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DETERMINANTS

** All Mathematical truths are relative and conditional. — C.P. STEINMETZ
**

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 = c 1

a2 x + b2 y = c 2

⎡ a b ⎤ ⎡ x ⎤ ⎡c ⎤

can be represented as ⎢ 1 1 ⎥ ⎢ ⎥ = ⎢ 1 ⎥ . Now, this

⎣ a2 b2 ⎦ ⎣ y ⎦ ⎣ c2 ⎦

system of equations has a unique solution or not, is

determined by the number a1 b2 – a2 b1. (Recall that if

a1 b1 P.S. Laplace

≠ or, a1 b2 – a2 b1 ≠ 0, then the system of linear

a2 b2 (1749-1827)

equations has a unique solution). The number a1 b2 – a2 b1

⎡a b ⎤

which determines uniqueness of solution is associated with the matrix A = ⎢ 1 1 ⎥

⎣ a2 b2 ⎦

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.

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.

104 MATHEMATICS

**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 Δ.

⎡a b ⎤ a b

If A = ⎢ ⎥ , then determinant of A is written as | A| = = det (A)

⎣c d ⎦ c d

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 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

⎡ a11 a12 ⎤

Let A= ⎢ ⎥ be a matrix of order 2 × 2,

⎣ a21 a22 ⎦

then the determinant of A is defined as:

det (A) = |A| = Δ = = a11a22 – a21a12

2 4

Example 1 Evaluate .

–1 2

2 4

Solution We have = 2 (2) – 4(–1) = 4 + 4 = 8.

–1 2

x x +1

Example 2 Evaluate

x –1 x

Solution We have

x x +1

= x (x) – (x + 1) (x – 1) = x2 – (x2 – 1) = x2 – x2 + 1 = 1

x –1 x

**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). There are six ways of expanding a determinant of order

DETERMINANTS 105

**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

a 11 a12 a13

i.e., | A | = a21 a22 a23

a31 a32 a33

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,

a22 a23

i.e., (–1)1 + 1 a11

a32 a33

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,

a21 a23

i.e., (–1)1 + 2 a12

a31 a33

Step 3 Multiply third element a13 of R1 by (–1)1 + 3 [(–1)sum of suffixes in a ] and the second

13

**order determinant obtained by deleting elements of first row (R1) and third column (C3)
**

of | A | as a13 lies in R1 and C3,

a21 a22

i.e., (–1)1 + 3 a13 a a32

31

**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

a22 a23 a21 a23

det A = |A| = (–1)1 + 1 a11 a + (–1)1 + 2 a12

32 a33 a31 a33

1+ 3 a21 a22

+ (–1) a13

a31 a32

or |A| = a11 (a22 a33 – a32 a23) – a12 (a21 a33 – a31 a23)

+ a13 (a21 a32 – a31 a22)

106 MATHEMATICS

= a11 a22 a33 – a11 a32 a23 – a12 a21 a33 + a12 a31 a23 + a13 a21 a32

– a13 a31 a22 ... (1)

**$Note We shall apply all four steps together.
**

Expansion along second row (R2)

a 11 a 12 a 13

| A | = a 21 a 22 a 23

a 31 a 32 a 33

Expanding along R2, we get

**2+1 a12 a13 a11 a13
**

| A | = (–1) a21 + (–1)2 + 2 a22

a32 a33 a31 a33

a11 a12

+ (–1) 2 + 3 a23

a31 a32

= – 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)

a11 a12 a13

| A | = a21 a22 a23

a31 a32 a33

By expanding along C1, we get

**1 + 1 a22 a23 a12 a13
**

| A | = a11 (–1) + a21 ( −1) 2 + 1

a32 a33 a32 a33

3 + 1 a 12 a13

+ a31 (–1) a22 a23

= a11 (a22 a33 – a23 a32) – a21 (a12 a33 – a13 a32) + a31 (a12 a23 – a13 a22)

3 1 2 4 Example 3 Evaluate the determinant Δ = –1 3 0 . Observe that. instead of multiplying by (–1)i + j. In general. | A | = 4 (– 2) = 22 | B | or | A | = 2n | B |. So expanding along third column (C3). 4 1 0 Solution Note that in the third column. we shall expand the determinant along that row or column which contains maximum number of zeros. C2 and C3 are equal to the value of | A | obtained in (1). then | A| = kn | B |. (2) and (3) are equal. (3) Clearly. if A = kB where A and B are square matrices of order n. values of | A | in (1). we can multiply by +1 or –1 according as (i + j) is even or odd.. where n = 2 is the order of square matrices A and B. 2. two entries are zero. DETERMINANTS 107 | 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 . expanding a determinant along any row or column gives same value. cos α – sin β 0 . (2) or (3). ⎡2 2⎤ ⎡1 1⎤ (iii) Let A = ⎢ ⎥ and B = ⎢ ⎥ . It is left as an exercise to the reader to verify that the values of |A| by expanding along R3. Hence. it is easy to verify that A = 2B. Then. where n = 1. (ii) While expanding. Also ⎣4 0⎦ ⎣2 0⎦ | A | = 0 – 8 = – 8 and | B | = 0 – 2 = – 2. we get –1 3 1 2 1 2 Δ= 4 –0 +0 4 1 4 1 –1 3 = 4 (–1 – 12) – 0 + 0 = – 52 0 sin α – cos α Example 4 Evaluate Δ = – sin α 0 sin β . Remarks (i) For easier calculations..

x 1 4 1 3 x 3 2 Solution We have = x 1 4 1 i. Evaluate the determinants 3 –1 –2 3 –4 5 (i) 0 0 –1 (ii) 1 1 –2 3 –5 0 2 3 1 .1 Evaluate the determinants in Exercises 1 and 2. 3 – x2 = 3 – 8 i. x2 = 8 Hence x= ±2 2 EXERCISE 4.e. –5 –1 cos θ – sin θ x2 – x + 1 x – 1 2.108 MATHEMATICS Solution Expanding along R1. then show that | 2A | = 4 | A | ⎣ 4 2⎦ ⎡1 0 1⎤ ⎢ ⎥ 4. we get 0 sin β – sin α sin β – sin α 0 Δ= 0 – sin α – cos α – sin β 0 cos α 0 cos α – sin 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 3 x 3 2 Example 5 Find values of x for which = . then show that | 3 A | = 27 | A | ⎣⎢ 0 0 4 ⎥⎦ 5.e. If A= ⎢ ⎥ . (i) (ii) sin θ cos θ x +1 x +1 ⎡1 2⎤ 3. 2 4 1. If A = ⎢ 0 1 2 ⎥ .

Property 1 The value of the determinant remains unchanged if its rows and columns are interchanged. These properties are true for determinants of any order. if 2 4 2x 4 2 3 x 3 (i) = (ii) = 5 1 6 x 4 5 2x 5 x 2 6 2 8. However. In this section. DETERMINANTS 109 0 1 2 2 –1 –2 (iii) –1 0 –3 (iv) 0 2 –1 –2 3 0 3 –5 0 ⎡ 1 1 –2 ⎤ ⎢ ⎥ 6. we get b2 b3 b1 b3 b1 b2 Δ = a1 − a2 + a3 c2 c3 c1 c3 c1 c2 = a1 (b2 c3 – b3 c2) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1) By interchanging the rows and columns of Δ. we shall restrict ourselves upto determinants of order 3 only. we get the determinant a1 b1 c1 Δ1 = a2 b2 c2 a3 b3 c3 . a1 a2 a3 Verification Let Δ = b1 b2 b3 c1 c2 c3 Expanding along first row. we will study some properties of determinants which simplifies its evaluation by obtaining maximum number of zeros in a row or a column. we have learnt how to expand the determinants.3 Properties of Determinants In the previous section. Find values of x. If A = ⎢ 2 1 –3 ⎥ . find | A | ⎢⎣ 5 4 –9 ⎥⎦ 7. If = . then x is equal to 18 x 18 6 (A) 6 (B) ± 6 (C) – 6 (D) 0 4.

then sign of determinant changes. a1 a2 a3 Verification Let Δ = b1 b2 b3 c1 c2 c3 . Property 2 If any two rows (or columns) of a determinant are interchanged.110 MATHEMATICS Expanding Δ1 along first column. we get 2 6 1 Δ1 = –3 0 5 (Expanding along first column) 5 4 –7 0 5 6 1 6 1 = 2 – (–3) +5 4 –7 4 –7 0 5 = 2 (0 – 20) + 3 (– 42 – 4) + 5 (30 – 0) = – 40 – 138 + 150 = – 28 Clearly Δ = Δ1 Hence. 2 –3 5 Example 6 Verify Property 1 for Δ = 6 0 4 1 5 –7 Solution Expanding the determinant along first row. we have 0 4 6 4 6 0 Δ= 2 – (–3) +5 5 –7 1 –7 1 5 = 2 (0 – 20) + 3 (– 42 – 4) + 5 (30 – 0) = – 40 – 138 + 150 = – 28 By interchanging rows and columns. we get Δ1 = a1 (b2 c3 – c2 b3) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1) Hence Δ = Δ 1 Remark It follows from above property that if A is a square matrix. $ Note If R = ith row and C = ith column. we will symbolically write C ↔ R i i i Let us verify the above property by example. Property 1 is verified. then for interchange of row and i columns. then det (A) = det (A′). where A′ = transpose of A.

the new determinant obtained is given by c1 c2 c3 Δ1 = b1 b2 b3 a1 a2 a3 Expanding along third row. R2 ↔ R3. $Note We can denote the interchange of rows by R ↔ R and interchange of i j columns by Ci ↔ Cj. we have 5 –7 1 –7 1 5 Δ1 = 2 – (–3) +5 0 4 6 4 6 0 = 2 (20 – 0) + 3 (4 + 42) + 5 (0 – 30) = 40 + 138 – 150 = 28 . we get Δ1 = a1 (c2 b3 – b2 c3) – a2 (c1 b3 – c3 b1) + a3 (b2 c1 – b1 c2) = – [a1 (b2 c3 – b3 c2) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1)] Clearly Δ1 = – Δ Similarly. we have 2 –3 5 Δ1 = 1 5 –7 6 0 4 Expanding the determinant Δ1 along first row.e. we can verify the result by interchanging any two columns. we get Δ = a1 (b2 c3 – b3 c2) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1) Interchanging first and third rows. DETERMINANTS 111 Expanding along first row. 2 –3 5 Example 7 Verify Property 2 for Δ = 6 0 4 . 1 5 –7 2 –3 5 Solution Δ = 6 0 4 = – 28 (See Example 6) 1 5 –7 Interchanging rows R2 and R3 i..

Property 3 If any two rows (or columns) of a determinant are identical (all corresponding elements are same). we get Δ = 3 (6 – 6) – 2 (6 – 9) + 3 (4 – 6) = 0 – 2 (–3) + 3 (–2) = 6 – 6 = 0 Here R1 and R3 are identical. we get Δ1 = k a1 (b2 c3 – b3 c2) – k b1 (a2 c3 – c2 a3) + k c1 (a2 b3 – b2 a3) = k [a1 (b2 c3 – b3 c2) – b1 (a2 c3 – c2 a3) + c1 (a2 b3 – b2 a3)] =k Δ . then value of determinant is zero. However. Then k a1 k b1 k c1 Δ1 = a2 b2 c2 a3 b3 c3 Expanding along first row. a1 b1 c1 Verification Let Δ = a2 b2 c2 a3 b3 c3 and Δ1 be the determinant obtained by multiplying the elements of the first row by k.112 MATHEMATICS Clearly Δ1 = – Δ Hence. Property 2 is verified. by Property 2. it follows that Δ has changed its sign Therefore Δ=– Δ or Δ=0 Let us verify the above property by an example. then Δ does not change. then its value gets multiplied by k. 3 2 3 Example 8 Evaluate Δ = 2 2 3 3 2 3 Solution Expanding along first row. Property 4 If each element of a row (or a column) of a determinant is multiplied by a constant k. Proof If we interchange the identical rows (or columns) of the determinant Δ.

(ii) If corresponding elements of any two rows (or columns) of a determinant are proportional (in the same ratio). b1 b2 b3 = b1 b2 b3 + b1 b2 b3 c1 c2 c3 c1 c2 c3 c1 c2 c3 a1 + λ1 a2 + λ 2 a3 + λ 3 Verification L. = b1 b2 b3 c1 c2 c3 . then its value is zero. a1 + λ1 a2 + λ 2 a3 + λ 3 a1 a2 a3 λ1 λ 2 λ3 For example.S. DETERMINANTS 113 k a1 k b1 k c1 a1 b1 c1 Hence a2 b2 c2 = k a2 b2 c2 a3 b3 c3 a3 b3 c3 Remarks (i) By this property. then the determinant can be expressed as sum of two (or more) determinants.H. For example a1 a2 a3 Δ= b1 b2 b3 = 0 (rows R1 and R2 are proportional) k a1 k a2 k a3 102 18 36 Example 9 Evaluate 1 3 4 17 3 6 102 18 36 6(17) 6(3) 6(6) 17 3 6 Solution Note that 1 3 4 = 1 3 4 =6 1 3 4 =0 17 3 6 17 3 6 17 3 6 (Using Properties 3 and 4) Property 5 If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms. we can take out any common factor from any one row or any one column of a given determinant.

Here. Symbolically. c1 c2 c3 c1 c2 c3 Similarly.e. we may verify Property 5 for other rows or columns. c1 c2 c3 c1 c2 c3 where Δ1 is obtained by the operation R1 → R1 + kR3 . the equimultiples of corresponding elements of other row (or column) are added. we write this operation as R1 → R1 + k R3. then value of determinant remains the same.. we get Δ = (a1 + λ1) (b2 c3 – c2 b3) – (a2 + λ2) (b1 c3 – b3 c1) + (a3 + λ3) (b1 c2 – b2 c1) = a1 (b2 c3 – c2 b3) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1) + λ1 (b2 c3 – c2 b3) – λ2 (b1 c3 – b3 c1) + λ3 (b1 c2 – b2 c1) (by rearranging terms) a1 a2 a3 λ1 λ 2 λ3 = b1 b2 b3 + b1 b2 b3 = R. the value of determinant remain same if we apply the operation Ri → Ri + kRj or Ci → Ci + k Cj .114 MATHEMATICS Expanding the determinants along the first row. i. to each element of any row or column of a determinant. a b c Example 10 Show that a + 2 x b + 2 y c + 2 z = 0 x y z a b c a b c a b c Solution We have a + 2 x b + 2 y c + 2 z = a b c + 2x 2 y 2z x y z x y z x y z (by Property 5) =0+0=0 (Using Property 3 and Property 4) Property 6 If. . we have multiplied the elements of the third row (R3) by a constant k and added them to the corresponding elements of the first row (R1).S.H. Verification a1 a2 a3 a1 + k c1 a2 + k c2 a3 + k c3 Let Δ = b1 b2 b3 and Δ1 = b1 b2 b3 .

again a1 a2 a3 k c1 k c2 k c3 Δ1 = b1 b2 b3 + b1 b2 b3 (Using Property 5) c1 c2 c3 c1 c2 c3 =Δ+0 (since R1 and R3 are proportional) Hence Δ = Δ1 Remarks (i) If Δ1 is the determinant obtained by applying Ri → kRi or Ci → kCi to the determinant Δ. we obtain a 2a + b Δ= a +0+0 0 a = a (a2 – 0) = a (a2) = a3 . we have a a+b a+b+c Δ= 0 a 2a + b 0 3a 7a + 3b Now applying R3 → R3 – 3R2 . we get a a+b a+b+c Δ= 0 a 2a + b 0 0 a Expanding along C1. then Δ1 = kΔ. A similar remark applies to column operations. (ii) If more than one operation like Ri → Ri + kRj is done in one step. 3a 6a + 3b 10a + 6b + 3c Solution Applying operations R2 → R2 – 2R1 and R3 → R3 – 3R1 to the given determinant Δ. DETERMINANTS 115 Now. care should be taken to see that a row that is affected in one operation should not be used in another operation. a a+b a+b+c Example 11 Prove that 2a 3a + 2b 4a + 3b + 2c = a 3 .

116 MATHEMATICS Example 12 Without expanding. Example 13 Evaluate 1 a bc Δ = 1 b ca 1 c ab Solution Applying R2 → R2 – R1 and R3 → R3 – R1. we get 1 a bc Δ = (b − a ) (c − a) 0 1 –c 0 1 –b = (b – a) (c – a) [(– b + c)] (Expanding along first column) = (a – b) (b – c) (c – a) b+c a a Example 14 Prove that b c+a b = 4 abc c c a+b b+c a a Solution Let Δ = b c+a b c c a+b . Δ = 0. we get 1 a bc Δ = 0 b − a c ( a − b) 0 c − a b (a − c) Taking factors (b – a) and (c – a) common from R2 and R3. we get x+y+z x+ y+z x+ y+z Δ= z x y 1 1 1 Since the elements of R1 and R3 are proportional. prove that x+y y+z z+x Δ= z x y =0 1 1 1 Solution Applying R1 → R1 + R2 to Δ. respectively.

we obtain c+a b b b b c+a Δ= 0 – (–2 c ) + (–2b) c a+b c a+b c c = 2 c (a b + b2 – bc) – 2 b (b c – c2 – ac) = 2 a b c + 2 cb2 – 2 bc2 – 2 b2c + 2 bc2 + 2 abc = 4 abc x x2 1 + x3 Example 15 If x. z are different and Δ = y y 2 1 + y 3 = 0 . we get 0 –2c –2b Δ= b c+a b c c a+b Expanding along R1. DETERMINANTS 117 Applying R1 → R1 – R2 – R3 to Δ. then z z2 1 + z3 show that 1 + xyz = 0 Solution We have x x2 1 + x3 Δ= y y 2 1 + y3 z z2 1 + z3 x x2 1 x x2 x3 = y y2 1 + y y2 y 3 (Using Property 5) z z2 1 z z2 z3 1 x x2 1 x x2 = (−1) 1 y y 2 + xyz 1 y 2 y2 (Using C3 ↔ C2 and then C1 ↔ C2) 2 2 1 z z 1 z z 1 x x2 = 1 y y 2 (1+ xyz ) 1 z z2 . y.

118 MATHEMATICS 1 x x2 = (1 + xyz ) 0 y−x y2 − x2 (Using R2 → R2–R1 and R3 → R3– R1) 0 z−x z −x 2 2 Taking out common factor (y – x) from R2 and (z – x) from R3. we get 1 x x2 Δ = (1+xyz ) (y –x ) (z –x) 0 1 y+x 0 1 z+x = (1 + xyz) (y – x) (z – x) (z – y) (on expanding along C1) Since Δ = 0 and x.H. we get 1 + xyz = 0 Example 16 Show that 1+ a 1 1 ⎛ 1 1 1⎞ 1 1+ b 1 = abc ⎜ 1 + + + ⎟ = abc + bc + ca + ab ⎝ a b c⎠ 1 1 1+ c Solution Taking out factors a. x – y ≠ 0. y – z ≠ 0.e.c common from R1. R2 and R3.S. = abc +1 b b b 1 1 1 +1 c c c Applying R1→ R1 + R2 + R3. y. i. z are all different. z – x ≠ 0.. we get 1 1 1 +1 a a a 1 1 1 L. we have 1 1 1 1 1 1 1 1 1 1+ + + 1+ + + 1+ + + a b c a b c a b c 1 1 1 Δ = abc +1 b b b 1 1 1 +1 c c c .b.

2 Using the property of determinants and without expanding in Exercises 1 to 7. c+a r+ p z+x = 2 b q y a+b p+q x+ y c r z . then apply C →C –aC. we get 1 0 0 ⎛ 1 1 1⎞ 1 Δ = abc ⎜ 1+ + + ⎟ 1 0 ⎝ a b c⎠ b 1 0 1 c ⎛ 1 1 1⎞ = abc ⎜1 + + + ⎟ ⎡⎣1(1 – 0 )⎤⎦ ⎝ a b c⎠ ⎛ 1 1 1⎞ = abc ⎜1+ + + ⎟ = abc + bc + ca + ab = R. y b y +b = 0 2. DETERMINANTS 119 1 1 1 ⎛ 1 1 1⎞ 1 1 1 = abc ⎜ 1+ + + ⎟ +1 ⎝ a b c⎠ b b b 1 1 1 +1 c c c Now applying C2 → C2 – C1. C3 → C3 – C1. 1 ca b ( c + a ) = 0 5 9 86 1 ab c ( a + b ) b+c q+r y+z a p x 5. ⎝ a b c⎠ $ Note Alternately try by applying C → C – C and C → C – C .H. 3 8 75 = 0 4. b−c c−a a −b = 0 z c z+c c−a a−b b−c 2 7 65 1 bc a ( b + c ) 3. 1 1 2 3 3 2 1 1 3 EXERCISE 4. prove that: x a x+a a −b b −c c − a 1.S.

− a 0 −c = 0 7. ba −b 2 bc = 4 a 2 b 2 c 2 b c 0 ca cb −c 2 By using properties of determinants.120 MATHEMATICS 0 a −b −a2 ab ac 6. (i) 2 x x + 4 2x = ( 5 x + 4 )( 4 − x ) 2 2x 2x x + 4 y+k y y (ii) y y+k y = k 2 (3y + k ) y y y+k a −b −c 2a 2a 2b = ( a + b + c ) 3 11. y y2 zx = (x – y) (y – z) (z – x) (xy + yz + zx) z z2 xy x + 4 2x 2x 10. show that: 1 a a2 8. (i) 1 b b = ( a − b )( b − c )( c − a ) 2 1 c c2 1 1 1 (ii) a b c = ( a − b )( b − c )( c − a )( a + b + c ) a3 b3 c3 x x2 yz 9. (i) 2b b−c−a 2c 2c c−a −b x + y + 2z x y = 2( x + y + z ) 3 (ii) z y + z + 2x y z x z + x + 2y . in Exercises 8 to 14.

. . 15. (B) Determinant is a number associated to a matrix. y3). we always take the absolute value of the determinant in (1). Let A be a square matrix of order 3 × 3. (x2. (D) None of these 4. Which of the following is correct (A) Determinant is a square matrix. is given by the expression [x (y –y ) + x2 (y3–y1) + 2 1 2 3 x3 (y1–y2)]. x2 1 x = 1 − x3 x x2 1 1 + a 2 − b2 2ab −2b ( ) 3 13. y1). (1) 2 x3 y3 1 Remarks (i) Since area is a positive quantity. ab b +1 2 bc =1 + a 2 + b2 + c 2 ca cb c2 + 1 Choose the correct answer in Exercises 15 and 16. then | kA | is equal to (A) k| A | (B) k 2 | A | (C) k 3 | A | (D) 3k | A | 16. we have studied that the area of a triangle whose vertices are 1 (x1. Now this expression can be written in the form of a determinant as x1 y1 1 1 Δ= x2 y2 1 . 2ab 1− a + b 2 2 2a = 1 + a2 + b2 2b −2a 1 − a2 − b2 a2 + 1 ab ac 14.4 Area of a Triangle In earlier classes. y2) and (x3. DETERMINANTS 121 1 x x2 ( ) 2 12.. (C) Determinant is a number associated to a square matrix.

1). k = ∓ 2. So 0 0 1 1 1 3 1 =0 2 x y 1 1 This gives ( y – 3 x ) = 0 or y = 3x. Example 17 Find the area of the triangle whose vertices are (3. (6. –3). = ± 3 . 7). since the area of the triangle ABD is 3 sq. (4. 0). (1. 2) and (5. Also. y) be any point on AB. we have 1 3 1 1 0 0 1 =±3 2 k 0 1 − 3k This gives. Then. (–1. 1). –8) . 2 which is the equation of required line AB.. 0) is a point such that area of triangle ABD is 3sq units.3 1.122 MATHEMATICS (ii) If area is given.e. 2 EXERCISE 4. Solution The area of triangle is given by 3 8 1 1 Δ= –4 2 1 2 5 1 1 1 ⎡3 ( 2 – 1) – 8 ( – 4 – 5 ) + 1( – 4 – 10 ) ⎤⎦ 2⎣ = 1 61 = ( 3 + 72 – 14 ) = 2 2 Example 18 Find the equation of the line joining A(1. (3. 3) and B (0. (10. 0). i. Solution Let P (x. 8) (iii) (–2. area of triangle ABP is zero (Why?). (iii) The area of the triangle formed by three collinear points is zero. units. 3) (ii) (2. 0) using determinants and find k if D(k. 2). 8). Find area of the triangle with vertices at the point given in each of the following : (i) (1. use both positive and negative values of the determinant for calculation. (– 4.

–2 (D) 12. k) 4. B (b.5 Minors and Cofactors In this section. 6) using determinants. (0. Show that points A (a. 5. So M11 = Minor of a11= 3 M12 = Minor of the element a12 = 4 M21 = Minor of the element a21 = –2 . 3. Remark Minor of an element of a determinant of order n(n ≥ 2) is a determinant of order n – 1. 0). its minor M23 is given by 1 2 M23 = = 8 – 14 = – 6 (obtained by deleting R2 and C3 in Δ). – 6). 1 2 3 Example 19 Find the minor of element 6 in the determinant Δ = 4 5 6 7 8 9 Solution Since 6 lies in the second row and third column. (0. a + b) are collinear. (i) Find equation of line joining (1. (ii) Find equation of line joining (3. Find values of k if area of triangle is 4 sq. b + c). denoted by Aij is defined by Aij = (–1)i + j Mij . If area of triangle is 35 sq units with vertices (2. Then k is (A) 12 (B) –2 (C) –12. 7 8 Definition 2 Cofactor of an element aij . 4). –2 4. C (c. 3) using determinants. 0). (4. 2) and (3. DETERMINANTS 123 2. 1) and (9. 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. we will learn to write the expansion of a determinant in compact form using minors and cofactors. where Mij is minor of aij . (0. 1 –2 Example 20 Find minors and cofactors of all the elements of the determinant 4 3 Solution Minor of the element aij is Mij Here a11 = 1. (5. c + a). 4). 0). Minor of an element aij is denoted by Mij. 2) (ii) (–2. 4) and (k. units and vertices are (i) (k.

For example. we have a22 a23 a21 a23 a21 a22 Δ = (–1) a11 a 1+1 1+2 a + (–1) a12 a 1+3 a + (–1) a13 a31 a32 32 33 31 33 = a11 A11 + a12 A12 + a13 A13. R3.124 MATHEMATICS M22 = Minor of the element a22 = 1 Now. C1. Δ can be calculated by other five ways of expansion that is along R2. along R1. . then their sum is zero. we have a22 a23 Minor of a11 = M11 = = a22 a33– a23 a32 a32 a33 Cofactor of a11 = A11 = (–1)1+1 M11 = a22 a33 – a23 a32 a12 a13 Minor of a21 = M21 = = a12 a33 – a13 a32 a32 a33 Cofactor of a21 = A21 = (–1)2+1 M21 = (–1) (a12 a33 – a13 a32) = – a12 a33 + a13 a32 Remark Expanding the determinant Δ. where Aij is cofactor of aij = sum of product of elements of R1 with their corresponding cofactors Similarly. 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 Example 21 Find minors and cofactors of the elements a11. $Note If elements of a row (or column) are multiplied with cofactors of any other row (or column). cofactor of aij is Aij. Hence Δ = sum of the product of elements of any row (or column) with their corresponding cofactors. C2 and C3. in Example 21. a21 in the determinant a11 a12 a13 Δ = a21 a22 a23 a31 a32 a33 Solution By definition of minors and cofactors.

A13 = (–1)1+3 (30) = 30 1 5 –3 5 M21 = = 21 – 25 = – 4. Example 22 Find minors and cofactors of the elements of the determinant 2 –3 5 6 0 4 and verify that a11 A31 + a12 A32 + a13 A33= 0 1 5 –7 0 4 Solution We have M11 = = 0 –20 = –20. A21 = (–1)2+1 (– 4) = 4 5 –7 2 5 M22 = = –14 – 5 = –19. we can try for other rows and columns. A12 = (–1)1+2 (– 46) = 46 1 –7 6 0 M13 = = 30 – 0 = 30. A11 = (–1)1+1 (–20) = –20 5 –7 6 4 M12 = = – 42 – 4 = – 46. DETERMINANTS 125 Δ = a11 A21 + a12 A22 + a13 A23 a12 a13 a11 a13 a11 a12 = a11 (–1)1+1 + a12 (–1)1+2 + a13 (–1)1+3 a32 a33 a31 a33 a31 a32 a11 a12 a13 = a11 a12 a13 = 0 (since R and R are identical) 1 2 a31 a32 a33 Similarly. A31 = (–1)3+1 (–12) = –12 0 4 . A23 = (–1)2+3 (13) = –13 1 5 –3 5 M31 = = –12 – 0 = –12. A22 = (–1)2+2 (–19) = –19 1 –7 2 –3 M23 = = 10 + 3 = 13.

A32 = (–1)3+2 (–22) = 22 6 4 2 –3 and M33 = = 0 + 18 = 18.4 Write Minors and Cofactors of the elements of following determinants: 2 –4 a c 1.6 Adjoint and Inverse of a Matrix In the previous chapter. (i) (ii) 0 3 b d 1 0 0 1 0 4 2. A33 = (–1)3+3 (18) = 18 6 0 Now a11 = 2.. A32 = 22. Using Cofactors of elements of second row. a13 = 5.e. If Δ = a21 a22 a23 and Aij is Cofactors of aij . Using Cofactors of elements of third column. then value of Δ is given by a31 a32 a33 (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 z xy a11 a12 a13 5. (i) 0 1 0 (ii) 3 5 –1 0 0 1 0 1 2 5 3 8 3. i. we shall discuss the condition for existence of inverse of a matrix. we have studied inverse of a matrix. A–1 we shall first define adjoint of a matrix. A31 = –12. evaluate Δ = 2 0 1 . 1 2 3 1 x yz 4. a12 = –3. A33 = 18 So a11 A31 + a12 A32 + a13 A33 = 2 (–12) + (–3) (22) + 5 (18) = –24 – 66 + 90 = 0 EXERCISE 4.126 MATHEMATICS 2 5 M32 = = 8 – 30 = –22. evaluate Δ = 1 y zx . . In this section. To find inverse of a matrix A.

6.. A21 = –3. given by ⎡ a11 a12 ⎤ A= ⎢ ⎥ ⎣ a21 a22 ⎦ The adj A can also be obtained by interchanging a11 and a22 and by changing signs of a12 and a21. ⎡ a11 a12 a13 ⎤ Let A = ⎢⎢ a21 a22 a23 ⎥⎥ ⎢⎣ a31 a32 a33 ⎥⎦ ⎡ A11 A12 A13 ⎤ ⎡ A11 A 21 A 31 ⎤ Then adj A = Transpose of ⎢⎢A 21 A 22 A 23 ⎥⎥ = ⎢ A12 A 22 A 32 ⎥⎥ ⎢ ⎢⎣ A31 A32 A 33 ⎥⎦ ⎢⎣ A13 A 23 A33 ⎦⎥ ⎡ 2 3⎤ Example 23 Find adj A for A = ⎢ ⎥ ⎣1 4⎦ Solution We have A11 = 4.e.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. i. A12 = –1. We state the following theorem without proof. where Aij is the cofactor of the element aij . where I is the identity matrix of order n . then A(adj A) = (adj A) A = A I . Adjoint of the matrix A is denoted by adj A. Theorem 1 If A be any given square matrix of order n. DETERMINANTS 127 4. A22 = 2 ⎡ A11 A 21 ⎤ ⎡ 4 –3⎤ Hence adj A = ⎢ ⎥ =⎢ ⎥ ⎣ A12 A 22 ⎦ ⎣ –1 2 ⎦ Remark For a square matrix of order 2.

where A and B are square matrices of the same order ⎡A 0 0⎤ ⎢ ⎥ Remark We know that (adj A) A = A I = ⎢ 0 A 0⎥ ⎢⎣ 0 0 A ⎥⎦ . then adj A = ⎢ A12 A 22 A32 ⎥⎥ ⎢⎣ a31 a32 a33 ⎥⎦ ⎢⎣ A13 A 23 A 33 ⎥⎦ Since sum of product of elements of a row (or a column) with corresponding cofactors is equal to | A | and otherwise zero.128 MATHEMATICS Verification ⎡ a11 a12 a13 ⎤ ⎡ A11 A 21 A 31 ⎤ ⎢a ⎥ ⎢ Let A = ⎢ 21 a22 a23 ⎥ . Theorem 2 If A and B are nonsingular matrices of the same order. Theorem 3 The determinant of the product of matrices is equal to product of their respective determinants. AB = A B . the determinant of matrix A = ⎢ 4 8 ⎥ is zero ⎣ ⎦ Hence A is a singular matrix. ⎣3 4 ⎦ 3 4 Hence A is a nonsingular matrix We state the following theorems without proof. 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 2⎤ For example. Then A = = 4 – 6 = – 2 ≠ 0. that is. we have ⎡A 0 0⎤ ⎡ 1 0 0⎤ ⎢ ⎥ A (adj A) = ⎢ 0 A 0 ⎥ = A ⎢⎢ 0 1 0⎥⎥ = A I ⎢⎣ 0 0 A ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ Similarly. then AB and BA are also nonsingular matrices of the same order. Definition 5 A square matrix A is said to be non-singular if A ≠ 0 ⎡1 2 ⎤ 1 2 Let A= ⎢ ⎥ .

then verify that A adj A = | A | I. Then. then | adj (A) | = | A |n – 1. if A is a square matrix of order n. |(adj A)| |A| = | A |3 (1) i. Also find A–1. we have A 0 0 (adj A) A = 0 A 0 0 0 A 1 0 0 3 i. |(adj A)| |A| = A 0 1 0 (Why?) 0 0 1 i. AB = A B ) This gives A ≠ 0. Conversely. Then A ≠ 0 Now A (adj A) = (adj A) A = A I (Theorem 1) ⎛ 1 ⎞ ⎛ 1 ⎞ or A⎜ adj A ⎟ = ⎜ adj A ⎟ A = I ⎝|A| ⎠ ⎝ |A| ⎠ 1 or AB = BA = I. Hence A is nonsingular. ⎢⎣1 3 4⎥⎦ Solution We have A = 1 (16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1 ≠ 0 . Theorem 4 A square matrix A is invertible if and only if A is nonsingular matrix. there exists a square matrix B of order n such that AB = BA = I Now AB = I. where B = adj A |A| 1 Thus A is invertible and A–1 = adj A |A| ⎡1 3 3⎤ ⎢ ⎥ Example 24 If A = ⎢1 4 3⎥ .e.e. let A be nonsingular. DETERMINANTS 129 Writing determinants of matrices on both sides. |(adj A)| = | A | 2 In general. So AB = I or A B =1 (since I =1.e. Proof Let A be invertible matrix of order n and I be the identity matrix of order n.

AB = –11 ≠ 0. A31 = –3.I ⎣⎢ 0 0 1 ⎥⎦ ⎣⎢ 0 0 1 ⎥⎦ ⎡ 7 −3 −3⎤ ⎡ 7 −3 −3⎤ −1 1 1⎢ ⎥ ⎢ ⎥ Also A = adj A = ⎢ −1 1 0 ⎥ = ⎢ −1 1 0 ⎥ A 1 ⎢⎣ −1 0 1 ⎥⎦ ⎢⎣ −1 0 1 ⎥⎦ ⎡2 3 ⎤ ⎡ 1 −2 ⎤ Example 25 If A = ⎢ ⎥ and B = ⎢ ⎥ . A21 = –3. (AB)–1 exists and is given by 1 1 ⎡ −14 −5⎤ = 1 ⎡14 5⎤ adj (AB) = − ⎢ ⎢ ⎥ 11 ⎣ −5 −1⎥⎦ (AB)–1 = 11 ⎣ 5 1⎦ AB Further.B = ⎢ 11 ⎢⎣ −1 2 ⎥⎦ ⎥ ⎣1 1⎦ . then verify that (AB) = B A . A22 = 1. –1 –1 –1 ⎣ 1 − 4 ⎦ ⎣ − 1 3 ⎦ ⎡2 3⎤ ⎡ 1 −2 ⎤ ⎡ −1 5 ⎤ Solution We have AB = ⎢ ⎥ ⎢ −1 3 ⎥ = ⎢ 5 −14 ⎥ ⎣1 − 4⎦ ⎣ ⎦ ⎣ ⎦ Since. A33 = 1 ⎡ 7 −3 −3⎤ ⎢ ⎥ Therefore adj A = ⎢ −1 1 0 ⎥ ⎢⎣ −1 0 1 ⎥⎦ ⎡1 3 3⎤ ⎡ 7 −3 −3⎤ ⎢ ⎥⎢ ⎥ Now A (adj A) = ⎢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 ⎤ ⎡1 0 0 ⎤ ⎢ ⎥ ⎢0 1 0 ⎥ = ⎢ 0 1 0 ⎥ = (1) ⎢ ⎥ = A .130 MATHEMATICS Now A11 = 7. A32 = 0. Therefore. A = –11 ≠ 0 and B = 1 ≠ 0. A–1 and B–1 both exist and are given by 1 ⎡ − 4 −3⎤ −1 ⎡3 2⎤ A–1 = − . A13 = –1.A23 = 0. A12 = –1.

⎢ 2 3 5⎥ ⎣ ⎦ ⎢⎣ −2 0 1 ⎥⎦ Verify A (adj A) = (adj A) A = | A | I in Exercises 3 and 4 ⎡1 −1 2 ⎤ ⎡2 3⎤ ⎢ ⎥ 3. Using this equation. find A–1. ⎡ 2 3 ⎤ ⎡ 2 3 ⎤ ⎡ 7 12⎤ Solution We have A 2 = A. ⎢ −4 −6 ⎥ 4.5 Find adjoint of each of the matrices in Exercises 1 and 2. ⎣ 1 2⎦ where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix. ⎡ 1 −1 2⎤ ⎡ 1 2⎤ ⎢ ⎥ 1. ⎢ 3 4⎥ 2. DETERMINANTS 131 1 ⎡3 2⎤ ⎡ −4 −3⎤ 1 ⎡ −14 −5⎤ 1 ⎡14 5⎤ Therefore B−1 A −1 = − ⎢ ⎥ ⎢ ⎥ =− ⎢ ⎥ = ⎢ 11 ⎣1 1 ⎦ ⎣ −1 2 ⎦ 11 ⎣ −5 −1⎦ 11 ⎣ 5 1⎥⎦ Hence (AB)–1 = B–1 A–1 ⎡ 2 3⎤ Example 26 Show that the matrix A = ⎢ ⎥ satisfies the equation A2 – 4A + I = O.A = ⎢ ⎥⎢ ⎥ =⎢ ⎥ ⎣1 2 ⎦ ⎣1 2 ⎦ ⎣ 4 7 ⎦ ⎡ 7 12⎤ ⎡ 8 12⎤ ⎡ 1 0 ⎤ ⎡ 0 0⎤ Hence A 2 − 4A + I = ⎢ ⎥− ⎢ ⎥+⎢ ⎥ =⎢ ⎥=O ⎣ 4 7 ⎦ ⎣ 4 8 ⎦ ⎣ 0 1 ⎦ ⎣ 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 ⎡ 4 0 ⎤ ⎡2 3 ⎤ ⎡ 2 −3 ⎤ or A–1 = 4I – A = ⎢ ⎥ −⎢ ⎥ = ⎢ ⎥ ⎣0 4 ⎦ ⎣1 2 ⎦ ⎣ −1 2 ⎦ ⎡ 2 −3 ⎤ Hence A −1 = ⎢ ⎥ ⎣ −1 2 ⎦ EXERCISE 4. ⎢ 3 0 −2⎥ ⎣ ⎦ ⎢⎣1 0 3 ⎥⎦ .

Then | adj A | is equal to (A) | A | (B) | A | 2 (C) | A | 3 (D) 3 | A | 18. show that A – 5A + 7I = O. 2 –1 ⎣ −1 2 ⎦ ⎡3 2 ⎤ 14. Let A = ⎢ ⎥ and B = ⎢ ⎥ . –1 –1 –1 ⎣2 5⎦ ⎣ 7 9⎦ ⎡ 3 1⎤ 13. Verify that (AB) = B A . Hence find A . find the numbers a and b such that A + aA + bI = O. ⎢ 0 2 −3⎥ ⎢⎣ 5 2 −1⎥⎦ ⎢⎣ −7 2 1⎥⎦ ⎣⎢ 3 −2 4 ⎥⎦ ⎡1 0 0 ⎤ ⎢ 0 cos α sin α ⎥ 11. then det (A–1) is equal to 1 (A) det (A) (B) det (A) (C) 1 (D) 0 . Hence. Let A be a nonsingular square matrix of order 3 × 3. If A = ⎢ ⎥ . find A–1. 2 ⎣ 1 1 ⎦ ⎡1 1 1 ⎤ ⎢ ⎥ 15. ⎡ 2 −1 1 ⎤ ⎢ ⎥ 16. ⎢ 0 2 4⎥ ⎣ ⎦ ⎣ −3 2 ⎦ ⎢⎣ 0 0 5⎥⎦ ⎡1 0 0 ⎤ ⎡ 2 1 3⎤ ⎡ 1 −1 2 ⎤ ⎢3 3 0 ⎥ ⎢ ⎥ ⎢ ⎥ 8. If A is an invertible matrix of order 2. ⎡ 1 2 3⎤ ⎡ 2 −2 ⎤ ⎡ −1 5 ⎤ ⎢ ⎥ 5. ⎢ ⎥ 7. ⎢ 4 −1 0⎥ 10. ⎢4 3 ⎥ 6. If A = ⎢ −1 2 −1⎥ ⎣⎢ 1 −1 2 ⎥⎦ Verify that A3 – 6A2 + 9A – 4I = O and hence find A–1 17.132 MATHEMATICS Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11. For the matrix A = ⎢ ⎥ . For the matrix A = ⎢ 1 2 −3⎥ ⎢⎣ 2 −1 3 ⎥⎦ Show that A3– 6A2 + 5A + 11 I = O. ⎢ ⎥ 9. ⎢ ⎥ ⎢⎣ 0 sin α − cos α ⎥⎦ ⎡3 7⎤ ⎡ 6 8⎤ 12.

e. then its inverse exists. AX = B. Consider the system of equations a1 x + b1 y + c 1 z = d1 a2 x + b2 y + c 2 z = d 2 a3 x + b3 y + c 3 z = d 3 ⎡ a1 b1 c1 ⎤ ⎡x⎤ ⎡ d1 ⎤ ⎢ a b c ⎥ . 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. ⎡ a1 b1 c1 ⎤ ⎡ x⎤ ⎡ d1 ⎤ ⎢a b2 c2 ⎥⎥ ⎢ y⎥ ⎢ ⎥ ⎢ 2 ⎢ ⎥ = ⎢ d2 ⎥ ⎢⎣ a3 b3 c3 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ d3 ⎥⎦ Case I If A is a nonsingular matrix. This method of solving system of equations is known as Matrix Method. Consistent system A system of equations is said to be consistent if its solution (one or more) exists. . $ Note In this chapter. the system of equations can be written as. Now AX = B or A (AX) = A–1 B –1 (premultiplying by A–1) or (A–1A) X = A–1 B (by associative property) –1 or IX=A 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.7. DETERMINANTS 133 4. X = ⎢ y ⎥ and B = ⎢ d ⎥ Let A = ⎢ 2 2 2⎥ ⎢ ⎥ ⎢ 2⎥ ⎢⎣ a3 b3 c3 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ d 3 ⎥⎦ Then.7 Applications of Determinants and Matrices In this section. 4. Inconsistent system A system of equations is said to be inconsistent if its solution does not exist.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.. i. we restrict ourselves to the system of linear equations having unique solutions only.

e. then solution does not exist and the system of equations is called inconsistent. where ⎡ 3 −2 3 ⎤ ⎡ x⎤ ⎡ 8⎤ A = ⎢ 2 1 −1⎥ . then system may be either consistent or inconsistent according as the system have either infinitely many solutions or no solution. y = – 1 Example 28 Solve the following system of equations by matrix method. A is nonsingular matrix and so has a unique solution. X = ⎢⎢ y ⎥⎥ and B = ⎢ ⎥ ⎢ 1⎥ ⎢ ⎥ ⎢⎣ 4 −3 2 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ 4⎥⎦ We see that A = 3 (2 – 3) + 2(4 + 4) + 3 (– 6 – 4) = – 17 ≠ 0 . 1 ⎡ 2 −5 ⎤ A–1 = − 11 ⎢⎣ −3 2 ⎥⎦ Note that 1 ⎡ 2 −5 ⎤ ⎡ 1 ⎤ Therefore X = A–1B = – 11 ⎢⎣ −3 2 ⎥⎦ ⎢⎣ 7⎥⎦ ⎡ x⎤ 1 ⎡ −33⎤ ⎡ 3 ⎤ ⎢ y⎥ = − ⎢ = 11 ⎣ 11 ⎥⎦ ⎢⎣ −1⎥⎦ i. In this case. A = –11 ≠ 0.134 MATHEMATICS Case II If A is a singular matrix. If (adj A) B ≠ O. then | A | = 0. Hence. (O being zero matrix). we calculate (adj A) B. X = ⎢ ⎥ and B = ⎢ ⎥ ⎣3 2⎦ ⎣ y⎦ ⎣7 ⎦ Now. ⎣ ⎦ Hence x = 3. If (adj A) B = O. where ⎡2 5⎤ ⎡x⎤ ⎡1 ⎤ A= ⎢ ⎥ . 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. Example 27 Solve the system of equations 2x + 5y = 1 3x + 2y = 7 Solution The system of equations can be written in the form AX = B.

A13 = – 1 A21 = – (1 + 2) = – 3. Now we find adj A A11 = 1 (1 + 6) = 7. A13 = –10 A21 = –5. where ⎡1 1 1⎤ ⎡ x⎤ ⎡6⎤ ⎢ ⎥ ⎢ y⎥ ⎢11⎥ A = ⎢0 1 3⎥ . according to given conditions. ⎢ ⎥ = − 17 ⎢ −34 ⎥ = ⎢ 2 ⎥ ⎢⎣ z ⎥⎦ ⎢⎣ −51⎥⎦ ⎢⎣ 3 ⎥⎦ Hence x = 1. A12 = – 8. A33 = (1 – 0) = 1 . If we multiply third number by 3 and add second number to it. DETERMINANTS 135 Hence. A22 = 0. A12 = – (0 – 3) = 3. By adding first and third numbers. A23 = – (– 2 – 1) = 3 A31 = (3 – 1) = 2. we get double of the second number.e. Then. A32 = 9. Now A11 = –1. A33 = 7 ⎡ −1 − 5 −1⎤ 1 ⎢ ⎥ Therefore A = − ⎢ −8 − 6 9 ⎥ –1 17 ⎢⎣ −10 1 7 ⎥⎦ ⎡ −1 − 5 −1⎤ ⎡ 8 ⎤ 1 ⎢ ⎥ ⎢ ⎥ X = A B = − ⎢ −8 − 6 9 ⎥ ⎢ 1 ⎥ –1 So 17 ⎣⎢ −10 1 7 ⎥⎦ ⎢⎣ 4⎥⎦ ⎡ x⎤ ⎡ −17 ⎤ ⎡ 1 ⎤ ⎢ y⎥ 1 ⎢ ⎥ ⎢ ⎥ i. X = ⎢ ⎥ and B = ⎢ ⎥ ⎢⎣1 –2 1⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ 0 ⎥⎦ Here A = 1 (1 + 6) – (0 – 3) + ( 0 – 1) = 9 ≠ 0 . Solution Let first. y = 2 and z = 3. Represent it algebraically and find the numbers using matrix method. A23 = 1 A31 = –1. we get 11. y and z. 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. respectively. A22 = – 6. Example 29 The sum of three numbers is 6. second and third numbers be denoted by x. A is nonsingular and so its inverse exists. A32 = – (3 – 0) = – 3.

6 Examine the consistency of the system of equations in Exercises 1 to 6. x + 3y = 5 2x + 3y = 3 x+y=4 2x + 6y = 8 4. using matrix method. x – y + z = 4 3 3x + 2y = 5 x – 2y – z = 2x + y – 3z = 0 2 3y – 5z = 9 x+y+z=2 13. z = 3 EXERCISE 4. 2x – y = 5 3. in Exercises 7 to 14. 4x – 3y = 3 7x + 3y = 5 3x + 4y = 3 3x – 5y = 7 10. 5x + 2y = 4 8. 7. 3x–y – 2z = 2 6. y = 2. 5x + 2y = 3 11.136 MATHEMATICS ⎡ 7 –3 2 ⎤ ⎢ ⎥ Hence adj A = ⎢ 3 0 –3⎥ ⎢⎣ –1 3 1 ⎥⎦ ⎡ 7 –3 2 ⎤ 1 1 ⎢ ⎥ Thus A –1 = adj (A) = ⎢ 3 0 –3⎥ A 9 ⎢⎣ –1 3 1 ⎥⎦ Since X = A–1 B ⎡ 7 –3 2 ⎤ ⎡ 6 ⎤ 1⎢ ⎥⎢ ⎥ X = ⎢ 3 0 –3⎥ ⎢11⎥ 9 ⎢⎣ –1 3 1 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎡ x⎤ ⎡ 42 − 33 + 0 ⎤ ⎡9⎤ ⎡ 1⎤ ⎢ y⎥ ⎢ 1 18 + 0 + 0 ⎥ ⎢ 1 18 ⎥ ⎢ ⎥ or ⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ = ⎢ 2⎥ ⎢⎣ z ⎥⎦ 9 ⎢ −6 + 33 + 0⎥ 9 ⎢ 27 ⎥ ⎢ 3⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Thus x = 1. 1. x – y + 2z = 7 x – 2y + z = – 4 3x + 4y – 5z = – 5 3x – y – 2z = 3 2x – y + 3z = 12 . 2x + y + z = 1 12. 2x – y = –2 9. x + 2y = 2 2. x + y + z = 1 5. 2x + 3y +3 z = 5 14. 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 A solve the system of equations ⎢⎣ 1 1 –2 ⎥⎦ 2x – 3y + 5z = 11 3x + 2y – 4z = – 5 x + y – 2z = – 3 16. Find cost of each item per kg by matrix method. c are positive and unequal. The cost of 4 kg onion. we get a+b+c b c 1 b c Δ = a + b + c c a = (a + b + c) 1 c a a+b+c a b 1 a b 1 b c = (a + b + c) 0 c – b a – c (Applying R2→ R2–R1. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. find A . 3 kg wheat and 2 kg rice is Rs 60.and R3 →R3 –R1) 0 a–b b–c = (a + b + c) [(c – b) (b – c) – (a – c) (a – b)] (Expanding along C1) = (a + b + c)(– a2 – b2 – c2 + ab + bc + ca) –1 = (a + b + c) (2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca) 2 –1 = (a + b + c) [(a – b)2 + (b – c)2 + (c – a)2] 2 which is negative (since a + b + c > 0 and (a – b)2 + (b – c)2 + (c – a)2 > 0) . show that value of the determinant a b c Δ = b c a is negative. c a b Solution Applying C1 → C1 + C2 + C3 to the given determinant. DETERMINANTS 137 ⎡ 2 –3 5⎤ ⎢3 2 – 4⎥ –1 –1 15. If A = ⎢ ⎥ . The cost of 2 kg onion. b. Miscellaneous Examples Example 30 If a. 4 kg wheat and 6 kg rice is Rs 90.

respectively. y. we get ( y+ z) 2 x2 x2 xyz (x+ z) 2 Δ= y2 y2 xyz ( x+ y) 2 z2 z2 Applying C2 → C2– C1. we obtain 0 0 0 3y + 5 6 y + 8 9 y + b = 0 (Since 2b = a + c) 4 y + 6 7 y + 9 10 y + c Example 32 Show that ( y+ z ) 2 xy zx ( x+ z ) 2 Δ= xy yz = 2xyz (x + y + z)3 ( x+ y ) 2 xz yz Solution Applying R1 → xR1. R3 → z R3 to Δ and dividing by xyz.138 MATHEMATICS Example 31 If a. b. z from C1 C2 and C3. C3 → C3– C1. c.P. we get x ( y+ z) 2 x2 y x2 z 1 y ( x+ z ) 2 Δ= xy 2 y2 z xyz z ( x+ y ) 2 xz 2 yz 2 Taking common factors x. R2 → yR2 . find value of 2y + 4 5y + 7 8y + a 3y + 5 6 y + 8 9 y + b 4 y + 6 7 y + 9 10 y + c Solution Applying R1 → R1 + R3 – 2R2 to the given determinant. we have ( y + z )2 x2 – ( y + z ) 2 x2 − ( y + z ) 2 Δ= y2 ( x + z )2 − y 2 0 z 2 0 ( x + y )2 – z 2 . are in A.

we have (y + z) x – ( y + z) x – ( y + z) 2 Δ = (x + y + z)2 y 2 (x+ z) – y 0 z2 0 ( x + y) – z Applying R1 → R1 – (R2 + R3). we get y ⎝ z ⎠ 2 yz 0 0 y2 Δ = (x + y + z)2 y2 x+ z z z2 z2 x+ y y Finally expanding along R1. we have Δ = (x + y + z)2 (2yz) [(x + z) (x + y) – yz] = (x + y + z)2 (2yz) (x2 + xy + xz) = (x + y + z)3 (2xyz) ⎡1 –1 2 ⎤ ⎡ –2 0 1 ⎤ ⎢ ⎥ ⎢ ⎥ Example 33 Use product ⎢0 2 –3⎥ ⎢ 9 2 –3⎥ to solve the system of equations ⎢⎣ 3 –2 4 ⎥⎦ ⎢⎣ 6 1 –2⎥⎦ x – y + 2z = 1 2y – 3z = 1 3x – 2y + 4z = 2 ⎡1 –1 2⎤ ⎡ –2 0 1 ⎤ ⎢ –3⎥ ⎢ 9 –3⎥ Solution Consider the product ⎢ 0 2 ⎥ ⎢ 2 ⎥ ⎢⎣ 3 –2 4 ⎥⎦ ⎢⎣ 6 1 – 2 ⎥⎦ . DETERMINANTS 139 Taking common factor (x + y + z) from C2 and C3. we have 2 yz –2z –2y Δ = (x + y + z)2 y2 x− y+z 0 z2 0 x+ y –z 1 ⎛ 1 ⎞ Applying C2 → (C2 + C1) and C3 → ⎜ C3 + C1 ⎟ .

y = 5 and z = 3 Example 34 Prove that a + bx c + dx p + qx a c p Δ = ax + b cx + d px + q = (1 − x ) b d2 q u v w u v w Solution Applying R1 → R1 – x R2 to Δ. in matrix form. given system of equations can be written.140 MATHEMATICS ⎡ − 2 − 9 + 12 0 − 2 + 2 1 + 3 − 4⎤ ⎡ 1 0 0⎤ ⎢ ⎥ ⎢ ⎥ = ⎢ 0 + 18 − 18 0 + 4 − 3 0 − 6 + 6⎥ = ⎢ 0 1 0⎥ ⎢⎣ − 6 − 18 + 24 0 − 4 + 4 3 + 6 − 8⎥⎦ ⎢⎣ 0 0 1⎥⎦ ⎡ 1 –1 2 ⎤ ⎡ –2 0 1 ⎤ –1 Hence ⎢ 0 2 –3 = ⎢ 9 2 –3⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 3 –2 4 ⎥⎦ ⎢⎣ 6 1 –2⎥⎦ Now. we get a (1 − x 2 ) c (1 − x 2 ) p (1 − x 2 ) Δ= ax + b cx + d px + q u v w a c p = (1 − x ) ax + b cx + d px + q 2 u v w . as follows ⎡ 1 –1 2 ⎤ ⎡ x ⎤ ⎡ 1 ⎤ ⎢ 0 2 –3⎥ ⎢ y ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢1 ⎥ ⎢⎣ 3 –2 4 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ 2⎥⎦ −1 ⎡ x ⎤ ⎡1 −1 2 ⎤ ⎡ 1 ⎤ ⎡ –2 0 1 ⎤ ⎡1⎤ ⎢ y⎥ ⎢ or ⎢ ⎥ = ⎢0 2 −3⎥⎥ ⎢⎢ 1 ⎥⎥ = ⎢⎢ 9 2 –3⎥⎥ ⎢⎢1⎥⎥ ⎢⎣ z ⎥⎦ ⎢⎣ 3 −2 4 ⎥⎦ ⎢⎣ 2 ⎥⎦ ⎢⎣ 6 1 –2⎥⎦ ⎢⎣ 2⎥⎦ ⎡ −2 + 0 + 2⎤ ⎡ 0 ⎤ ⎢ ⎥ ⎢ ⎥ = ⎢ 9 + 2 − 6 ⎥ = ⎢5 ⎥ ⎢⎣ 6 + 1 − 4 ⎥⎦ ⎢⎣ 3⎥⎦ Hence x = 0.

Prove that the determinant – sin θ – x 1 is independent of θ. prove that b b 2 ca = 1 b 2 b3 . c c2 ab 1 c2 c3 cos α cos β cos α sin β – sin α 3. find AB –1 7. cos θ 1 x a a2 bc 1 a2 a3 2. a ≠ 0 x x x+a a 2 bc ac + c 2 6. Solve the equation x x+a x = 0. Without expanding the determinant. Prove that a + ab 2 b2 ac = 4a2b2c2 ab b 2 + bc c2 ⎡ 3 –1 1 ⎤ ⎡ 1 2 –2 ⎤ ⎢ –15 6 –5⎥ and B = ⎢ –1 3 0 ⎥ . b and c are real numbers. If a. DETERMINANTS 141 Applying R2 → R2 – x R1. a+ b b+ c c+ a Show that either a + b + c = 0 or a = b = c. sin α cos β sin α sin β cos α 4. x+a x x 5. we get a c p Δ = (1 − x ) b d q 2 u v w Miscellaneous Exercises on Chapter 4 x sin θ cos θ 1. If A = ⎢ –1 ⎥ ⎢ ⎥ ( ) ⎢⎣ 5 –2 2 ⎥⎦ ⎢⎣ 0 –2 1 ⎥⎦ . Evaluate – sin β cos β 0 . and b+ c c+ a a+ b Δ = c + a a + b b + c = 0.

sin β cos β cos ( β + δ ) = 0 3 6 + 3 p 10 + 6 p + 3 q sin γ cos γ cos ( γ + δ ) 16. –b+ a 3b – b + c = 3(a + b + c) (ab + bc + ca) –c+ a – c+ b 3c 1 1+ p 1+ p+ q sin α cos α cos ( α + δ ) 14. Solve the system of equations 2 3 10 + + =4 x y z . Let A = ⎢ –2 3 ⎥ ⎢⎣ 1 1 5⎥⎦ (i) [adj A]–1 = adj (A–1) (ii) (A–1)–1 = A x y x+ y 9. z z2 1 + pz 3 3a – a+ b – a+ c 13. Verify that 8. 2 3+ 2 p 4 + 3 p + 2q = 1 15.142 MATHEMATICS ⎡ 1 –2 1⎤ ⎢ 1⎥ . prove that: α α2 β+γ 11. Evaluate 1 x + y y 1 x x+ y Using properties of determinants in Exercises 11 to 15. β β 2 γ + α = (β – γ) (γ – α) (α – β) (α + β + γ) γ γ 2 α +β x x 2 1 + px 3 12. y y 2 1 + py 3 = (1 + pxyz) (x – y) (y – z) (z – x). where p is any scalar. Evaluate y x+ y x x+ y x y 1 x y 10.

4) (D) Det (A) ∈ [2. y. b.P. ∞) (C) Det (A) ∈ (2. Let A = ⎢ ⎢⎣ −1 − sin θ 1 ⎥⎦ (A) Det (A) = 0 (B) Det (A) ∈ (2. c. then the determinant x + 2 x + 3 x + 2a x + 3 x + 4 x + 2b is x + 4 x + 5 x + 2c (A) 0 (B) 1 (C) x (D) 2x ⎡ x 0 0⎤ 18. 17. If x. 4] . Then 19. If a. are in A. z are nonzero real numbers. DETERMINANTS 143 4 6 5 – + =1 x y z 6 9 20 + – =2 x y z Choose the correct answer in Exercise 17 to 19. then the inverse of matrix A = ⎢ 0 y 0⎥ is ⎢ ⎥ ⎢⎣ 0 0 z ⎥⎦ ⎡ x −1 0 0 ⎤ ⎡ x −1 0 0 ⎤ ⎢ ⎥ ⎢ ⎥ (A) ⎢ 0 y −1 0 ⎥ (B) xyz ⎢ 0 y −1 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 0 z −1 ⎦ ⎣ 0 0 z −1 ⎦ ⎡ x 0 0⎤ ⎡1 0 0 ⎤ 1 ⎢ 0 y 0 ⎥⎥ 1 ⎢ (C) (D) 0 1 0 ⎥⎥ xyz ⎢ xyz ⎢ ⎢⎣ 0 0 z ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎡ 1 sin θ 1 ⎤ ⎢ − sin θ 1 sin θ⎥⎥ . where 0 ≤ θ ≤ 2π.

then k . then the given determinant can be expressed as sum of two or more determinants. If any two rows or any two columns are identical or proportional. then value of determinant is multiplied by k. then sign of determinant changes. If A = [aij ]3×3 .144 MATHEMATICS Summary Determinant of a matrix A = [a11]1× 1 is given by | a11| = a11 ⎡a a12 ⎤ Determinant of a matrix A = ⎢ 11 a22 ⎥⎦ is given by ⎣ a21 a11 a12 A = a21 a22 = a11 a22 – a12 a21 ⎡ a1 b1 c1 ⎤ ⎢ c2 ⎥⎥ is given by (expanding along R1) Determinant of a matrix A = ⎢ a2 b2 ⎢⎣ a3 b3 c3 ⎥⎦ a1 b1 c1 b c2 a2 c2 a2 b2 A = a2 b2 c2 = a1 2 − b1 + c1 b3 c3 a3 c3 a3 b3 a3 b3 c3 For any square matrix A. Multiplying a determinant by k means multiply elements of only one row (or one column) by k. the |A| satisfy following properties. where A′ = transpose of A. . If we interchange any two rows (or columns). If to each element of a row or a column of a determinant the equimultiples of corresponding elements of other rows or columns are added. If we multiply each element of a row or a column of a determinant by constant k. then value of determinant is zero. then value of determinant remains same.A = k 3 A If elements of a row or a column in a determinant can be expressed as sum of two or more elements. |A′| = |A|.

then adj A = ⎢ A12 A 22 A32 ⎥ . A square matrix A is said to be singular or non-singular according as | A | = 0 or | A | ≠ 0. If elements of one row (or column) are multiplied with cofactors of elements of any other row (or column). then B is called inverse of A. where A ij is ⎢ ⎥ ⎢⎣ a31 a32 a33 ⎥⎦ ⎢⎣ A13 A 23 A33 ⎥⎦ cofactor of aij A (adj A) = (adj A) A = | A | I. y3) is given by x1 y1 1 1 Δ= x2 y2 1 2 x3 y3 1 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. X = ⎢ y ⎥ and B= ⎢⎢ d 2 ⎥⎥ ⎥ ⎢ ⎥ ⎢⎣ a3 b3 c3 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ d3 ⎥⎦ . A = a11 A11 + a12 A12 + a13 A13. 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. where ⎡ a1 b1 c1 ⎤ ⎡ x⎤ ⎡ d1 ⎤ A = ⎢⎢ a2 b2 c2 ⎥ . where A is square matrix of order n. where B is square matrix. (x2. then these equations can be written as A X = B. then their sum is zero. 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. 1 A –1 = ( adj A) A If a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d2 a3 x + b3 y + c3 z = d3. y1). y2) and (x3. If AB = BA = I. a11 A21 + a12 A22 + a13 A23 = 0 ⎡ a11 a12 a13 ⎤ ⎡ A11 A 21 A31 ⎤ ⎢ ⎥ If A = ⎢ a21 a22 a23 ⎥ . DETERMINANTS 145 Area of a triangle with vertices (x1. For example.

pp 30. The arrangement of rods was precisely that of the numbers in a determinant. 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.” in the proc.. Also on the same day. He may be called the formal founder. Hayashi. The Chinese. In 1773 Lagrange treated determinants of the second and third orders and used them for purpose other than the solution of equations. which for the special case of m = n reduces to the multiplication theorem. Soc. He gave the proof of multiplication theorem more satisfactory than Binet’s.Marie Binet. then system may or may not be consistent. where A ≠ 0 . then there exists no solution (iii) | A | = 0 and (adj A) B = 0. after this the word determinant received its final acceptance. 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.146 MATHEMATICS Unique solution of equation AX = B is given by X = A–1 B. ‘T. there exists unique solution (ii) | A | = 0 and (adj A) B ≠ 0. “The Fakudoi and Determinants in Japanese Mathematics. V. (1812) who stated the theorem relating to the product of two matrices of m-columns and n- rows.Philippe . early developed the idea of subtracting columns and rows as in simplification of a determinant ‘Mikami. 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. . of the Tokyo Math. The greatest contributor to the theory was Carl Gustav Jacob Jacobi. Seki Kowa. therefore. A system of equation is consistent or inconsistent according as its solution exists or not. Cauchy (1812) presented one on the same subject. The next great contributor was Jacques . Vendermonde was the first to recognise determinants as independent functions. In 1801. gave general method of expanding a determinant in terms of its complementary minors. Laplace (1772). 93. Gauss used determinants in his theory of numbers. For a square matrix A in matrix equation AX = B (i) | A | ≠ 0. He used the word ‘determinant’ in its present sense. China.

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- Chapter 8- Em WavesUploaded byAnshuman Singh
- Ch 1-Relations and FunctionsUploaded byAnshuman Singh
- Ch 6-Applications of DerivativesUploaded byAnshuman Singh
- Appendix 1-Proofs in MathematicsUploaded byAnshuman Singh
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- Chapter 2-Potential and CapacitanceUploaded byAnshuman Singh
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- Chapter 1-Electric Chrges and FieldsUploaded byAnshuman Singh
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