0
0
0
1
0
0
0
1
Hence
1
1
2
0
2
3
3
2
4
2
0
1
9
2
3
6
1
2
1
–
–
–
–
–
–
–
=
Now, given system of equations can be written, in matrix form, as follows
1
–1
2
0
2
–3
3
–2
4
x
y
z
=
1
1
2
Rationalised 2023-24
DETERMINANTS 99
or
x
y
z
=
1
1
1
2
1
0
2
3
1
3
2
4
2
−
−
−
−
=
�
�
�
2
0
1
9
2
3
6
1
2
1
1
2
=
2
0
2
0
9
2
6
5
6
1
4
3
− +
+
+
−
=
+ −
Hence
x = 0, y = 5 and z = 3
Miscellaneous Exercises on Chapter 4
1 |
1 | 1660-1663 | The cost of 6 kg onion 2 kg wheat and 3 kg rice
is ` 70 Find cost of each item per kg by matrix method Miscellaneous Examples
Example 19 Use product
1
1
2
0
2
3
3
2
4
2
0
1
9
2
3
6
1
2
�
�
�
�
�
�
to solve the system of equations
x – y + 2z = 1
2y – 3z = 1
3x – 2y + 4z = 2
Solution Consider the product
1
1
2
2
0
1
0
2
3
9
2
3
3
2
4
6
1
2
–
–
–
–
–
–
=
2
9
12
0
2
2
1
3
4
0
18 18
0
4
3
0
6
6
6 18
24
0
4
4
3
6
8
− −
+
−
+
+
−
+
−
+
−
−
+
− −
+
−
+
+
−
=
1
0
0
0
1
0
0
0
1
Hence
1
1
2
0
2
3
3
2
4
2
0
1
9
2
3
6
1
2
1
–
–
–
–
–
–
–
=
Now, given system of equations can be written, in matrix form, as follows
1
–1
2
0
2
–3
3
–2
4
x
y
z
=
1
1
2
Rationalised 2023-24
DETERMINANTS 99
or
x
y
z
=
1
1
1
2
1
0
2
3
1
3
2
4
2
−
−
−
−
=
�
�
�
2
0
1
9
2
3
6
1
2
1
1
2
=
2
0
2
0
9
2
6
5
6
1
4
3
− +
+
+
−
=
+ −
Hence
x = 0, y = 5 and z = 3
Miscellaneous Exercises on Chapter 4
1 Prove that the determinant
sin
cos
–sin
–
1
cos
1
x
x
x
θ
θ
θ
θ
is independent of θ |
1 | 1661-1664 | Find cost of each item per kg by matrix method Miscellaneous Examples
Example 19 Use product
1
1
2
0
2
3
3
2
4
2
0
1
9
2
3
6
1
2
�
�
�
�
�
�
to solve the system of equations
x – y + 2z = 1
2y – 3z = 1
3x – 2y + 4z = 2
Solution Consider the product
1
1
2
2
0
1
0
2
3
9
2
3
3
2
4
6
1
2
–
–
–
–
–
–
=
2
9
12
0
2
2
1
3
4
0
18 18
0
4
3
0
6
6
6 18
24
0
4
4
3
6
8
− −
+
−
+
+
−
+
−
+
−
−
+
− −
+
−
+
+
−
=
1
0
0
0
1
0
0
0
1
Hence
1
1
2
0
2
3
3
2
4
2
0
1
9
2
3
6
1
2
1
–
–
–
–
–
–
–
=
Now, given system of equations can be written, in matrix form, as follows
1
–1
2
0
2
–3
3
–2
4
x
y
z
=
1
1
2
Rationalised 2023-24
DETERMINANTS 99
or
x
y
z
=
1
1
1
2
1
0
2
3
1
3
2
4
2
−
−
−
−
=
�
�
�
2
0
1
9
2
3
6
1
2
1
1
2
=
2
0
2
0
9
2
6
5
6
1
4
3
− +
+
+
−
=
+ −
Hence
x = 0, y = 5 and z = 3
Miscellaneous Exercises on Chapter 4
1 Prove that the determinant
sin
cos
–sin
–
1
cos
1
x
x
x
θ
θ
θ
θ
is independent of θ 2 |
1 | 1662-1665 | Miscellaneous Examples
Example 19 Use product
1
1
2
0
2
3
3
2
4
2
0
1
9
2
3
6
1
2
�
�
�
�
�
�
to solve the system of equations
x – y + 2z = 1
2y – 3z = 1
3x – 2y + 4z = 2
Solution Consider the product
1
1
2
2
0
1
0
2
3
9
2
3
3
2
4
6
1
2
–
–
–
–
–
–
=
2
9
12
0
2
2
1
3
4
0
18 18
0
4
3
0
6
6
6 18
24
0
4
4
3
6
8
− −
+
−
+
+
−
+
−
+
−
−
+
− −
+
−
+
+
−
=
1
0
0
0
1
0
0
0
1
Hence
1
1
2
0
2
3
3
2
4
2
0
1
9
2
3
6
1
2
1
–
–
–
–
–
–
–
=
Now, given system of equations can be written, in matrix form, as follows
1
–1
2
0
2
–3
3
–2
4
x
y
z
=
1
1
2
Rationalised 2023-24
DETERMINANTS 99
or
x
y
z
=
1
1
1
2
1
0
2
3
1
3
2
4
2
−
−
−
−
=
�
�
�
2
0
1
9
2
3
6
1
2
1
1
2
=
2
0
2
0
9
2
6
5
6
1
4
3
− +
+
+
−
=
+ −
Hence
x = 0, y = 5 and z = 3
Miscellaneous Exercises on Chapter 4
1 Prove that the determinant
sin
cos
–sin
–
1
cos
1
x
x
x
θ
θ
θ
θ
is independent of θ 2 Evaluate
cos
cos
cos
sin
–sin
–sin
cos
0
sin
cos
sin
sin
cos
α
β
α
β
α
β
β
α
β
α
β
α |
1 | 1663-1666 | Prove that the determinant
sin
cos
–sin
–
1
cos
1
x
x
x
θ
θ
θ
θ
is independent of θ 2 Evaluate
cos
cos
cos
sin
–sin
–sin
cos
0
sin
cos
sin
sin
cos
α
β
α
β
α
β
β
α
β
α
β
α 3 |
1 | 1664-1667 | 2 Evaluate
cos
cos
cos
sin
–sin
–sin
cos
0
sin
cos
sin
sin
cos
α
β
α
β
α
β
β
α
β
α
β
α 3 If A–1 =
(
)
1
3
1
1
1
2
2
15
6
5 and B
1
3
0
, find AB
5
2
2
0
2
1
–
–
–
–
–
–
–
–
=
4 |
1 | 1665-1668 | Evaluate
cos
cos
cos
sin
–sin
–sin
cos
0
sin
cos
sin
sin
cos
α
β
α
β
α
β
β
α
β
α
β
α 3 If A–1 =
(
)
1
3
1
1
1
2
2
15
6
5 and B
1
3
0
, find AB
5
2
2
0
2
1
–
–
–
–
–
–
–
–
=
4 Let A =
1
2
1
2
3
1
1
1
5
�
�
|
1 | 1666-1669 | 3 If A–1 =
(
)
1
3
1
1
1
2
2
15
6
5 and B
1
3
0
, find AB
5
2
2
0
2
1
–
–
–
–
–
–
–
–
=
4 Let A =
1
2
1
2
3
1
1
1
5
�
�
Verify that
(i) [adj A]–1 = adj (A–1)
(ii) (A–1)–1 = A
5 |
1 | 1667-1670 | If A–1 =
(
)
1
3
1
1
1
2
2
15
6
5 and B
1
3
0
, find AB
5
2
2
0
2
1
–
–
–
–
–
–
–
–
=
4 Let A =
1
2
1
2
3
1
1
1
5
�
�
Verify that
(i) [adj A]–1 = adj (A–1)
(ii) (A–1)–1 = A
5 Evaluate
x
y
x
y
y
x
y
x
x
y
x
y
+
+
+
6 |
1 | 1668-1671 | Let A =
1
2
1
2
3
1
1
1
5
�
�
Verify that
(i) [adj A]–1 = adj (A–1)
(ii) (A–1)–1 = A
5 Evaluate
x
y
x
y
y
x
y
x
x
y
x
y
+
+
+
6 Evaluate
1
1
1
x
y
x
y
y
x
x+ y
+
Rationalised 2023-24
100
MATHEMATICS
Using properties of determinants in Exercises 11 to 15, prove that:
7 |
1 | 1669-1672 | Verify that
(i) [adj A]–1 = adj (A–1)
(ii) (A–1)–1 = A
5 Evaluate
x
y
x
y
y
x
y
x
x
y
x
y
+
+
+
6 Evaluate
1
1
1
x
y
x
y
y
x
x+ y
+
Rationalised 2023-24
100
MATHEMATICS
Using properties of determinants in Exercises 11 to 15, prove that:
7 Solve the system of equations
2
3
10
4
+
+
=
x
y
z
4
6
5
1
+
=
x–
y
z
6
9
20
2
+
=
–
x
y
z
Choose the correct answer in Exercise 17 to 19 |
1 | 1670-1673 | Evaluate
x
y
x
y
y
x
y
x
x
y
x
y
+
+
+
6 Evaluate
1
1
1
x
y
x
y
y
x
x+ y
+
Rationalised 2023-24
100
MATHEMATICS
Using properties of determinants in Exercises 11 to 15, prove that:
7 Solve the system of equations
2
3
10
4
+
+
=
x
y
z
4
6
5
1
+
=
x–
y
z
6
9
20
2
+
=
–
x
y
z
Choose the correct answer in Exercise 17 to 19 8 |
1 | 1671-1674 | Evaluate
1
1
1
x
y
x
y
y
x
x+ y
+
Rationalised 2023-24
100
MATHEMATICS
Using properties of determinants in Exercises 11 to 15, prove that:
7 Solve the system of equations
2
3
10
4
+
+
=
x
y
z
4
6
5
1
+
=
x–
y
z
6
9
20
2
+
=
–
x
y
z
Choose the correct answer in Exercise 17 to 19 8 If x, y, z are nonzero real numbers, then the inverse of matrix
0
0
A
0
0
0
0
x
y
z
=
is
(A)
1
1
1
0
0
0
0
0
0
x
y
z
−
−
−
(B)
1
1
1
0
0
0
0
0
0
x
xyz
y
z
−
−
−
(C)
0
0
1
0
0
0
0
x
y
xyz
z
(D)
1
0
0
1
0
1
0
0
0
1
xyz
9 |
1 | 1672-1675 | Solve the system of equations
2
3
10
4
+
+
=
x
y
z
4
6
5
1
+
=
x–
y
z
6
9
20
2
+
=
–
x
y
z
Choose the correct answer in Exercise 17 to 19 8 If x, y, z are nonzero real numbers, then the inverse of matrix
0
0
A
0
0
0
0
x
y
z
=
is
(A)
1
1
1
0
0
0
0
0
0
x
y
z
−
−
−
(B)
1
1
1
0
0
0
0
0
0
x
xyz
y
z
−
−
−
(C)
0
0
1
0
0
0
0
x
y
xyz
z
(D)
1
0
0
1
0
1
0
0
0
1
xyz
9 Let A =
1
sin
1
sin
1
sin
1
sin
1
θ
−
θ
θ
−
−
θ
, where 0 ≤ θ ≤ 2π |
1 | 1673-1676 | 8 If x, y, z are nonzero real numbers, then the inverse of matrix
0
0
A
0
0
0
0
x
y
z
=
is
(A)
1
1
1
0
0
0
0
0
0
x
y
z
−
−
−
(B)
1
1
1
0
0
0
0
0
0
x
xyz
y
z
−
−
−
(C)
0
0
1
0
0
0
0
x
y
xyz
z
(D)
1
0
0
1
0
1
0
0
0
1
xyz
9 Let A =
1
sin
1
sin
1
sin
1
sin
1
θ
−
θ
θ
−
−
θ
, where 0 ≤ θ ≤ 2π Then
(A) Det (A) = 0
(B) Det (A) ∈ (2, ∞)
(C) Det (A) ∈ (2, 4)
(D) Det (A) ∈ [2, 4]
Rationalised 2023-24
DETERMINANTS 101
Summary
® Determinant of a matrix A = [a11]1×1 is given by |a11| = a11
® Determinant of a matrix A =
a
a
a
a
11
12
21
22
is given by
11
12
21
22
A
a
a
a
a
=
= a11 a22 – a12 a21
® Determinant of a matrix A =
a
b
c
a
b
c
a
b
c
1
1
1
2
2
2
3
3
3
is given by (expanding along R1)
1
1
1
2
2
2
2
2
2
2
2
2
1
1
1
3
3
3
3
3
3
3
3
3
A
a
b
c
b
c
a
c
a
b
a
b
c
a
b
c
b
c
a
c
a
b
a
b
c
=
=
−
+
For any square matrix A, the |A| satisfy following properties |
1 | 1674-1677 | If x, y, z are nonzero real numbers, then the inverse of matrix
0
0
A
0
0
0
0
x
y
z
=
is
(A)
1
1
1
0
0
0
0
0
0
x
y
z
−
−
−
(B)
1
1
1
0
0
0
0
0
0
x
xyz
y
z
−
−
−
(C)
0
0
1
0
0
0
0
x
y
xyz
z
(D)
1
0
0
1
0
1
0
0
0
1
xyz
9 Let A =
1
sin
1
sin
1
sin
1
sin
1
θ
−
θ
θ
−
−
θ
, where 0 ≤ θ ≤ 2π Then
(A) Det (A) = 0
(B) Det (A) ∈ (2, ∞)
(C) Det (A) ∈ (2, 4)
(D) Det (A) ∈ [2, 4]
Rationalised 2023-24
DETERMINANTS 101
Summary
® Determinant of a matrix A = [a11]1×1 is given by |a11| = a11
® Determinant of a matrix A =
a
a
a
a
11
12
21
22
is given by
11
12
21
22
A
a
a
a
a
=
= a11 a22 – a12 a21
® Determinant of a matrix A =
a
b
c
a
b
c
a
b
c
1
1
1
2
2
2
3
3
3
is given by (expanding along R1)
1
1
1
2
2
2
2
2
2
2
2
2
1
1
1
3
3
3
3
3
3
3
3
3
A
a
b
c
b
c
a
c
a
b
a
b
c
a
b
c
b
c
a
c
a
b
a
b
c
=
=
−
+
For any square matrix A, the |A| satisfy following properties ® Area of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is given by
1
1
2
2
3
3
1
1
1
2
1
x
y
x
y
x
y
∆=
® Minor of an element aij of the determinant of matrix A is the determinant
obtained by deleting ith row and jth column and denoted by Mij |
1 | 1675-1678 | Let A =
1
sin
1
sin
1
sin
1
sin
1
θ
−
θ
θ
−
−
θ
, where 0 ≤ θ ≤ 2π Then
(A) Det (A) = 0
(B) Det (A) ∈ (2, ∞)
(C) Det (A) ∈ (2, 4)
(D) Det (A) ∈ [2, 4]
Rationalised 2023-24
DETERMINANTS 101
Summary
® Determinant of a matrix A = [a11]1×1 is given by |a11| = a11
® Determinant of a matrix A =
a
a
a
a
11
12
21
22
is given by
11
12
21
22
A
a
a
a
a
=
= a11 a22 – a12 a21
® Determinant of a matrix A =
a
b
c
a
b
c
a
b
c
1
1
1
2
2
2
3
3
3
is given by (expanding along R1)
1
1
1
2
2
2
2
2
2
2
2
2
1
1
1
3
3
3
3
3
3
3
3
3
A
a
b
c
b
c
a
c
a
b
a
b
c
a
b
c
b
c
a
c
a
b
a
b
c
=
=
−
+
For any square matrix A, the |A| satisfy following properties ® Area of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is given by
1
1
2
2
3
3
1
1
1
2
1
x
y
x
y
x
y
∆=
® Minor of an element aij of the determinant of matrix A is the determinant
obtained by deleting ith row and jth column and denoted by Mij ® Cofactor of aij of given by Aij = (– 1)i+ j Mij
® Value of determinant of a matrix A is obtained by sum of product of elements
of a row (or a column) with corresponding cofactors |
1 | 1676-1679 | Then
(A) Det (A) = 0
(B) Det (A) ∈ (2, ∞)
(C) Det (A) ∈ (2, 4)
(D) Det (A) ∈ [2, 4]
Rationalised 2023-24
DETERMINANTS 101
Summary
® Determinant of a matrix A = [a11]1×1 is given by |a11| = a11
® Determinant of a matrix A =
a
a
a
a
11
12
21
22
is given by
11
12
21
22
A
a
a
a
a
=
= a11 a22 – a12 a21
® Determinant of a matrix A =
a
b
c
a
b
c
a
b
c
1
1
1
2
2
2
3
3
3
is given by (expanding along R1)
1
1
1
2
2
2
2
2
2
2
2
2
1
1
1
3
3
3
3
3
3
3
3
3
A
a
b
c
b
c
a
c
a
b
a
b
c
a
b
c
b
c
a
c
a
b
a
b
c
=
=
−
+
For any square matrix A, the |A| satisfy following properties ® Area of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is given by
1
1
2
2
3
3
1
1
1
2
1
x
y
x
y
x
y
∆=
® Minor of an element aij of the determinant of matrix A is the determinant
obtained by deleting ith row and jth column and denoted by Mij ® Cofactor of aij of given by Aij = (– 1)i+ j Mij
® Value of determinant of a matrix A is obtained by sum of product of elements
of a row (or a column) with corresponding cofactors For example,
A = a11 A11 + a12 A12 + a13 A13 |
1 | 1677-1680 | ® Area of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is given by
1
1
2
2
3
3
1
1
1
2
1
x
y
x
y
x
y
∆=
® Minor of an element aij of the determinant of matrix A is the determinant
obtained by deleting ith row and jth column and denoted by Mij ® Cofactor of aij of given by Aij = (– 1)i+ j Mij
® Value of determinant of a matrix A is obtained by sum of product of elements
of a row (or a column) with corresponding cofactors For example,
A = a11 A11 + a12 A12 + a13 A13 ® If elements of one row (or column) are multiplied with cofactors of elements
of any other row (or column), then their sum is zero |
1 | 1678-1681 | ® Cofactor of aij of given by Aij = (– 1)i+ j Mij
® Value of determinant of a matrix A is obtained by sum of product of elements
of a row (or a column) with corresponding cofactors For example,
A = a11 A11 + a12 A12 + a13 A13 ® If elements of one row (or column) are multiplied with cofactors of elements
of any other row (or column), then their sum is zero For example, a11 A21 + a12
A22 + a13 A23 = 0
Rationalised 2023-24
102
MATHEMATICS
® If
11
12
13
21
22
23
31
32
33
A
,
a
a
a
a
a
a
a
a
a
=
then
11
21
31
12
22
32
13
23
33
A
A
A
A
A
A
A
A
A
A
adj
=
, where Aij is
cofactor of aij
® A (adj A) = (adj A) A = |A| I, where A is square matrix of order n |
1 | 1679-1682 | For example,
A = a11 A11 + a12 A12 + a13 A13 ® If elements of one row (or column) are multiplied with cofactors of elements
of any other row (or column), then their sum is zero For example, a11 A21 + a12
A22 + a13 A23 = 0
Rationalised 2023-24
102
MATHEMATICS
® If
11
12
13
21
22
23
31
32
33
A
,
a
a
a
a
a
a
a
a
a
=
then
11
21
31
12
22
32
13
23
33
A
A
A
A
A
A
A
A
A
A
adj
=
, where Aij is
cofactor of aij
® A (adj A) = (adj A) A = |A| I, where A is square matrix of order n ® A square matrix A is said to be singular or non-singular according as
|A| = 0 or |A| ≠ 0 |
1 | 1680-1683 | ® If elements of one row (or column) are multiplied with cofactors of elements
of any other row (or column), then their sum is zero For example, a11 A21 + a12
A22 + a13 A23 = 0
Rationalised 2023-24
102
MATHEMATICS
® If
11
12
13
21
22
23
31
32
33
A
,
a
a
a
a
a
a
a
a
a
=
then
11
21
31
12
22
32
13
23
33
A
A
A
A
A
A
A
A
A
A
adj
=
, where Aij is
cofactor of aij
® A (adj A) = (adj A) A = |A| I, where A is square matrix of order n ® A square matrix A is said to be singular or non-singular according as
|A| = 0 or |A| ≠ 0 ® If AB = BA = I, where B is square matrix, then B is called inverse of A |
1 | 1681-1684 | For example, a11 A21 + a12
A22 + a13 A23 = 0
Rationalised 2023-24
102
MATHEMATICS
® If
11
12
13
21
22
23
31
32
33
A
,
a
a
a
a
a
a
a
a
a
=
then
11
21
31
12
22
32
13
23
33
A
A
A
A
A
A
A
A
A
A
adj
=
, where Aij is
cofactor of aij
® A (adj A) = (adj A) A = |A| I, where A is square matrix of order n ® A square matrix A is said to be singular or non-singular according as
|A| = 0 or |A| ≠ 0 ® If AB = BA = I, where B is square matrix, then B is called inverse of A Also A–1 = B or B–1 = A and hence (A–1)–1 = A |
1 | 1682-1685 | ® A square matrix A is said to be singular or non-singular according as
|A| = 0 or |A| ≠ 0 ® If AB = BA = I, where B is square matrix, then B is called inverse of A Also A–1 = B or B–1 = A and hence (A–1)–1 = A ® A square matrix A has inverse if and only if A is non-singular |
1 | 1683-1686 | ® If AB = BA = I, where B is square matrix, then B is called inverse of A Also A–1 = B or B–1 = A and hence (A–1)–1 = A ® A square matrix A has inverse if and only if A is non-singular ®
–1
1
A
(
A adjA)
=
® If
a1 x + b1 y + c1 z = d1
a2 x + b2 y + c2 z = d2
a3 x + b3 y + c3 z = d3,
then these equations can be written as A X = B, where
1
1
1
1
2
2
2
2
3
3
3
3
A
,X=
and B=
a
b
c
x
d
a
b
c
y
d
a
b
c
z
d
=
® Unique solution of equation AX = B is given by X = A–1 B, where A
≠0 |
1 | 1684-1687 | Also A–1 = B or B–1 = A and hence (A–1)–1 = A ® A square matrix A has inverse if and only if A is non-singular ®
–1
1
A
(
A adjA)
=
® If
a1 x + b1 y + c1 z = d1
a2 x + b2 y + c2 z = d2
a3 x + b3 y + c3 z = d3,
then these equations can be written as A X = B, where
1
1
1
1
2
2
2
2
3
3
3
3
A
,X=
and B=
a
b
c
x
d
a
b
c
y
d
a
b
c
z
d
=
® Unique solution of equation AX = B is given by X = A–1 B, where A
≠0 ® A system of equation is consistent or inconsistent according as its solution
exists or not |
1 | 1685-1688 | ® A square matrix A has inverse if and only if A is non-singular ®
–1
1
A
(
A adjA)
=
® If
a1 x + b1 y + c1 z = d1
a2 x + b2 y + c2 z = d2
a3 x + b3 y + c3 z = d3,
then these equations can be written as A X = B, where
1
1
1
1
2
2
2
2
3
3
3
3
A
,X=
and B=
a
b
c
x
d
a
b
c
y
d
a
b
c
z
d
=
® Unique solution of equation AX = B is given by X = A–1 B, where A
≠0 ® A system of equation is consistent or inconsistent according as its solution
exists or not ® For a square matrix A in matrix equation AX = B
(i) |A| ≠ 0, there exists unique solution
(ii) |A| = 0 and (adj A) B ≠ 0, then there exists no solution
(iii) |A| = 0 and (adj A) B = 0, then system may or may not be consistent |
1 | 1686-1689 | ®
–1
1
A
(
A adjA)
=
® If
a1 x + b1 y + c1 z = d1
a2 x + b2 y + c2 z = d2
a3 x + b3 y + c3 z = d3,
then these equations can be written as A X = B, where
1
1
1
1
2
2
2
2
3
3
3
3
A
,X=
and B=
a
b
c
x
d
a
b
c
y
d
a
b
c
z
d
=
® Unique solution of equation AX = B is given by X = A–1 B, where A
≠0 ® A system of equation is consistent or inconsistent according as its solution
exists or not ® For a square matrix A in matrix equation AX = B
(i) |A| ≠ 0, there exists unique solution
(ii) |A| = 0 and (adj A) B ≠ 0, then there exists no solution
(iii) |A| = 0 and (adj A) B = 0, then system may or may not be consistent Rationalised 2023-24
DETERMINANTS 103
Historical Note
The Chinese method of representing the coefficients of the unknowns of
several linear equations by using rods on a calculating board naturally led to the
discovery of simple method of elimination |
1 | 1687-1690 | ® A system of equation is consistent or inconsistent according as its solution
exists or not ® For a square matrix A in matrix equation AX = B
(i) |A| ≠ 0, there exists unique solution
(ii) |A| = 0 and (adj A) B ≠ 0, then there exists no solution
(iii) |A| = 0 and (adj A) B = 0, then system may or may not be consistent Rationalised 2023-24
DETERMINANTS 103
Historical Note
The Chinese method of representing the coefficients of the unknowns of
several linear equations by using rods on a calculating board naturally led to the
discovery of simple method of elimination The arrangement of rods was precisely
that of the numbers in a determinant |
1 | 1688-1691 | ® For a square matrix A in matrix equation AX = B
(i) |A| ≠ 0, there exists unique solution
(ii) |A| = 0 and (adj A) B ≠ 0, then there exists no solution
(iii) |A| = 0 and (adj A) B = 0, then system may or may not be consistent Rationalised 2023-24
DETERMINANTS 103
Historical Note
The Chinese method of representing the coefficients of the unknowns of
several linear equations by using rods on a calculating board naturally led to the
discovery of simple method of elimination The arrangement of rods was precisely
that of the numbers in a determinant The Chinese, therefore, early developed the
idea of subtracting columns and rows as in simplification of a determinant
Mikami, China, pp 30, 93 |
1 | 1689-1692 | Rationalised 2023-24
DETERMINANTS 103
Historical Note
The Chinese method of representing the coefficients of the unknowns of
several linear equations by using rods on a calculating board naturally led to the
discovery of simple method of elimination The arrangement of rods was precisely
that of the numbers in a determinant The Chinese, therefore, early developed the
idea of subtracting columns and rows as in simplification of a determinant
Mikami, China, pp 30, 93 Seki Kowa, the greatest of the Japanese Mathematicians of seventeenth
century in his work ‘Kai Fukudai no Ho’ in 1683 showed that he had the idea of
determinants and of their expansion |
1 | 1690-1693 | The arrangement of rods was precisely
that of the numbers in a determinant The Chinese, therefore, early developed the
idea of subtracting columns and rows as in simplification of a determinant
Mikami, China, pp 30, 93 Seki Kowa, the greatest of the Japanese Mathematicians of seventeenth
century in his work ‘Kai Fukudai no Ho’ in 1683 showed that he had the idea of
determinants and of their expansion But he used this device only in eliminating a
quantity from two equations and not directly in the solution of a set of simultaneous
linear equations |
1 | 1691-1694 | The Chinese, therefore, early developed the
idea of subtracting columns and rows as in simplification of a determinant
Mikami, China, pp 30, 93 Seki Kowa, the greatest of the Japanese Mathematicians of seventeenth
century in his work ‘Kai Fukudai no Ho’ in 1683 showed that he had the idea of
determinants and of their expansion But he used this device only in eliminating a
quantity from two equations and not directly in the solution of a set of simultaneous
linear equations T |
1 | 1692-1695 | Seki Kowa, the greatest of the Japanese Mathematicians of seventeenth
century in his work ‘Kai Fukudai no Ho’ in 1683 showed that he had the idea of
determinants and of their expansion But he used this device only in eliminating a
quantity from two equations and not directly in the solution of a set of simultaneous
linear equations T Hayashi, “The Fakudoi and Determinants in Japanese
Mathematics,” in the proc |
1 | 1693-1696 | But he used this device only in eliminating a
quantity from two equations and not directly in the solution of a set of simultaneous
linear equations T Hayashi, “The Fakudoi and Determinants in Japanese
Mathematics,” in the proc of the Tokyo Math |
1 | 1694-1697 | T Hayashi, “The Fakudoi and Determinants in Japanese
Mathematics,” in the proc of the Tokyo Math Soc |
1 | 1695-1698 | Hayashi, “The Fakudoi and Determinants in Japanese
Mathematics,” in the proc of the Tokyo Math Soc , V |
1 | 1696-1699 | of the Tokyo Math Soc , V Vendermonde was the first to recognise determinants as independent functions |
1 | 1697-1700 | Soc , V Vendermonde was the first to recognise determinants as independent functions He may be called the formal founder |
1 | 1698-1701 | , V Vendermonde was the first to recognise determinants as independent functions He may be called the formal founder Laplace (1772), gave general method of
expanding a determinant in terms of its complementary minors |
1 | 1699-1702 | Vendermonde was the first to recognise determinants as independent functions He may be called the formal founder Laplace (1772), gave general method of
expanding a determinant in terms of its complementary minors In 1773 Lagrange
treated determinants of the second and third orders and used them for purpose
other than the solution of equations |
1 | 1700-1703 | He may be called the formal founder Laplace (1772), gave general method of
expanding a determinant in terms of its complementary minors In 1773 Lagrange
treated determinants of the second and third orders and used them for purpose
other than the solution of equations In 1801, Gauss used determinants in his
theory of numbers |
1 | 1701-1704 | Laplace (1772), gave general method of
expanding a determinant in terms of its complementary minors In 1773 Lagrange
treated determinants of the second and third orders and used them for purpose
other than the solution of equations In 1801, Gauss used determinants in his
theory of numbers The next great contributor was Jacques - Philippe - Marie Binet, (1812) who
stated the theorem relating to the product of two matrices of m-columns and n-
rows, which for the special case of m = n reduces to the multiplication theorem |
1 | 1702-1705 | In 1773 Lagrange
treated determinants of the second and third orders and used them for purpose
other than the solution of equations In 1801, Gauss used determinants in his
theory of numbers The next great contributor was Jacques - Philippe - Marie Binet, (1812) who
stated the theorem relating to the product of two matrices of m-columns and n-
rows, which for the special case of m = n reduces to the multiplication theorem Also on the same day, Cauchy (1812) presented one on the same subject |
1 | 1703-1706 | In 1801, Gauss used determinants in his
theory of numbers The next great contributor was Jacques - Philippe - Marie Binet, (1812) who
stated the theorem relating to the product of two matrices of m-columns and n-
rows, which for the special case of m = n reduces to the multiplication theorem Also on the same day, Cauchy (1812) presented one on the same subject He
used the word ‘determinant’ in its present sense |
1 | 1704-1707 | The next great contributor was Jacques - Philippe - Marie Binet, (1812) who
stated the theorem relating to the product of two matrices of m-columns and n-
rows, which for the special case of m = n reduces to the multiplication theorem Also on the same day, Cauchy (1812) presented one on the same subject He
used the word ‘determinant’ in its present sense He gave the proof of multiplication
theorem more satisfactory than Binet’s |
1 | 1705-1708 | Also on the same day, Cauchy (1812) presented one on the same subject He
used the word ‘determinant’ in its present sense He gave the proof of multiplication
theorem more satisfactory than Binet’s The greatest contributor to the theory was Carl Gustav Jacob Jacobi, after
this the word determinant received its final acceptance |
1 | 1706-1709 | He
used the word ‘determinant’ in its present sense He gave the proof of multiplication
theorem more satisfactory than Binet’s The greatest contributor to the theory was Carl Gustav Jacob Jacobi, after
this the word determinant received its final acceptance Rationalised 2023-24
MATHEMATICS
104
vThe whole of science is nothing more than a refinement
of everyday thinking |
1 | 1707-1710 | He gave the proof of multiplication
theorem more satisfactory than Binet’s The greatest contributor to the theory was Carl Gustav Jacob Jacobi, after
this the word determinant received its final acceptance Rationalised 2023-24
MATHEMATICS
104
vThe whole of science is nothing more than a refinement
of everyday thinking ” — ALBERT EINSTEIN v
5 |
1 | 1708-1711 | The greatest contributor to the theory was Carl Gustav Jacob Jacobi, after
this the word determinant received its final acceptance Rationalised 2023-24
MATHEMATICS
104
vThe whole of science is nothing more than a refinement
of everyday thinking ” — ALBERT EINSTEIN v
5 1 Introduction
This chapter is essentially a continuation of our study of
differentiation of functions in Class XI |
1 | 1709-1712 | Rationalised 2023-24
MATHEMATICS
104
vThe whole of science is nothing more than a refinement
of everyday thinking ” — ALBERT EINSTEIN v
5 1 Introduction
This chapter is essentially a continuation of our study of
differentiation of functions in Class XI We had learnt to
differentiate certain functions like polynomial functions and
trigonometric functions |
1 | 1710-1713 | ” — ALBERT EINSTEIN v
5 1 Introduction
This chapter is essentially a continuation of our study of
differentiation of functions in Class XI We had learnt to
differentiate certain functions like polynomial functions and
trigonometric functions In this chapter, we introduce the
very important concepts of continuity, differentiability and
relations between them |
1 | 1711-1714 | 1 Introduction
This chapter is essentially a continuation of our study of
differentiation of functions in Class XI We had learnt to
differentiate certain functions like polynomial functions and
trigonometric functions In this chapter, we introduce the
very important concepts of continuity, differentiability and
relations between them We will also learn differentiation
of inverse trigonometric functions |
1 | 1712-1715 | We had learnt to
differentiate certain functions like polynomial functions and
trigonometric functions In this chapter, we introduce the
very important concepts of continuity, differentiability and
relations between them We will also learn differentiation
of inverse trigonometric functions Further, we introduce a
new class of functions called exponential and logarithmic
functions |
1 | 1713-1716 | In this chapter, we introduce the
very important concepts of continuity, differentiability and
relations between them We will also learn differentiation
of inverse trigonometric functions Further, we introduce a
new class of functions called exponential and logarithmic
functions These functions lead to powerful techniques of
differentiation |
1 | 1714-1717 | We will also learn differentiation
of inverse trigonometric functions Further, we introduce a
new class of functions called exponential and logarithmic
functions These functions lead to powerful techniques of
differentiation We illustrate certain geometrically obvious
conditions through differential calculus |
1 | 1715-1718 | Further, we introduce a
new class of functions called exponential and logarithmic
functions These functions lead to powerful techniques of
differentiation We illustrate certain geometrically obvious
conditions through differential calculus In the process, we
will learn some fundamental theorems in this area |
1 | 1716-1719 | These functions lead to powerful techniques of
differentiation We illustrate certain geometrically obvious
conditions through differential calculus In the process, we
will learn some fundamental theorems in this area 5 |
1 | 1717-1720 | We illustrate certain geometrically obvious
conditions through differential calculus In the process, we
will learn some fundamental theorems in this area 5 2 Continuity
We start the section with two informal examples to get a feel of continuity |
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