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Chapter 4

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)

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

If A= ⎢ ⎥ .e. 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 = .e. –5 –1 cos θ – sin θ x2 – x + 1 x – 1 2. 3 – x2 = 3 – 8 i.108 MATHEMATICS Solution Expanding along R1. Evaluate the determinants 3 –1 –2 3 –4 5 (i) 0 0 –1 (ii) 1 1 –2 3 –5 0 2 3 1 . x2 = 8 Hence x= ±2 2 EXERCISE 4. then show that | 3 A | = 27 | A | ⎣⎢ 0 0 4 ⎥⎦ 5. 2 4 1.1 Evaluate the determinants in Exercises 1 and 2. x 1 4 1 3 x 3 2 Solution We have = x 1 4 1 i. If A = ⎢ 0 1 2 ⎥ . (i) (ii) sin θ cos θ x +1 x +1 ⎡1 2⎤ 3.

we will study some properties of determinants which simplifies its evaluation by obtaining maximum number of zeros in a row or a column. 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. then x is equal to 18 x 18 6 (A) 6 (B) ± 6 (C) – 6 (D) 0 4. If A = ⎢ 2 1 –3 ⎥ . If = . 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. find | A | ⎢⎣ 5 4 –9 ⎥⎦ 7. In this section. a1 a2 a3 Verification Let Δ = b1 b2 b3 c1 c2 c3 Expanding along first row. we shall restrict ourselves upto determinants of order 3 only.3 Properties of Determinants In the previous section. Find values of x. 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. we have learnt how to expand the determinants. 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 get the determinant a1 b1 c1 Δ1 = a2 b2 c2 a3 b3 c3 .

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. then det (A) = det (A′). 2 –3 5 Example 6 Verify Property 1 for Δ = 6 0 4 1 5 –7 Solution Expanding the determinant along first row. Property 1 is verified. a1 a2 a3 Verification Let Δ = b1 b2 b3 c1 c2 c3 .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. then sign of determinant changes. 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. $ Note If R = ith row and C = ith column. where A′ = transpose of A. we will symbolically write C ↔ R i i i Let us verify the above property by example. Property 2 If any two rows (or columns) of a determinant are interchanged. then for interchange of row and i columns.

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

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

then its value is zero. we can take out any common factor from any one row or any one column of a given determinant. (ii) If corresponding elements of any two rows (or columns) of a determinant are proportional (in the same ratio). 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. 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.H. a1 + λ1 a2 + λ 2 a3 + λ 3 a1 a2 a3 λ1 λ 2 λ3 For example.S. then the determinant can be expressed as sum of two (or more) determinants. 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. = b1 b2 b3 c1 c2 c3 .

the value of determinant remain same if we apply the operation Ri → Ri + kRj or Ci → Ci + k Cj . 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 write this operation as R1 → R1 + k R3. Symbolically. Verification a1 a2 a3 a1 + k c1 a2 + k c2 a3 + k c3 Let Δ = b1 b2 b3 and Δ1 = b1 b2 b3 . Here. c1 c2 c3 c1 c2 c3 Similarly.114 MATHEMATICS Expanding the determinants along the first row.e. then value of determinant remains the same.. i. we may verify Property 5 for other rows or columns. 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. c1 c2 c3 c1 c2 c3 where Δ1 is obtained by the operation R1 → R1 + kR3 . to each element of any row or column of a determinant. the equimultiples of corresponding elements of other row (or column) are added. 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.

A similar remark applies to column operations. then Δ1 = kΔ. 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. a a+b a+b+c Example 11 Prove that 2a 3a + 2b 4a + 3b + 2c = a 3 . (ii) If more than one operation like Ri → Ri + kRj is done in one step. DETERMINANTS 115 Now. 3a 6a + 3b 10a + 6b + 3c Solution Applying operations R2 → R2 – 2R1 and R3 → R3 – 3R1 to the given determinant Δ. care should be taken to see that a row that is affected in one operation should not be used in another operation. 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. we obtain a 2a + b Δ= a +0+0 0 a = a (a2 – 0) = a (a2) = a3 .

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 .116 MATHEMATICS Example 12 Without expanding. 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. Δ = 0. respectively. prove that x+y y+z z+x Δ= z x y =0 1 1 1 Solution Applying R1 → R1 + R2 to Δ. Example 13 Evaluate 1 a bc Δ = 1 b ca 1 c ab Solution Applying R2 → R2 – R1 and R3 → R3 – R1. 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.

z are different and Δ = y y 2 1 + y 3 = 0 . DETERMINANTS 117 Applying R1 → R1 – R2 – R3 to Δ. we get 0 –2c –2b Δ= b c+a b c c a+b Expanding along R1. 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 . 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. y.

. 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. z are all different.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.b. we get 1 1 1 +1 a a a 1 1 1 L. 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. z – x ≠ 0. y. x – y ≠ 0.c common from R1. R2 and R3. y – z ≠ 0.S.H. = abc +1 b b b 1 1 1 +1 c c c Applying R1→ R1 + R2 + R3. 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 .e. i.

H. 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. 1 1 2 3 3 2 1 1 3 EXERCISE 4. 1 ca b ( c + a ) = 0 5 9 86 1 ab c ( a + b ) b+c q+r y+z a p x 5. 3 8 75 = 0 4. then apply C →C –aC. y b y +b = 0 2. c+a r+ p z+x = 2 b q y a+b p+q x+ y c r z . ⎝ a b c⎠ $ Note Alternately try by applying C → C – C and C → C – C . prove that: x a x+a a −b b −c c − a 1. 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 – C1.2 Using the property of determinants and without expanding in Exercises 1 to 7.S. 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. C3 → C3 – C1.

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. ba −b 2 bc = 4 a 2 b 2 c 2 b c 0 ca cb −c 2 By using properties of determinants. (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. in Exercises 8 to 14. (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 .120 MATHEMATICS 0 a −b −a2 ab ac 6. − a 0 −c = 0 7. (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.

(B) Determinant is a number associated to a matrix. (C) Determinant is a number associated to a square matrix.4 Area of a Triangle In earlier classes. y1). 15. (1) 2 x3 y3 1 Remarks (i) Since area is a positive quantity. 2ab 1− a + b 2 2 2a = 1 + a2 + b2 2b −2a 1 − a2 − b2 a2 + 1 ab ac 14. ab b +1 2 bc =1 + a 2 + b2 + c 2 ca cb c2 + 1 Choose the correct answer in Exercises 15 and 16. (D) None of these 4.. . is given by the expression [x (y –y ) + x2 (y3–y1) + 2 1 2 3 x3 (y1–y2)]. (x2. Let A be a square matrix of order 3 × 3. Which of the following is correct (A) Determinant is a square matrix. y3). we have studied that the area of a triangle whose vertices are 1 (x1. x2 1 x = 1 − x3 x x2 1 1 + a 2 − b2 2ab −2b ( ) 3 13. y2) and (x3. then | kA | is equal to (A) k| A | (B) k 2 | A | (C) k 3 | A | (D) 3k | A | 16.. Now this expression can be written in the form of a determinant as x1 y1 1 1 Δ= x2 y2 1 . DETERMINANTS 121 1 x x2 ( ) 2 12. we always take the absolute value of the determinant in (1).

e. 8) (iii) (–2. 1). = ± 3 . 0). (3. (10.3 1. we have 1 3 1 1 0 0 1 =±3 2 k 0 1 − 3k This gives. Example 17 Find the area of the triangle whose vertices are (3. k = ∓ 2. (6. since the area of the triangle ABD is 3 sq. y) be any point on AB. 0) is a point such that area of triangle ABD is 3sq units. (–1. Also. 8). use both positive and negative values of the determinant for calculation. 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. Then. 2). 0) using determinants and find k if D(k. 3) and B (0. units. area of triangle ABP is zero (Why?). 2 which is the equation of required line AB. 3) (ii) (2. 1). 0). Find area of the triangle with vertices at the point given in each of the following : (i) (1. i.122 MATHEMATICS (ii) If area is given. 2) and (5. Solution Let P (x. So 0 0 1 1 1 3 1 =0 2 x y 1 1 This gives ( y – 3 x ) = 0 or y = 3x. (4. 7). (iii) The area of the triangle formed by three collinear points is zero. –8) .. (1. –3). 2 EXERCISE 4. (– 4.

6) using determinants. 5. –2 4. (5. (ii) Find equation of line joining (3. Show that points A (a. 4). c + a). (0. Find values of k if area of triangle is 4 sq. (0. where Mij is minor of aij . 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. 1) and (9. 2) and (3. 0). denoted by Aij is defined by Aij = (–1)i + j Mij . 2) (ii) (–2. 0). Remark Minor of an element of a determinant of order n(n ≥ 2) is a determinant of order n – 1. 7 8 Definition 2 Cofactor of an element aij . DETERMINANTS 123 2. (4. b + c). Minor of an element aij is denoted by Mij. So M11 = Minor of a11= 3 M12 = Minor of the element a12 = 4 M21 = Minor of the element a21 = –2 . If area of triangle is 35 sq units with vertices (2. a + b) are collinear. 0). – 6). units and vertices are (i) (k. 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. B (b. 3) using determinants. k) 4. 3. –2 (D) 12. 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. 4). its minor M23 is given by 1 2 M23 = = 8 – 14 = – 6 (obtained by deleting R2 and C3 in Δ). Then k is (A) 12 (B) –2 (C) –12.5 Minors and Cofactors In this section. (0. we will learn to write the expansion of a determinant in compact form using minors and cofactors. 4) and (k. C (c. (i) Find equation of line joining (1.

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

A31 = (–1)3+1 (–12) = –12 0 4 . A21 = (–1)2+1 (– 4) = 4 5 –7 2 5 M22 = = –14 – 5 = –19. 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. A13 = (–1)1+3 (30) = 30 1 5 –3 5 M21 = = 21 – 25 = – 4. we can try for other rows and columns. 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. 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. A12 = (–1)1+2 (– 46) = 46 1 –7 6 0 M13 = = 30 – 0 = 30.

To find inverse of a matrix A. (i) 0 1 0 (ii) 3 5 –1 0 0 1 0 1 2 5 3 8 3. A33 = (–1)3+3 (18) = 18 6 0 Now a11 = 2. In this section. . A31 = –12. (i) (ii) 0 3 b d 1 0 0 1 0 4 2. evaluate Δ = 1 y zx .4 Write Minors and Cofactors of the elements of following determinants: 2 –4 a c 1. A32 = 22. a12 = –3. a13 = 5. evaluate Δ = 2 0 1 . A–1 we shall first define adjoint of a matrix. A32 = (–1)3+2 (–22) = 22 6 4 2 –3 and M33 = = 0 + 18 = 18. 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. we have studied inverse of a matrix.e. Using Cofactors of elements of third column. we shall discuss the condition for existence of inverse of a matrix. 1 z xy a11 a12 a13 5.126 MATHEMATICS 2 5 M32 = = 8 – 30 = –22. i. Using Cofactors of elements of second row. 1 2 3 1 x yz 4.. If Δ = a21 a22 a23 and Aij is Cofactors of aij .6 Adjoint and Inverse of a Matrix In the previous chapter. A33 = 18 So a11 A31 + a12 A32 + a13 A33 = 2 (–12) + (–3) (22) + 5 (18) = –24 – 66 + 90 = 0 EXERCISE 4.

Theorem 1 If A be any given square matrix of order n. where I is the identity matrix of order n . A22 = 2 ⎡ A11 A 21 ⎤ ⎡ 4 –3⎤ Hence adj A = ⎢ ⎥ =⎢ ⎥ ⎣ A12 A 22 ⎦ ⎣ –1 2 ⎦ Remark For a square matrix of order 2. A21 = –3. We state the following theorem without proof. DETERMINANTS 127 4.6.1 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. A12 = –1.. Adjoint of the matrix A is denoted by adj A. then A(adj A) = (adj A) A = A I . where Aij is the cofactor of the element aij . 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. i.

AB = A B . that is. ⎣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. Then A = = 4 – 6 = – 2 ≠ 0. 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. ⎡1 2⎤ For example. 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. 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 ⎥⎦ .128 MATHEMATICS Verification ⎡ a11 a12 a13 ⎤ ⎡ A11 A 21 A 31 ⎤ ⎢a ⎥ ⎢ Let A = ⎢ 21 a22 a23 ⎥ . Definition 5 A square matrix A is said to be non-singular if A ≠ 0 ⎡1 2 ⎤ 1 2 Let A= ⎢ ⎥ . the determinant of matrix A = ⎢ 4 8 ⎥ is zero ⎣ ⎦ Hence A is a singular matrix. 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. then AB and BA are also nonsingular matrices of the same order.

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

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

5 Find adjoint of each of the matrices in Exercises 1 and 2. ⎡ 2 3 ⎤ ⎡ 2 3 ⎤ ⎡ 7 12⎤ Solution We have A 2 = A. 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. ⎣ 1 2⎦ where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix. ⎢ 3 0 −2⎥ ⎣ ⎦ ⎢⎣1 0 3 ⎥⎦ . find A–1. ⎡ 1 −1 2⎤ ⎡ 1 2⎤ ⎢ ⎥ 1.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. ⎢ 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. ⎢ −4 −6 ⎥ 4. ⎢ 3 4⎥ 2. Using this equation.

⎢ 4 −1 0⎥ 10.132 MATHEMATICS Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11. If A is an invertible matrix of order 2. For the matrix A = ⎢ 1 2 −3⎥ ⎢⎣ 2 −1 3 ⎥⎦ Show that A3– 6A2 + 5A + 11 I = O. ⎢ ⎥ 9. 2 ⎣ 1 1 ⎦ ⎡1 1 1 ⎤ ⎢ ⎥ 15. If A = ⎢ −1 2 −1⎥ ⎣⎢ 1 −1 2 ⎥⎦ Verify that A3 – 6A2 + 9A – 4I = O and hence find A–1 17. Then | adj A | is equal to (A) | A | (B) | A | 2 (C) | A | 3 (D) 3 | A | 18. ⎢ 0 2 4⎥ ⎣ ⎦ ⎣ −3 2 ⎦ ⎢⎣ 0 0 5⎥⎦ ⎡1 0 0 ⎤ ⎡ 2 1 3⎤ ⎡ 1 −1 2 ⎤ ⎢3 3 0 ⎥ ⎢ ⎥ ⎢ ⎥ 8. For the matrix A = ⎢ ⎥ . Hence. ⎡ 2 −1 1 ⎤ ⎢ ⎥ 16. ⎢ ⎥ ⎢⎣ 0 sin α − cos α ⎥⎦ ⎡3 7⎤ ⎡ 6 8⎤ 12. If A = ⎢ ⎥ . Let A be a nonsingular square matrix of order 3 × 3. ⎢ 0 2 −3⎥ ⎢⎣ 5 2 −1⎥⎦ ⎢⎣ −7 2 1⎥⎦ ⎣⎢ 3 −2 4 ⎥⎦ ⎡1 0 0 ⎤ ⎢ 0 cos α sin α ⎥ 11. Let A = ⎢ ⎥ and B = ⎢ ⎥ . Hence find A . find A–1. ⎢ ⎥ 7. 2 –1 ⎣ −1 2 ⎦ ⎡3 2 ⎤ 14. find the numbers a and b such that A + aA + bI = O. ⎢4 3 ⎥ 6. –1 –1 –1 ⎣2 5⎦ ⎣ 7 9⎦ ⎡ 3 1⎤ 13. Verify that (AB) = B A . ⎡ 1 2 3⎤ ⎡ 2 −2 ⎤ ⎡ −1 5 ⎤ ⎢ ⎥ 5. then det (A–1) is equal to 1 (A) det (A) (B) det (A) (C) 1 (D) 0 . show that A – 5A + 7I = O.

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

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

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

2x – y = –2 9. x – y + z = 4 3 3x + 2y = 5 x – 2y – z = 2x + y – 3z = 0 2 3y – 5z = 9 x+y+z=2 13. 2x + 3y +3 z = 5 14. 2x – y = 5 3. 4x – 3y = 3 7x + 3y = 5 3x + 4y = 3 3x – 5y = 7 10.136 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. x + 2y = 2 2. x + y + z = 1 5. x + 3y = 5 2x + 3y = 3 x+y=4 2x + 6y = 8 4. y = 2. 2x + y + z = 1 12. 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. 5x + 2y = 3 11. 3x–y – 2z = 2 6. 1. z = 3 EXERCISE 4. x – y + 2z = 7 x – 2y + z = – 4 3x + 4y – 5z = – 5 3x – y – 2z = 3 2x – y + 3z = 12 . 5x + 2y = 4 8. in Exercises 7 to 14.6 Examine the consistency of the system of equations in Exercises 1 to 6. 7. using matrix method.

The cost of 2 kg onion. Find cost of each item per kg by matrix method. Miscellaneous Examples Example 30 If a. 3 kg wheat and 2 kg rice is Rs 60. 4 kg wheat and 6 kg rice is Rs 90. find A . DETERMINANTS 137 ⎡ 2 –3 5⎤ ⎢3 2 – 4⎥ –1 –1 15. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. c are positive and unequal. The cost of 4 kg onion. c a b Solution Applying C1 → C1 + C2 + C3 to the given determinant. show that value of the determinant a b c Δ = b c a is negative. If A = ⎢ ⎥ . Using A solve the system of equations ⎢⎣ 1 1 –2 ⎥⎦ 2x – 3y + 5z = 11 3x + 2y – 4z = – 5 x + y – 2z = – 3 16. b.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) . 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.

R3 → z R3 to Δ and dividing by xyz. C3 → C3– C1. y. respectively. 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. R2 → yR2 .P.138 MATHEMATICS Example 31 If a. z from C1 C2 and C3. are in A. 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. b. 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. 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 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 . c.

DETERMINANTS 139 Taking common factor (x + y + z) from C2 and C3. 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 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 ⎟ . 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 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 ⎥⎦ .

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 . in matrix form. given system of equations can be written. 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. 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.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.

If a. b and c are real numbers. x+a x x 5. and b+ c c+ a a+ b Δ = c + a a + b b + c = 0. sin α cos β sin α sin β cos α 4. find AB –1 7. a+ b b+ c c+ a Show that either a + b + c = 0 or a = b = c. 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. c c2 ab 1 c2 c3 cos α cos β cos α sin β – sin α 3. Solve the equation x x+a x = 0. Evaluate – sin β cos β 0 . 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 ⎥ . Without expanding the determinant. DETERMINANTS 141 Applying R2 → R2 – x R1. cos θ 1 x a a2 bc 1 a2 a3 2. prove that b b 2 ca = 1 b 2 b3 . a ≠ 0 x x x+a a 2 bc ac + c 2 6. If A = ⎢ –1 ⎥ ⎢ ⎥ ( ) ⎢⎣ 5 –2 2 ⎥⎦ ⎢⎣ 0 –2 1 ⎥⎦ . Prove that the determinant – sin θ – x 1 is independent of θ.

Let A = ⎢ –2 3 ⎥ ⎢⎣ 1 1 5⎥⎦ (i) [adj A]–1 = adj (A–1) (ii) (A–1)–1 = A x y x+ y 9. prove that: α α2 β+γ 11. y y 2 1 + py 3 = (1 + pxyz) (x – y) (y – z) (z – x). Evaluate y x+ y x x+ y x y 1 x y 10. 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. Evaluate 1 x + y y 1 x x+ y Using properties of determinants in Exercises 11 to 15. Verify that 8. –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 . 2 3+ 2 p 4 + 3 p + 2q = 1 15. z z2 1 + pz 3 3a – a+ b – a+ c 13. 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. where p is any scalar.142 MATHEMATICS ⎡ 1 –2 1⎤ ⎢ 1⎥ .

17. Let A = ⎢ ⎢⎣ −1 − sin θ 1 ⎥⎦ (A) Det (A) = 0 (B) Det (A) ∈ (2. c. If a. where 0 ≤ θ ≤ 2π. are in A. 4] . y. If x. 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. 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. ∞) (C) Det (A) ∈ (2. b.P. Then 19. 4) (D) Det (A) ∈ [2. 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 A′ = transpose of A.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.  |A′| = |A|.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. then value of determinant remains same. then the given determinant can be expressed as sum of two or more determinants. then k .  If we interchange any two rows (or columns). then sign of determinant changes.  If we multiply each element of a row or a column of a determinant by constant k. then value of determinant is zero.  If A = [aij ]3×3 . . the |A| satisfy following properties.  If any two rows or any two columns are identical or proportional.  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.  Multiplying a determinant by k means multiply elements of only one row (or one column) by k. then value of determinant is multiplied by k.

then their sum is zero. where B is square matrix. where ⎡ a1 b1 c1 ⎤ ⎡ x⎤ ⎡ d1 ⎤ A = ⎢⎢ a2 b2 c2 ⎥ . DETERMINANTS 145  Area of a triangle with vertices (x1. a11 A21 + a12 A22 + a13 A23 = 0 ⎡ a11 a12 a13 ⎤ ⎡ A11 A 21 A31 ⎤ ⎢ ⎥  If A = ⎢ a21 a22 a23 ⎥ . Also A–1 = B or B–1 = A and hence (A–1)–1 = A. then these equations can be written as A X = B. 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.  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.  If AB = BA = I. y1).  A square matrix A is said to be singular or non-singular according as | A | = 0 or | A | ≠ 0. (x2. y2) and (x3. For example. 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 ⎥⎦ . then adj A = ⎢ A12 A 22 A32 ⎥ .  If elements of one row (or column) are multiplied with cofactors of elements of any other row (or column). where A is square matrix of order n. A = a11 A11 + a12 A12 + a13 A13. For example.  A square matrix A has inverse if and only if A is non-singular. 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.

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