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1 | 1418-1421 | — C P STEINMETZ v
4 1 Introduction
In the previous chapter, we have studied about matrices
and algebra of matrices |
1 | 1419-1422 | P STEINMETZ v
4 1 Introduction
In the previous chapter, we have studied about matrices
and algebra of matrices We have also learnt that a system
of algebraic equations can be expressed in the form of
matrices |
1 | 1420-1423 | STEINMETZ v
4 1 Introduction
In the previous chapter, we have studied about matrices
and algebra of matrices We have also learnt that a system
of algebraic equations can be expressed in the form of
matrices This means, a system of linear equations like
a1 x + b1 y = c1
a2 x + b2 y = c2
can be represented as
1
1
1
2
2
2
a
b
c
x
a
b
c
y
=
|
1 | 1421-1424 | 1 Introduction
In the previous chapter, we have studied about matrices
and algebra of matrices We have also learnt that a system
of algebraic equations can be expressed in the form of
matrices This means, a system of linear equations like
a1 x + b1 y = c1
a2 x + b2 y = c2
can be represented as
1
1
1
2
2
2
a
b
c
x
a
b
c
y
=
Now, this
system of equations has a unique solution or not, is
determined by the number a1 b2 – a2 b1 |
1 | 1422-1425 | We have also learnt that a system
of algebraic equations can be expressed in the form of
matrices This means, a system of linear equations like
a1 x + b1 y = c1
a2 x + b2 y = c2
can be represented as
1
1
1
2
2
2
a
b
c
x
a
b
c
y
=
Now, this
system of equations has a unique solution or not, is
determined by the number a1 b2 – a2 b1 (Recall that if
1
1
2
2
a
b
a
≠b
or, a1 b2 – a2 b1 ≠ 0, then the system of linear
equations has a unique solution) |
1 | 1423-1426 | This means, a system of linear equations like
a1 x + b1 y = c1
a2 x + b2 y = c2
can be represented as
1
1
1
2
2
2
a
b
c
x
a
b
c
y
=
Now, this
system of equations has a unique solution or not, is
determined by the number a1 b2 – a2 b1 (Recall that if
1
1
2
2
a
b
a
≠b
or, a1 b2 – a2 b1 ≠ 0, then the system of linear
equations has a unique solution) The number a1 b2 – a2 b1
which determines uniqueness of solution is associated with the matrix
1
1
2
2
A
a
b
a
b
=
and is called the determinant of A or det A |
1 | 1424-1427 | Now, this
system of equations has a unique solution or not, is
determined by the number a1 b2 – a2 b1 (Recall that if
1
1
2
2
a
b
a
≠b
or, a1 b2 – a2 b1 ≠ 0, then the system of linear
equations has a unique solution) The number a1 b2 – a2 b1
which determines uniqueness of solution is associated with the matrix
1
1
2
2
A
a
b
a
b
=
and is called the determinant of A or det A Determinants have wide applications in
Engineering, Science, Economics, Social Science, etc |
1 | 1425-1428 | (Recall that if
1
1
2
2
a
b
a
≠b
or, a1 b2 – a2 b1 ≠ 0, then the system of linear
equations has a unique solution) The number a1 b2 – a2 b1
which determines uniqueness of solution is associated with the matrix
1
1
2
2
A
a
b
a
b
=
and is called the determinant of A or det A Determinants have wide applications in
Engineering, Science, Economics, Social Science, etc In this chapter, we shall study determinants up to order three only with real entries |
1 | 1426-1429 | The number a1 b2 – a2 b1
which determines uniqueness of solution is associated with the matrix
1
1
2
2
A
a
b
a
b
=
and is called the determinant of A or det A Determinants have wide applications in
Engineering, Science, Economics, Social Science, etc In this chapter, we shall study determinants up to order three only with real entries Also, we will study various properties of determinants, minors, cofactors and applications
of determinants in finding the area of a triangle, adjoint and inverse of a square matrix,
consistency and inconsistency of system of linear equations and solution of linear
equations in two or three variables using inverse of a matrix |
1 | 1427-1430 | Determinants have wide applications in
Engineering, Science, Economics, Social Science, etc In this chapter, we shall study determinants up to order three only with real entries Also, we will study various properties of determinants, minors, cofactors and applications
of determinants in finding the area of a triangle, adjoint and inverse of a square matrix,
consistency and inconsistency of system of linear equations and solution of linear
equations in two or three variables using inverse of a matrix 4 |
1 | 1428-1431 | In this chapter, we shall study determinants up to order three only with real entries Also, we will study various properties of determinants, minors, cofactors and applications
of determinants in finding the area of a triangle, adjoint and inverse of a square matrix,
consistency and inconsistency of system of linear equations and solution of linear
equations in two or three variables using inverse of a matrix 4 2 Determinant
To every square matrix A = [aij] of order n, we can associate a number (real or
complex) called determinant of the square matrix A, where aij = (i, j)th element of A |
1 | 1429-1432 | Also, we will study various properties of determinants, minors, cofactors and applications
of determinants in finding the area of a triangle, adjoint and inverse of a square matrix,
consistency and inconsistency of system of linear equations and solution of linear
equations in two or three variables using inverse of a matrix 4 2 Determinant
To every square matrix A = [aij] of order n, we can associate a number (real or
complex) called determinant of the square matrix A, where aij = (i, j)th element of A Chapter 4
DETERMINANTS
P |
1 | 1430-1433 | 4 2 Determinant
To every square matrix A = [aij] of order n, we can associate a number (real or
complex) called determinant of the square matrix A, where aij = (i, j)th element of A Chapter 4
DETERMINANTS
P S |
1 | 1431-1434 | 2 Determinant
To every square matrix A = [aij] of order n, we can associate a number (real or
complex) called determinant of the square matrix A, where aij = (i, j)th element of A Chapter 4
DETERMINANTS
P S Laplace
(1749-1827)
Rationalised 2023-24
DETERMINANTS 77
This may be thought of as a function which associates each square matrix with a
unique number (real or complex) |
1 | 1432-1435 | Chapter 4
DETERMINANTS
P S Laplace
(1749-1827)
Rationalised 2023-24
DETERMINANTS 77
This may be thought of as a function which associates each square matrix with a
unique number (real or complex) If M is the set of square matrices, K is the set of
numbers (real or complex) and f : M → K is defined by f (A) = k, where A ∈ M and
k ∈ K, then f (A) is called the determinant of A |
1 | 1433-1436 | S Laplace
(1749-1827)
Rationalised 2023-24
DETERMINANTS 77
This may be thought of as a function which associates each square matrix with a
unique number (real or complex) If M is the set of square matrices, K is the set of
numbers (real or complex) and f : M → K is defined by f (A) = k, where A ∈ M and
k ∈ K, then f (A) is called the determinant of A It is also denoted by |A| or det A or ∆ |
1 | 1434-1437 | Laplace
(1749-1827)
Rationalised 2023-24
DETERMINANTS 77
This may be thought of as a function which associates each square matrix with a
unique number (real or complex) If M is the set of square matrices, K is the set of
numbers (real or complex) and f : M → K is defined by f (A) = k, where A ∈ M and
k ∈ K, then f (A) is called the determinant of A It is also denoted by |A| or det A or ∆ If A =
a
b
c
d
, then determinant of A is written as |A| =
a
b
c
d = det (A)
Remarks
(i)
For matrix A, |A| is read as determinant of A and not modulus of A |
1 | 1435-1438 | If M is the set of square matrices, K is the set of
numbers (real or complex) and f : M → K is defined by f (A) = k, where A ∈ M and
k ∈ K, then f (A) is called the determinant of A It is also denoted by |A| or det A or ∆ If A =
a
b
c
d
, then determinant of A is written as |A| =
a
b
c
d = det (A)
Remarks
(i)
For matrix A, |A| is read as determinant of A and not modulus of A (ii)
Only square matrices have determinants |
1 | 1436-1439 | It is also denoted by |A| or det A or ∆ If A =
a
b
c
d
, then determinant of A is written as |A| =
a
b
c
d = det (A)
Remarks
(i)
For matrix A, |A| is read as determinant of A and not modulus of A (ii)
Only square matrices have determinants 4 |
1 | 1437-1440 | If A =
a
b
c
d
, then determinant of A is written as |A| =
a
b
c
d = det (A)
Remarks
(i)
For matrix A, |A| is read as determinant of A and not modulus of A (ii)
Only square matrices have determinants 4 2 |
1 | 1438-1441 | (ii)
Only square matrices have determinants 4 2 1 Determinant of a matrix of order one
Let A = [a] be the matrix of order 1, then determinant of A is defined to be equal to a
4 |
1 | 1439-1442 | 4 2 1 Determinant of a matrix of order one
Let A = [a] be the matrix of order 1, then determinant of A is defined to be equal to a
4 2 |
1 | 1440-1443 | 2 1 Determinant of a matrix of order one
Let A = [a] be the matrix of order 1, then determinant of A is defined to be equal to a
4 2 2 Determinant of a matrix of order two
Let
A =
11
12
21
22
a
a
a
a
be a matrix of order 2 × 2,
then the determinant of A is defined as:
det (A) = |A| = ∆ =
= a11a22 – a21a12
Example 1 Evaluate
2
4
–1
2 |
1 | 1441-1444 | 1 Determinant of a matrix of order one
Let A = [a] be the matrix of order 1, then determinant of A is defined to be equal to a
4 2 2 Determinant of a matrix of order two
Let
A =
11
12
21
22
a
a
a
a
be a matrix of order 2 × 2,
then the determinant of A is defined as:
det (A) = |A| = ∆ =
= a11a22 – a21a12
Example 1 Evaluate
2
4
–1
2 Solution We have
2
4
–1
2 = 2(2) – 4(–1) = 4 + 4 = 8 |
1 | 1442-1445 | 2 2 Determinant of a matrix of order two
Let
A =
11
12
21
22
a
a
a
a
be a matrix of order 2 × 2,
then the determinant of A is defined as:
det (A) = |A| = ∆ =
= a11a22 – a21a12
Example 1 Evaluate
2
4
–1
2 Solution We have
2
4
–1
2 = 2(2) – 4(–1) = 4 + 4 = 8 Example 2 Evaluate
1
x– 1
x
x
x
+
Solution We have
1
x– 1
x
x
x
+
= x (x) – (x + 1) (x – 1) = x2 – (x2 – 1) = x2 – x2 + 1 = 1
4 |
1 | 1443-1446 | 2 Determinant of a matrix of order two
Let
A =
11
12
21
22
a
a
a
a
be a matrix of order 2 × 2,
then the determinant of A is defined as:
det (A) = |A| = ∆ =
= a11a22 – a21a12
Example 1 Evaluate
2
4
–1
2 Solution We have
2
4
–1
2 = 2(2) – 4(–1) = 4 + 4 = 8 Example 2 Evaluate
1
x– 1
x
x
x
+
Solution We have
1
x– 1
x
x
x
+
= x (x) – (x + 1) (x – 1) = x2 – (x2 – 1) = x2 – x2 + 1 = 1
4 2 |
1 | 1444-1447 | Solution We have
2
4
–1
2 = 2(2) – 4(–1) = 4 + 4 = 8 Example 2 Evaluate
1
x– 1
x
x
x
+
Solution We have
1
x– 1
x
x
x
+
= x (x) – (x + 1) (x – 1) = x2 – (x2 – 1) = x2 – x2 + 1 = 1
4 2 3 Determinant of a matrix of order 3 × 3
Determinant of a matrix of order three can be determined by expressing it in terms of
second order determinants |
1 | 1445-1448 | Example 2 Evaluate
1
x– 1
x
x
x
+
Solution We have
1
x– 1
x
x
x
+
= x (x) – (x + 1) (x – 1) = x2 – (x2 – 1) = x2 – x2 + 1 = 1
4 2 3 Determinant of a matrix of order 3 × 3
Determinant of a matrix of order three can be determined by expressing it in terms of
second order determinants This is known as expansion of a determinant along
a row (or a column) |
1 | 1446-1449 | 2 3 Determinant of a matrix of order 3 × 3
Determinant of a matrix of order three can be determined by expressing it in terms of
second order determinants This is known as expansion of a determinant along
a row (or a column) There are six ways of expanding a determinant of order
Rationalised 2023-24
78
MATHEMATICS
3 corresponding to each of three rows (R1, R2 and R3) and three columns (C1, C2 and
C3) giving the same value as shown below |
1 | 1447-1450 | 3 Determinant of a matrix of order 3 × 3
Determinant of a matrix of order three can be determined by expressing it in terms of
second order determinants This is known as expansion of a determinant along
a row (or a column) There are six ways of expanding a determinant of order
Rationalised 2023-24
78
MATHEMATICS
3 corresponding to each of three rows (R1, R2 and R3) and three columns (C1, C2 and
C3) giving the same value as shown below Consider the determinant of square matrix A = [aij]3 × 3
i |
1 | 1448-1451 | This is known as expansion of a determinant along
a row (or a column) There are six ways of expanding a determinant of order
Rationalised 2023-24
78
MATHEMATICS
3 corresponding to each of three rows (R1, R2 and R3) and three columns (C1, C2 and
C3) giving the same value as shown below Consider the determinant of square matrix A = [aij]3 × 3
i e |
1 | 1449-1452 | There are six ways of expanding a determinant of order
Rationalised 2023-24
78
MATHEMATICS
3 corresponding to each of three rows (R1, R2 and R3) and three columns (C1, C2 and
C3) giving the same value as shown below Consider the determinant of square matrix A = [aij]3 × 3
i e ,
| A | =
21
22
23
31
32
33
a
a
a
a
a
a
11
12
13
a
a
a
Expansion along first Row (R1)
Step 1 Multiply first element a11 of R1 by (–1)(1 + 1) [(–1)sum of suffixes in a11] and with the
second order determinant obtained by deleting the elements of first row (R1) and first
column (C1) of | A | as a11 lies in R1 and C1,
i |
1 | 1450-1453 | Consider the determinant of square matrix A = [aij]3 × 3
i e ,
| A | =
21
22
23
31
32
33
a
a
a
a
a
a
11
12
13
a
a
a
Expansion along first Row (R1)
Step 1 Multiply first element a11 of R1 by (–1)(1 + 1) [(–1)sum of suffixes in a11] and with the
second order determinant obtained by deleting the elements of first row (R1) and first
column (C1) of | A | as a11 lies in R1 and C1,
i e |
1 | 1451-1454 | e ,
| A | =
21
22
23
31
32
33
a
a
a
a
a
a
11
12
13
a
a
a
Expansion along first Row (R1)
Step 1 Multiply first element a11 of R1 by (–1)(1 + 1) [(–1)sum of suffixes in a11] and with the
second order determinant obtained by deleting the elements of first row (R1) and first
column (C1) of | A | as a11 lies in R1 and C1,
i e ,
(–1)1 + 1 a11
22
23
32
33
a
a
a
a
Step 2 Multiply 2nd element a12 of R1 by (–1)1 + 2 [(–1)sum of suffixes in a12] and the second
order determinant obtained by deleting elements of first row (R1) and 2nd column (C2)
of | A | as a12 lies in R1 and C2,
i |
1 | 1452-1455 | ,
| A | =
21
22
23
31
32
33
a
a
a
a
a
a
11
12
13
a
a
a
Expansion along first Row (R1)
Step 1 Multiply first element a11 of R1 by (–1)(1 + 1) [(–1)sum of suffixes in a11] and with the
second order determinant obtained by deleting the elements of first row (R1) and first
column (C1) of | A | as a11 lies in R1 and C1,
i e ,
(–1)1 + 1 a11
22
23
32
33
a
a
a
a
Step 2 Multiply 2nd element a12 of R1 by (–1)1 + 2 [(–1)sum of suffixes in a12] and the second
order determinant obtained by deleting elements of first row (R1) and 2nd column (C2)
of | A | as a12 lies in R1 and C2,
i e |
1 | 1453-1456 | e ,
(–1)1 + 1 a11
22
23
32
33
a
a
a
a
Step 2 Multiply 2nd element a12 of R1 by (–1)1 + 2 [(–1)sum of suffixes in a12] and the second
order determinant obtained by deleting elements of first row (R1) and 2nd column (C2)
of | A | as a12 lies in R1 and C2,
i e ,
(–1)1 + 2 a12
21
23
31
33
a
a
a
a
Step 3 Multiply third element a13 of R1 by (–1)1 + 3 [(–1)sum of suffixes in a13] and the second
order determinant obtained by deleting elements of first row (R1) and third column (C3)
of | A | as a13 lies in R1 and C3,
i |
1 | 1454-1457 | ,
(–1)1 + 1 a11
22
23
32
33
a
a
a
a
Step 2 Multiply 2nd element a12 of R1 by (–1)1 + 2 [(–1)sum of suffixes in a12] and the second
order determinant obtained by deleting elements of first row (R1) and 2nd column (C2)
of | A | as a12 lies in R1 and C2,
i e ,
(–1)1 + 2 a12
21
23
31
33
a
a
a
a
Step 3 Multiply third element a13 of R1 by (–1)1 + 3 [(–1)sum of suffixes in a13] and the second
order determinant obtained by deleting elements of first row (R1) and third column (C3)
of | A | as a13 lies in R1 and C3,
i e |
1 | 1455-1458 | e ,
(–1)1 + 2 a12
21
23
31
33
a
a
a
a
Step 3 Multiply third element a13 of R1 by (–1)1 + 3 [(–1)sum of suffixes in a13] and the second
order determinant obtained by deleting elements of first row (R1) and third column (C3)
of | A | as a13 lies in R1 and C3,
i e ,
(–1)1 + 3 a13
21
22
31
32
a
a
a
a
Step 4 Now the expansion of determinant of A, that is, | A | written as sum of all three
terms obtained in steps 1, 2 and 3 above is given by
det A = |A| = (–1)1 + 1 a11
22
23
21
23
1
2
12
32
33
31
33
(–1)
a
a
a
a
a
a
a
a
a
+
+
+
21
22
1
3
13
31
32
(–1)
a
a
a
a
a
+
or
|A| = a11 (a22 a33 – a32 a23) – a12 (a21 a33 – a31 a23)
+ a13 (a21 a32 – a31 a22)
Rationalised 2023-24
DETERMINANTS 79
= a11 a22 a33 – a11 a32 a23 – a12 a21 a33 + a12 a31 a23 + a13 a21 a32
– a13 a31 a22 |
1 | 1456-1459 | ,
(–1)1 + 2 a12
21
23
31
33
a
a
a
a
Step 3 Multiply third element a13 of R1 by (–1)1 + 3 [(–1)sum of suffixes in a13] and the second
order determinant obtained by deleting elements of first row (R1) and third column (C3)
of | A | as a13 lies in R1 and C3,
i e ,
(–1)1 + 3 a13
21
22
31
32
a
a
a
a
Step 4 Now the expansion of determinant of A, that is, | A | written as sum of all three
terms obtained in steps 1, 2 and 3 above is given by
det A = |A| = (–1)1 + 1 a11
22
23
21
23
1
2
12
32
33
31
33
(–1)
a
a
a
a
a
a
a
a
a
+
+
+
21
22
1
3
13
31
32
(–1)
a
a
a
a
a
+
or
|A| = a11 (a22 a33 – a32 a23) – a12 (a21 a33 – a31 a23)
+ a13 (a21 a32 – a31 a22)
Rationalised 2023-24
DETERMINANTS 79
= a11 a22 a33 – a11 a32 a23 – a12 a21 a33 + a12 a31 a23 + a13 a21 a32
– a13 a31 a22 (1)
ANote We shall apply all four steps together |
1 | 1457-1460 | e ,
(–1)1 + 3 a13
21
22
31
32
a
a
a
a
Step 4 Now the expansion of determinant of A, that is, | A | written as sum of all three
terms obtained in steps 1, 2 and 3 above is given by
det A = |A| = (–1)1 + 1 a11
22
23
21
23
1
2
12
32
33
31
33
(–1)
a
a
a
a
a
a
a
a
a
+
+
+
21
22
1
3
13
31
32
(–1)
a
a
a
a
a
+
or
|A| = a11 (a22 a33 – a32 a23) – a12 (a21 a33 – a31 a23)
+ a13 (a21 a32 – a31 a22)
Rationalised 2023-24
DETERMINANTS 79
= a11 a22 a33 – a11 a32 a23 – a12 a21 a33 + a12 a31 a23 + a13 a21 a32
– a13 a31 a22 (1)
ANote We shall apply all four steps together Expansion along second row (R2)
| A | =
11
12
13
31
32
33
a
a
a
a
a
a
21
22
23
a
a
a
Expanding along R2, we get
| A | =
12
13
11
13
2
1
2
2
21
22
32
33
31
33
(–1)
(–1)
a
a
a
a
a
a
a
a
a
a
+
+
+
11
12
2
3
23
31
32
(–1)
a
a
a
a
a
+
+
= – a21 (a12 a33 – a32 a13) + a22 (a11 a33 – a31 a13)
– a23 (a11 a32 – a31 a12)
| A | = – a21 a12 a33 + a21 a32 a13 + a22 a11 a33 – a22 a31 a13 – a23 a11 a32
+ a23 a31 a12
= a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32
– a13 a31 a22 |
1 | 1458-1461 | ,
(–1)1 + 3 a13
21
22
31
32
a
a
a
a
Step 4 Now the expansion of determinant of A, that is, | A | written as sum of all three
terms obtained in steps 1, 2 and 3 above is given by
det A = |A| = (–1)1 + 1 a11
22
23
21
23
1
2
12
32
33
31
33
(–1)
a
a
a
a
a
a
a
a
a
+
+
+
21
22
1
3
13
31
32
(–1)
a
a
a
a
a
+
or
|A| = a11 (a22 a33 – a32 a23) – a12 (a21 a33 – a31 a23)
+ a13 (a21 a32 – a31 a22)
Rationalised 2023-24
DETERMINANTS 79
= a11 a22 a33 – a11 a32 a23 – a12 a21 a33 + a12 a31 a23 + a13 a21 a32
– a13 a31 a22 (1)
ANote We shall apply all four steps together Expansion along second row (R2)
| A | =
11
12
13
31
32
33
a
a
a
a
a
a
21
22
23
a
a
a
Expanding along R2, we get
| A | =
12
13
11
13
2
1
2
2
21
22
32
33
31
33
(–1)
(–1)
a
a
a
a
a
a
a
a
a
a
+
+
+
11
12
2
3
23
31
32
(–1)
a
a
a
a
a
+
+
= – a21 (a12 a33 – a32 a13) + a22 (a11 a33 – a31 a13)
– a23 (a11 a32 – a31 a12)
| A | = – a21 a12 a33 + a21 a32 a13 + a22 a11 a33 – a22 a31 a13 – a23 a11 a32
+ a23 a31 a12
= a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32
– a13 a31 a22 (2)
Expansion along first Column (C1)
| A | =
12
13
22
23
32
33
11
21
31
a
a
a
a
a
a
a
a
a
By expanding along C1, we get
| A | =
22
23
12
13
1
1
2
1
11
21
32
33
32
33
(–1)
( 1)
a
a
a
a
a
a
a
a
a
a
+
+
+
−
+
12
13
3
1
31
22
23
(–1)
a
a
a
a
a
+
= a11 (a22 a33 – a23 a32) – a21 (a12 a33 – a13 a32) + a31 (a12 a23 – a13 a22)
Rationalised 2023-24
80
MATHEMATICS
| A | = a11 a22 a33 – a11 a23 a32 – a21 a12 a33 + a21 a13 a32 + a31 a12 a23
– a31 a13 a22
= a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32
– a13 a31 a22 |
1 | 1459-1462 | (1)
ANote We shall apply all four steps together Expansion along second row (R2)
| A | =
11
12
13
31
32
33
a
a
a
a
a
a
21
22
23
a
a
a
Expanding along R2, we get
| A | =
12
13
11
13
2
1
2
2
21
22
32
33
31
33
(–1)
(–1)
a
a
a
a
a
a
a
a
a
a
+
+
+
11
12
2
3
23
31
32
(–1)
a
a
a
a
a
+
+
= – a21 (a12 a33 – a32 a13) + a22 (a11 a33 – a31 a13)
– a23 (a11 a32 – a31 a12)
| A | = – a21 a12 a33 + a21 a32 a13 + a22 a11 a33 – a22 a31 a13 – a23 a11 a32
+ a23 a31 a12
= a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32
– a13 a31 a22 (2)
Expansion along first Column (C1)
| A | =
12
13
22
23
32
33
11
21
31
a
a
a
a
a
a
a
a
a
By expanding along C1, we get
| A | =
22
23
12
13
1
1
2
1
11
21
32
33
32
33
(–1)
( 1)
a
a
a
a
a
a
a
a
a
a
+
+
+
−
+
12
13
3
1
31
22
23
(–1)
a
a
a
a
a
+
= a11 (a22 a33 – a23 a32) – a21 (a12 a33 – a13 a32) + a31 (a12 a23 – a13 a22)
Rationalised 2023-24
80
MATHEMATICS
| A | = a11 a22 a33 – a11 a23 a32 – a21 a12 a33 + a21 a13 a32 + a31 a12 a23
– a31 a13 a22
= a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32
– a13 a31 a22 (3)
Clearly, values of |A| in (1), (2) and (3) are equal |
1 | 1460-1463 | Expansion along second row (R2)
| A | =
11
12
13
31
32
33
a
a
a
a
a
a
21
22
23
a
a
a
Expanding along R2, we get
| A | =
12
13
11
13
2
1
2
2
21
22
32
33
31
33
(–1)
(–1)
a
a
a
a
a
a
a
a
a
a
+
+
+
11
12
2
3
23
31
32
(–1)
a
a
a
a
a
+
+
= – a21 (a12 a33 – a32 a13) + a22 (a11 a33 – a31 a13)
– a23 (a11 a32 – a31 a12)
| A | = – a21 a12 a33 + a21 a32 a13 + a22 a11 a33 – a22 a31 a13 – a23 a11 a32
+ a23 a31 a12
= a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32
– a13 a31 a22 (2)
Expansion along first Column (C1)
| A | =
12
13
22
23
32
33
11
21
31
a
a
a
a
a
a
a
a
a
By expanding along C1, we get
| A | =
22
23
12
13
1
1
2
1
11
21
32
33
32
33
(–1)
( 1)
a
a
a
a
a
a
a
a
a
a
+
+
+
−
+
12
13
3
1
31
22
23
(–1)
a
a
a
a
a
+
= a11 (a22 a33 – a23 a32) – a21 (a12 a33 – a13 a32) + a31 (a12 a23 – a13 a22)
Rationalised 2023-24
80
MATHEMATICS
| A | = a11 a22 a33 – a11 a23 a32 – a21 a12 a33 + a21 a13 a32 + a31 a12 a23
– a31 a13 a22
= a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32
– a13 a31 a22 (3)
Clearly, values of |A| in (1), (2) and (3) are equal It is left as an exercise to the
reader to verify that the values of |A| by expanding along R3, C2 and C3 are equal to the
value of |A| obtained in (1), (2) or (3) |
1 | 1461-1464 | (2)
Expansion along first Column (C1)
| A | =
12
13
22
23
32
33
11
21
31
a
a
a
a
a
a
a
a
a
By expanding along C1, we get
| A | =
22
23
12
13
1
1
2
1
11
21
32
33
32
33
(–1)
( 1)
a
a
a
a
a
a
a
a
a
a
+
+
+
−
+
12
13
3
1
31
22
23
(–1)
a
a
a
a
a
+
= a11 (a22 a33 – a23 a32) – a21 (a12 a33 – a13 a32) + a31 (a12 a23 – a13 a22)
Rationalised 2023-24
80
MATHEMATICS
| A | = a11 a22 a33 – a11 a23 a32 – a21 a12 a33 + a21 a13 a32 + a31 a12 a23
– a31 a13 a22
= a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32
– a13 a31 a22 (3)
Clearly, values of |A| in (1), (2) and (3) are equal It is left as an exercise to the
reader to verify that the values of |A| by expanding along R3, C2 and C3 are equal to the
value of |A| obtained in (1), (2) or (3) Hence, expanding a determinant along any row or column gives same value |
1 | 1462-1465 | (3)
Clearly, values of |A| in (1), (2) and (3) are equal It is left as an exercise to the
reader to verify that the values of |A| by expanding along R3, C2 and C3 are equal to the
value of |A| obtained in (1), (2) or (3) Hence, expanding a determinant along any row or column gives same value Remarks
(i)
For easier calculations, we shall expand the determinant along that row or column
which contains maximum number of zeros |
1 | 1463-1466 | It is left as an exercise to the
reader to verify that the values of |A| by expanding along R3, C2 and C3 are equal to the
value of |A| obtained in (1), (2) or (3) Hence, expanding a determinant along any row or column gives same value Remarks
(i)
For easier calculations, we shall expand the determinant along that row or column
which contains maximum number of zeros (ii)
While expanding, instead of multiplying by (–1)i + j, we can multiply by +1 or –1
according as (i + j) is even or odd |
1 | 1464-1467 | Hence, expanding a determinant along any row or column gives same value Remarks
(i)
For easier calculations, we shall expand the determinant along that row or column
which contains maximum number of zeros (ii)
While expanding, instead of multiplying by (–1)i + j, we can multiply by +1 or –1
according as (i + j) is even or odd (iii)
Let A = 2
2
4
0
and B =
1
1
2
0
|
1 | 1465-1468 | Remarks
(i)
For easier calculations, we shall expand the determinant along that row or column
which contains maximum number of zeros (ii)
While expanding, instead of multiplying by (–1)i + j, we can multiply by +1 or –1
according as (i + j) is even or odd (iii)
Let A = 2
2
4
0
and B =
1
1
2
0
Then, it is easy to verify that A = 2B |
1 | 1466-1469 | (ii)
While expanding, instead of multiplying by (–1)i + j, we can multiply by +1 or –1
according as (i + j) is even or odd (iii)
Let A = 2
2
4
0
and B =
1
1
2
0
Then, it is easy to verify that A = 2B Also
|A| = 0 – 8 = – 8 and |B| = 0 – 2 = – 2 |
1 | 1467-1470 | (iii)
Let A = 2
2
4
0
and B =
1
1
2
0
Then, it is easy to verify that A = 2B Also
|A| = 0 – 8 = – 8 and |B| = 0 – 2 = – 2 Observe that, |A| = 4(– 2) = 22|B| or |A| = 2n|B|, where n = 2 is the order of
square matrices A and B |
1 | 1468-1471 | Then, it is easy to verify that A = 2B Also
|A| = 0 – 8 = – 8 and |B| = 0 – 2 = – 2 Observe that, |A| = 4(– 2) = 22|B| or |A| = 2n|B|, where n = 2 is the order of
square matrices A and B In general, if A = kB where A and B are square matrices of order n, then | A| = kn
| B |, where n = 1, 2, 3
Example 3 Evaluate the determinant ∆ =
1
2
4
–1
3
0
4
1
0 |
1 | 1469-1472 | Also
|A| = 0 – 8 = – 8 and |B| = 0 – 2 = – 2 Observe that, |A| = 4(– 2) = 22|B| or |A| = 2n|B|, where n = 2 is the order of
square matrices A and B In general, if A = kB where A and B are square matrices of order n, then | A| = kn
| B |, where n = 1, 2, 3
Example 3 Evaluate the determinant ∆ =
1
2
4
–1
3
0
4
1
0 Solution Note that in the third column, two entries are zero |
1 | 1470-1473 | Observe that, |A| = 4(– 2) = 22|B| or |A| = 2n|B|, where n = 2 is the order of
square matrices A and B In general, if A = kB where A and B are square matrices of order n, then | A| = kn
| B |, where n = 1, 2, 3
Example 3 Evaluate the determinant ∆ =
1
2
4
–1
3
0
4
1
0 Solution Note that in the third column, two entries are zero So expanding along third
column (C3), we get
∆ =
–1
3
1
2
1
2
4
– 0
0
4
1
4
1
–1
3
+
= 4 (–1 – 12) – 0 + 0 = – 52
Example 4 Evaluate ∆ =
0
sin
–cos
–sin
0
sin
cos
–sin
0
α
α
α
β
α
β |
1 | 1471-1474 | In general, if A = kB where A and B are square matrices of order n, then | A| = kn
| B |, where n = 1, 2, 3
Example 3 Evaluate the determinant ∆ =
1
2
4
–1
3
0
4
1
0 Solution Note that in the third column, two entries are zero So expanding along third
column (C3), we get
∆ =
–1
3
1
2
1
2
4
– 0
0
4
1
4
1
–1
3
+
= 4 (–1 – 12) – 0 + 0 = – 52
Example 4 Evaluate ∆ =
0
sin
–cos
–sin
0
sin
cos
–sin
0
α
α
α
β
α
β Rationalised 2023-24
DETERMINANTS 81
Solution Expanding along R1, we get
∆ =
0
sin
–sin
sin
–sin
0
0
– sin
– cos
–sin
0
cos
0
cos
–sin
β
α
β
α
α
α
β
α
α
β
= 0 – sin α (0 – sin β cos α) – cos α (sin α sin β – 0)
= sin α sin β cos α – cos α sin α sin β = 0
Example 5 Find values of x for which 3
3
2
1
4
1
x
x
= |
1 | 1472-1475 | Solution Note that in the third column, two entries are zero So expanding along third
column (C3), we get
∆ =
–1
3
1
2
1
2
4
– 0
0
4
1
4
1
–1
3
+
= 4 (–1 – 12) – 0 + 0 = – 52
Example 4 Evaluate ∆ =
0
sin
–cos
–sin
0
sin
cos
–sin
0
α
α
α
β
α
β Rationalised 2023-24
DETERMINANTS 81
Solution Expanding along R1, we get
∆ =
0
sin
–sin
sin
–sin
0
0
– sin
– cos
–sin
0
cos
0
cos
–sin
β
α
β
α
α
α
β
α
α
β
= 0 – sin α (0 – sin β cos α) – cos α (sin α sin β – 0)
= sin α sin β cos α – cos α sin α sin β = 0
Example 5 Find values of x for which 3
3
2
1
4
1
x
x
= Solution We have 3
3
2
1
4
1
x
x
=
i |
1 | 1473-1476 | So expanding along third
column (C3), we get
∆ =
–1
3
1
2
1
2
4
– 0
0
4
1
4
1
–1
3
+
= 4 (–1 – 12) – 0 + 0 = – 52
Example 4 Evaluate ∆ =
0
sin
–cos
–sin
0
sin
cos
–sin
0
α
α
α
β
α
β Rationalised 2023-24
DETERMINANTS 81
Solution Expanding along R1, we get
∆ =
0
sin
–sin
sin
–sin
0
0
– sin
– cos
–sin
0
cos
0
cos
–sin
β
α
β
α
α
α
β
α
α
β
= 0 – sin α (0 – sin β cos α) – cos α (sin α sin β – 0)
= sin α sin β cos α – cos α sin α sin β = 0
Example 5 Find values of x for which 3
3
2
1
4
1
x
x
= Solution We have 3
3
2
1
4
1
x
x
=
i e |
1 | 1474-1477 | Rationalised 2023-24
DETERMINANTS 81
Solution Expanding along R1, we get
∆ =
0
sin
–sin
sin
–sin
0
0
– sin
– cos
–sin
0
cos
0
cos
–sin
β
α
β
α
α
α
β
α
α
β
= 0 – sin α (0 – sin β cos α) – cos α (sin α sin β – 0)
= sin α sin β cos α – cos α sin α sin β = 0
Example 5 Find values of x for which 3
3
2
1
4
1
x
x
= Solution We have 3
3
2
1
4
1
x
x
=
i e 3 – x2 = 3 – 8
i |
1 | 1475-1478 | Solution We have 3
3
2
1
4
1
x
x
=
i e 3 – x2 = 3 – 8
i e |
1 | 1476-1479 | e 3 – x2 = 3 – 8
i e x2 = 8
Hence
x =
2 2
±
EXERCISE 4 |
1 | 1477-1480 | 3 – x2 = 3 – 8
i e x2 = 8
Hence
x =
2 2
±
EXERCISE 4 1
Evaluate the determinants in Exercises 1 and 2 |
1 | 1478-1481 | e x2 = 8
Hence
x =
2 2
±
EXERCISE 4 1
Evaluate the determinants in Exercises 1 and 2 1 |
1 | 1479-1482 | x2 = 8
Hence
x =
2 2
±
EXERCISE 4 1
Evaluate the determinants in Exercises 1 and 2 1 2
4
–5
–1
2 |
1 | 1480-1483 | 1
Evaluate the determinants in Exercises 1 and 2 1 2
4
–5
–1
2 (i)
cos
–sin
sin
cos
θ
θ
θ
θ
(ii)
2 –
1
– 1
1
1
x
x
x
x
x
+
+
+
3 |
1 | 1481-1484 | 1 2
4
–5
–1
2 (i)
cos
–sin
sin
cos
θ
θ
θ
θ
(ii)
2 –
1
– 1
1
1
x
x
x
x
x
+
+
+
3 If
A =
1
2
4
2
, then show that | 2A | = 4 | A |
4 |
1 | 1482-1485 | 2
4
–5
–1
2 (i)
cos
–sin
sin
cos
θ
θ
θ
θ
(ii)
2 –
1
– 1
1
1
x
x
x
x
x
+
+
+
3 If
A =
1
2
4
2
, then show that | 2A | = 4 | A |
4 If
A =
1
0
1
0
1
2
0
0
4
, then show that | 3 A | = 27 | A |
5 |
1 | 1483-1486 | (i)
cos
–sin
sin
cos
θ
θ
θ
θ
(ii)
2 –
1
– 1
1
1
x
x
x
x
x
+
+
+
3 If
A =
1
2
4
2
, then show that | 2A | = 4 | A |
4 If
A =
1
0
1
0
1
2
0
0
4
, then show that | 3 A | = 27 | A |
5 Evaluate the determinants
(i)
3
–1
–2
0
0
–1
3
–5
0
(ii)
3
– 4
5
1
1
–2
2
3
1
Rationalised 2023-24
82
MATHEMATICS
(iii)
0
1
2
–1
0
–3
–2
3
0
(iv)
2
–1
–2
0
2
–1
3
–5
0
6 |
1 | 1484-1487 | If
A =
1
2
4
2
, then show that | 2A | = 4 | A |
4 If
A =
1
0
1
0
1
2
0
0
4
, then show that | 3 A | = 27 | A |
5 Evaluate the determinants
(i)
3
–1
–2
0
0
–1
3
–5
0
(ii)
3
– 4
5
1
1
–2
2
3
1
Rationalised 2023-24
82
MATHEMATICS
(iii)
0
1
2
–1
0
–3
–2
3
0
(iv)
2
–1
–2
0
2
–1
3
–5
0
6 If A =
1
1
–2
2
1
–3
5
4
–9
, find | A |
7 |
1 | 1485-1488 | If
A =
1
0
1
0
1
2
0
0
4
, then show that | 3 A | = 27 | A |
5 Evaluate the determinants
(i)
3
–1
–2
0
0
–1
3
–5
0
(ii)
3
– 4
5
1
1
–2
2
3
1
Rationalised 2023-24
82
MATHEMATICS
(iii)
0
1
2
–1
0
–3
–2
3
0
(iv)
2
–1
–2
0
2
–1
3
–5
0
6 If A =
1
1
–2
2
1
–3
5
4
–9
, find | A |
7 Find values of x, if
(i)
2
4
2
4
5
1
6
x
x
=
(ii)
2
3
3
4
5
2
5
x
x
=
8 |
1 | 1486-1489 | Evaluate the determinants
(i)
3
–1
–2
0
0
–1
3
–5
0
(ii)
3
– 4
5
1
1
–2
2
3
1
Rationalised 2023-24
82
MATHEMATICS
(iii)
0
1
2
–1
0
–3
–2
3
0
(iv)
2
–1
–2
0
2
–1
3
–5
0
6 If A =
1
1
–2
2
1
–3
5
4
–9
, find | A |
7 Find values of x, if
(i)
2
4
2
4
5
1
6
x
x
=
(ii)
2
3
3
4
5
2
5
x
x
=
8 If
2
6
2
18
18
6
x
x =
, then x is equal to
(A) 6
(B) ± 6
(C) – 6
(D) 0
4 |
1 | 1487-1490 | If A =
1
1
–2
2
1
–3
5
4
–9
, find | A |
7 Find values of x, if
(i)
2
4
2
4
5
1
6
x
x
=
(ii)
2
3
3
4
5
2
5
x
x
=
8 If
2
6
2
18
18
6
x
x =
, then x is equal to
(A) 6
(B) ± 6
(C) – 6
(D) 0
4 3 Area of a Triangle
In earlier classes, we have studied that the area of a triangle whose vertices are
(x1, y1), (x2, y2) and (x3, y3), is given by the expression 1
2 [x1(y2–y3) + x2 (y3–y1) +
x3 (y1–y2)] |
1 | 1488-1491 | Find values of x, if
(i)
2
4
2
4
5
1
6
x
x
=
(ii)
2
3
3
4
5
2
5
x
x
=
8 If
2
6
2
18
18
6
x
x =
, then x is equal to
(A) 6
(B) ± 6
(C) – 6
(D) 0
4 3 Area of a Triangle
In earlier classes, we have studied that the area of a triangle whose vertices are
(x1, y1), (x2, y2) and (x3, y3), is given by the expression 1
2 [x1(y2–y3) + x2 (y3–y1) +
x3 (y1–y2)] Now this expression can be written in the form of a determinant as
∆ =
1
1
2
2
3
3
1
1
1
2
1
x
y
x
y
x
y |
1 | 1489-1492 | If
2
6
2
18
18
6
x
x =
, then x is equal to
(A) 6
(B) ± 6
(C) – 6
(D) 0
4 3 Area of a Triangle
In earlier classes, we have studied that the area of a triangle whose vertices are
(x1, y1), (x2, y2) and (x3, y3), is given by the expression 1
2 [x1(y2–y3) + x2 (y3–y1) +
x3 (y1–y2)] Now this expression can be written in the form of a determinant as
∆ =
1
1
2
2
3
3
1
1
1
2
1
x
y
x
y
x
y (1)
Remarks
(i)
Since area is a positive quantity, we always take the absolute value of the
determinant in (1) |
1 | 1490-1493 | 3 Area of a Triangle
In earlier classes, we have studied that the area of a triangle whose vertices are
(x1, y1), (x2, y2) and (x3, y3), is given by the expression 1
2 [x1(y2–y3) + x2 (y3–y1) +
x3 (y1–y2)] Now this expression can be written in the form of a determinant as
∆ =
1
1
2
2
3
3
1
1
1
2
1
x
y
x
y
x
y (1)
Remarks
(i)
Since area is a positive quantity, we always take the absolute value of the
determinant in (1) (ii)
If area is given, use both positive and negative values of the determinant for
calculation |
1 | 1491-1494 | Now this expression can be written in the form of a determinant as
∆ =
1
1
2
2
3
3
1
1
1
2
1
x
y
x
y
x
y (1)
Remarks
(i)
Since area is a positive quantity, we always take the absolute value of the
determinant in (1) (ii)
If area is given, use both positive and negative values of the determinant for
calculation (iii)
The area of the triangle formed by three collinear points is zero |
1 | 1492-1495 | (1)
Remarks
(i)
Since area is a positive quantity, we always take the absolute value of the
determinant in (1) (ii)
If area is given, use both positive and negative values of the determinant for
calculation (iii)
The area of the triangle formed by three collinear points is zero Example 6 Find the area of the triangle whose vertices are (3, 8), (– 4, 2) and (5, 1) |
1 | 1493-1496 | (ii)
If area is given, use both positive and negative values of the determinant for
calculation (iii)
The area of the triangle formed by three collinear points is zero Example 6 Find the area of the triangle whose vertices are (3, 8), (– 4, 2) and (5, 1) Solution The area of triangle is given by
∆ =
3
8
1
1
4
2
1
2
5
1
1
–
Rationalised 2023-24
DETERMINANTS 83
=
(
)
(
)
(
)
1 3 2 –1 – 8 –4 – 5
1 –4 –10
2
+
=
(
)
1
61
3
72
14
2
2
–
+
=
Example 7 Find the equation of the line joining A(1, 3) and B (0, 0) using determinants
and find k if D(k, 0) is a point such that area of triangle ABD is 3sq units |
1 | 1494-1497 | (iii)
The area of the triangle formed by three collinear points is zero Example 6 Find the area of the triangle whose vertices are (3, 8), (– 4, 2) and (5, 1) Solution The area of triangle is given by
∆ =
3
8
1
1
4
2
1
2
5
1
1
–
Rationalised 2023-24
DETERMINANTS 83
=
(
)
(
)
(
)
1 3 2 –1 – 8 –4 – 5
1 –4 –10
2
+
=
(
)
1
61
3
72
14
2
2
–
+
=
Example 7 Find the equation of the line joining A(1, 3) and B (0, 0) using determinants
and find k if D(k, 0) is a point such that area of triangle ABD is 3sq units Solution Let P (x, y) be any point on AB |
1 | 1495-1498 | Example 6 Find the area of the triangle whose vertices are (3, 8), (– 4, 2) and (5, 1) Solution The area of triangle is given by
∆ =
3
8
1
1
4
2
1
2
5
1
1
–
Rationalised 2023-24
DETERMINANTS 83
=
(
)
(
)
(
)
1 3 2 –1 – 8 –4 – 5
1 –4 –10
2
+
=
(
)
1
61
3
72
14
2
2
–
+
=
Example 7 Find the equation of the line joining A(1, 3) and B (0, 0) using determinants
and find k if D(k, 0) is a point such that area of triangle ABD is 3sq units Solution Let P (x, y) be any point on AB Then, area of triangle ABP is zero (Why |
1 | 1496-1499 | Solution The area of triangle is given by
∆ =
3
8
1
1
4
2
1
2
5
1
1
–
Rationalised 2023-24
DETERMINANTS 83
=
(
)
(
)
(
)
1 3 2 –1 – 8 –4 – 5
1 –4 –10
2
+
=
(
)
1
61
3
72
14
2
2
–
+
=
Example 7 Find the equation of the line joining A(1, 3) and B (0, 0) using determinants
and find k if D(k, 0) is a point such that area of triangle ABD is 3sq units Solution Let P (x, y) be any point on AB Then, area of triangle ABP is zero (Why ) |
1 | 1497-1500 | Solution Let P (x, y) be any point on AB Then, area of triangle ABP is zero (Why ) So
0
0
1
1 1
3
1
2
1
x
y
= 0
This gives
(
)
1
2 y –3
x = 0 or y = 3x,
which is the equation of required line AB |
1 | 1498-1501 | Then, area of triangle ABP is zero (Why ) So
0
0
1
1 1
3
1
2
1
x
y
= 0
This gives
(
)
1
2 y –3
x = 0 or y = 3x,
which is the equation of required line AB Also, since the area of the triangle ABD is 3 sq |
1 | 1499-1502 | ) So
0
0
1
1 1
3
1
2
1
x
y
= 0
This gives
(
)
1
2 y –3
x = 0 or y = 3x,
which is the equation of required line AB Also, since the area of the triangle ABD is 3 sq units, we have
1
3
1
1 0
0
1
2
0
1
k
= ± 3
This gives,
3
3
2
−k
= ± , i |
1 | 1500-1503 | So
0
0
1
1 1
3
1
2
1
x
y
= 0
This gives
(
)
1
2 y –3
x = 0 or y = 3x,
which is the equation of required line AB Also, since the area of the triangle ABD is 3 sq units, we have
1
3
1
1 0
0
1
2
0
1
k
= ± 3
This gives,
3
3
2
−k
= ± , i e |
1 | 1501-1504 | Also, since the area of the triangle ABD is 3 sq units, we have
1
3
1
1 0
0
1
2
0
1
k
= ± 3
This gives,
3
3
2
−k
= ± , i e , k = ∓ 2 |
1 | 1502-1505 | units, we have
1
3
1
1 0
0
1
2
0
1
k
= ± 3
This gives,
3
3
2
−k
= ± , i e , k = ∓ 2 EXERCISE 4 |
1 | 1503-1506 | e , k = ∓ 2 EXERCISE 4 2
1 |
1 | 1504-1507 | , k = ∓ 2 EXERCISE 4 2
1 Find area of the triangle with vertices at the point given in each of the following :
(i) (1, 0), (6, 0), (4, 3)
(ii) (2, 7), (1, 1), (10, 8)
(iii) (–2, –3), (3, 2), (–1, –8)
2 |
1 | 1505-1508 | EXERCISE 4 2
1 Find area of the triangle with vertices at the point given in each of the following :
(i) (1, 0), (6, 0), (4, 3)
(ii) (2, 7), (1, 1), (10, 8)
(iii) (–2, –3), (3, 2), (–1, –8)
2 Show that points
A (a, b + c), B (b, c + a), C (c, a + b) are collinear |
1 | 1506-1509 | 2
1 Find area of the triangle with vertices at the point given in each of the following :
(i) (1, 0), (6, 0), (4, 3)
(ii) (2, 7), (1, 1), (10, 8)
(iii) (–2, –3), (3, 2), (–1, –8)
2 Show that points
A (a, b + c), B (b, c + a), C (c, a + b) are collinear 3 |
1 | 1507-1510 | Find area of the triangle with vertices at the point given in each of the following :
(i) (1, 0), (6, 0), (4, 3)
(ii) (2, 7), (1, 1), (10, 8)
(iii) (–2, –3), (3, 2), (–1, –8)
2 Show that points
A (a, b + c), B (b, c + a), C (c, a + b) are collinear 3 Find values of k if area of triangle is 4 sq |
1 | 1508-1511 | Show that points
A (a, b + c), B (b, c + a), C (c, a + b) are collinear 3 Find values of k if area of triangle is 4 sq units and vertices are
(i) (k, 0), (4, 0), (0, 2)
(ii) (–2, 0), (0, 4), (0, k)
4 |
1 | 1509-1512 | 3 Find values of k if area of triangle is 4 sq units and vertices are
(i) (k, 0), (4, 0), (0, 2)
(ii) (–2, 0), (0, 4), (0, k)
4 (i) Find equation of line joining (1, 2) and (3, 6) using determinants |
1 | 1510-1513 | Find values of k if area of triangle is 4 sq units and vertices are
(i) (k, 0), (4, 0), (0, 2)
(ii) (–2, 0), (0, 4), (0, k)
4 (i) Find equation of line joining (1, 2) and (3, 6) using determinants (ii) Find equation of line joining (3, 1) and (9, 3) using determinants |
1 | 1511-1514 | units and vertices are
(i) (k, 0), (4, 0), (0, 2)
(ii) (–2, 0), (0, 4), (0, k)
4 (i) Find equation of line joining (1, 2) and (3, 6) using determinants (ii) Find equation of line joining (3, 1) and (9, 3) using determinants 5 |
1 | 1512-1515 | (i) Find equation of line joining (1, 2) and (3, 6) using determinants (ii) Find equation of line joining (3, 1) and (9, 3) using determinants 5 If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4) |
1 | 1513-1516 | (ii) Find equation of line joining (3, 1) and (9, 3) using determinants 5 If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4) Then k is
(A) 12
(B) –2
(C) –12, –2
(D) 12, –2
Rationalised 2023-24
84
MATHEMATICS
4 |
1 | 1514-1517 | 5 If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4) Then k is
(A) 12
(B) –2
(C) –12, –2
(D) 12, –2
Rationalised 2023-24
84
MATHEMATICS
4 4 Minors and Cofactors
In this section, we will learn to write the expansion of a determinant in compact form
using minors and cofactors |
1 | 1515-1518 | If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4) Then k is
(A) 12
(B) –2
(C) –12, –2
(D) 12, –2
Rationalised 2023-24
84
MATHEMATICS
4 4 Minors and Cofactors
In this section, we will learn to write the expansion of a determinant in compact form
using minors and cofactors 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 |
1 | 1516-1519 | Then k is
(A) 12
(B) –2
(C) –12, –2
(D) 12, –2
Rationalised 2023-24
84
MATHEMATICS
4 4 Minors and Cofactors
In this section, we will learn to write the expansion of a determinant in compact form
using minors and cofactors 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 |
1 | 1517-1520 | 4 Minors and Cofactors
In this section, we will learn to write the expansion of a determinant in compact form
using minors and cofactors 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 |
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