Determinants

Top Careers & You _________________________________________________________________________________ ®

DETERMINANTS (1)

Definition The expression

a1 b1 a2 b2

is called a determinant of the second order and stands for ‘a1b2 – a2b1’. It

contains 4 numbers a1, b1, a2, b2 (called elements) which are arranged along two horizontal lines (called rows) and two vertical lines (called columns). a1 b1 c 1

Similarly a 2 b 2 a3 b3

c 2 is called a determinant of the third order. It consists of 9 elements which are c3

arranged in 3 rows and 3 columns. In general, a determinant of the nth order is denoted by a11

a12

a 21

a 22

.... ai 1

.... ai 2

.... a m1

.... am 2

...a1 j

...a1n

...a 2 j

...a 2n

.... ...a ij

. ... ...a in

.... a mj

. ... ...a mn

which is a block of n2 elements arranged in the form of a square along n rows and n columns. The diagonal through the left hand top corner which contains the elements a1, b2, c3, ……..ln is called the leading or principal diagonal. (2)

Minors, Cofactors: The minor of an element in a determinant is the determinant obtained by deleting the row and the column which intersect in that element. a1 b1 c 1

For instance, in ∆ = a 2 b 2 a3 b3

c 2 , minor of b2 = c3

a1 c 1 a2 c 2

and minor of c2 =

a1 b1 a3 b3

The cofactor of any element in a determinant is its minor with the proper sign. The sign of an element in the ith row and jth columns is (– 1)i + j. The cofactor of an element is usually denoted by the corresponding capital letter. Thus the cofactor of b3 = (– 1)3 + 2 × minor of b3 i.e. B3 = – Similarly, (3)

C2 = –

a1 c 1 a2 c 2

a1 b1 a3 b3

Laplace’s expansion. A determinant can be expanded in terms of any row (or column) as follow: Multiply each element of the row (or column) in terms of which we intend expanding the determinant, by its cofactor and then add up all these terms. ∴ Expanding by R1 (i.e. 1st row),

_________________________________________________________________________________________________ Page : 1 www.TCYonline.com

Top Careers & You _________________________________________________________________________________ ®

∆ = a1A1 + b1B1 + c1C1 = a1

b2 c 2 b3 c 3

– b1

a2 c 2 a3 c 3

+ c1

a2 b2 a3 b3

= a1(b2 c3 – b3 c2) – b1(a2 c3 – a3 c2) + c1(a2 b3 – a3 b2) Similarly expanding by C2 (i.e. 2nd column) ∆ = b1B1 + b2B2 + b3B3 = – b1

a2 c 2 a3 c 3

+ b2

a1 c 1 a3 c 3

– b3

a1 c 1 a2 c 2

= – b1(a2 c3 – a3c2) + b2(a1c3 – a3c1) – b3(a1c2 – a2c1) and expanding by R3 (i.e. 3rd row), ∆ = a3A3 + b3B3 + c3C3. Thus ∆ is the sum of the products of the elements of any row (or column) by the corresponding cofactors. If, however, the sum of the products of the elements of any row (or column) by the cofactors of another row (or column) be taken, the result is zero. 0 1 2 3 1 0 3 0 Exam. Find the values of ∆ = 2 3 0 1 3 0 1 2

Since there are two zeros in the second row, therefore, expanding by R2, we get 1 2 3 0 1 3 ∆=– 3 0 1 +0–3 2 3 1 +0 0 1 2 3 0 2

(Expand by C1)

(Expand by R1)

= – [1 (0 × 2 – 1 × 1) – 3(2 × 2 – 1 × 3) + 0] – 3[0 – (2 × 2 – 3 × 1) + 3(2 × 0 – 3 × 3)] = – (– 1 – 3) – 3(– 1 – 27) = 4 + 84 = 88. Properties of a Determinant. •

If all the elements of a row (column) are zero, then the value of the determinant is zero. 1 2 3

A=0 0 0

--Æ A = 0

2 3 4

•

If the rows (column) of a determinant are changed into columns (rows), the value of the determinant remains unaltered. a1 b1 c 1

Let ∆ = a 2 b 2

c2

a3 b3

c3

[Expand by R1]

= a1(b2 c3 – b3 c2) – b1(a2 c3 – a3 c2) + c1(a2 b3 – a3 b2) a1 a 2

a3

Then ∆’ = b 1 b 2

b3

c1 c 2

c3

[Expanding by R1]

= a1(b2c3 – b3c2) – a2(b1c3 – b3c1) + a3(b1c2 – b2c1) = a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2) = ∆.

_________________________________________________________________________________________________ Page : 2 www.TCYonline.com

Top Careers & You _________________________________________________________________________________ ®

•

If the elements of row (column) are (identical) proportional to the elements of any other row (column), then the determinant vanishes.

•

The interchange of a any two rows (columns) of a determinant changes its value in its sign only. a1 a 2

a3

∆ = b1 b 2

b3

c1 c 2

c3

[Expand by R1]

= a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2) Interchanging C2 and C3, we have a1 c 1 b1

∆’ = a 2 c 2 a3 c 3

b 2 [Expanding by R1] b3

= a1 (c2b3 – c3b2) – c1(a2b3 – a3b2) + b1(a2c3 – a3c2) = – {a1(b2c3 – b3c2) – b1 (a2c3 – a3c2) + c1(a2b3 – a3b2) = – ∆. Cor. If a line of ∆ be passed over two parallel lines, i.e., if the resulting determinant is like b1 c 1

a1

b2 c 2

2 a 2 then ∆’ = (–1) ∆.

b3 c 3

a3

In general, if any line of a determinant be passed over m parallel lines, the resulting determinant ∆’ = (– 1)m ∆. •

A determinant vanishes if two parallel lines are identical Consider a determinant ∆ in which two parallel line are identical Interchange of the identical lines leaves the determinant unaltered yet by the previous property, the interchanges of two parallel lines changes the sign of the determinant. Hence ∆ = ∆’ = – ∆ or 2∆ = 0, or ∆ = 0.

•

If all the elements of a row (column) of a determinant are multiplied by a constant k, then the value of the determinant gets multiplied by p. a1 pb 1 c 1

a1 b1 c 1

a 2 pb 2 c 2 = p a 2 b 2

c2

a 3 pb 3 c 3

c3

a3 b3

For on expanding by C2, L.H.S. = – pb1(a2c3 – a3c2) + pb2(a1c3 – a3c1) – pb3(a1c2 – a2c1) = p {– b1B1 + b2B2 – b3B3} = R.H.S. a1

Similarly

•

b1

c1

a1 b1 c 1

ka 2 kb 2

kc 2 = k a 2 b 2

c2

a3

c3

c3

b3

a3 b3

If the elements of row (column) of a determinant are expressed as the sum (difference) of two quantities, then the determinant can be expressed as the sum (difference) of two determinants of the same order.

_________________________________________________________________________________________________ Page : 3 www.TCYonline.com

Top Careers & You _________________________________________________________________________________ ®

a1 b1 c1 + d 1 − e1

Consider the determinant ∆ = a 2 b2

c 2 + d 2 − e2

a 3 b3

c 3 + d 3 − e3

each of whose third column elements consists of three terms. Expanding ∆ by C3, we have ∆ = (c1 + d1 – e1) (a2b3 – a3b2) – (c2 + d2 – e2) (a1b3 – a3b1) + (c3 + d3 – e3) (a1b2 – a2b1) = [c1(a2b3 – a3b2) – c2(a1b3 – a3b1) + c3(a1b2 – a2b1)] + [d1(a2b3 – a3b2) – d2(a1b3 – a3b1) + d3(a1b2 – a2b1)] – [e1(a2b3 – a3b2) – e2(a1b3 – a3b1) + e3(a1b2 – a2b1)] a1 b1 c 1

a1 b1

d1

a1 b1

e1

a2 b2

c 2 + a2 b2

d2 – a 2 b 2

e2

a3 b3

c3

d3

e3

a3 b3

a3 b3

Further, if the elements of three parallel lines consist of m, n and p terms respectively, the determinants can be expressed as the sum of m × n × p determinants. a1 b1 c 1

a1 b1 + β1 c 1 + γ 1

a1 b1

a1 β1 c 1

a2 b2 + β2 c 2 + γ 2 = a2 b2

c 2 + a2 β2 c 2 + a2 b2

a3 b3 + β3 c 3 + γ 3

c3

•

a3 b3

a3 β3 c 3

a3 b3

γ1

a1 β1 γ 1

γ 2 + a2 β2 γ 2 γ3

a3 β3 γ 3

If constant multiples of elements of any row (column) be added to (subtracted from) the corresponding elements of any other row(column) of adeterminant, then the determinant remains unaltered, i.e. if the operations Ri Æ Ri + mRj + nRk, j, k ≠ i and Ci Æ Ci + mCj + nCk, j, k ≠ i are performed on the determinant, it remains unaltered. a1 b1 c 1

Let ∆ = a 2 b 2

c2

a3 b3

c3

a 1 + pb 1 − qc 1 b 1

c1

Then ∆’ = a 2 + pb 2 − qc 2 b 2 c 2 a 3 + pb 3 − qc 3 b 3

c3

a1 b1 c 1

= a2 b2 a3 b3

pb 1 b 1 c 1

− qc 1 b 1

c1

c 2 + pb 2 b 2 c 2 + − qc 2 b 2 c 2 c3

pb 3 b 3 c 3

− qc 3 b 3 c 3

= ∆ + 0 + 0 = ∆. •

Factor Theorem. If the elements of a determinant ∆ are functions of x and two parallel lines become identical when x = a, then x – a is a factor of ∆.

Let

∆ = f (x)

Since

∆ = 0 when x = a,

∴ f (a) = 0.

i.e. (x – a) is a factor of f (x). Hence x – a is a factor of ∆. Obs: If k parallel lines of a determinant ∆ become identical when x = a, then (x – a)k – 1 is a factor of ∆. _________________________________________________________________________________________________ Page : 4 www.TCYonline.com

Top Careers & You _________________________________________________________________________________ ®

a3 a2 a 1

Exam.

Factorize ∆ =

b3 b2 c

3

c

2

d3 d2

b 1 c 1 d 1

Putting a = b, R1 ≡ R2 and hence ≡ = 0. ∴ a – b is a factor of ∆. Similarly, a – c and a – d are also factors of ∆. Again putting b = c, R2 ≡ R3 and hence ∆ = 0. ∴ b – c is a factor of ∆. Similarly b – d and c – d are also factors of ∆. Also ∆ is of the sixth degree in a, b, c, d and therefore, there cannot be any other algebraic factor of ∆. ∴ Suppose ∆ = k(a – b) (a – c) (a – d) (b – c) (b – d) (c – d), where k is a numerical constant. The leading term in ∆ = a3b2c. The corresponding term on R.H.S. = ka3b2c. ∴

k=1

Hence ∆ = (a – b) (a – c) (a – d) (b – c) (b – d) (c – d). 21 17 7 10

Evaluate

Exam.

24 22 6 10 6 8 2 0 5 7 1 2

Operating R1 – R2 – R4, – R2 – 3R3, – 2R4, the given determinant − 8 − 12 0 − 2

∆=

6 −2 −4 −6 5 7

0 1 0 −1 1 2

− 8 − 12 − 2 = 6 −2 1 =0 −4 − 6 −1

[Expand by C1]

(Q R1 = 2R2] a11 a12

Remark: If ∆1 is the 3 × 3 determinant of the co-factors of the elements of ∆ = a 21 a 22 a 31 a 32

∆1 =

C11 C12

C13

C 21 C 22

C 23

C 31 C 32

C 33

a11 a12

a13

C11 C12

C13

∆

0

0

⇒ ∆∆1 = a 21 a 22

a 23

C 21 C 22

C 23 =

0

∆

0 =∆

a 31 a 32

a 33

C 31 C 32

C 33

0

0

∆

a13 a 23 , then a 33

3

⇒ ∆1 = ∆2. In general, if ∆ is n × n determinant, then ∆ 1 = ∆ n–1

_________________________________________________________________________________________________ Page : 5 www.TCYonline.com

Top Careers & You _________________________________________________________________________________ ®

Product of Determinants: The product of two determinants is expressed as a1 b1 c1

α1 α2 α3

a2 b2

c2

× β1 β2

a3 b3

c3

γ1 γ2

a 1α 1 + b 1β 1 + c 1 γ 1

a 1α 2 − b 1β 2 + c 1γ 2

a 1α 3 + b 1β 3 + c 1γ 3

β3 = a 2 α 1 + b 2 β 1 + c 1γ 1

a 2 α 2 − b 2β 2 + c 2 γ 2

a 2 α 3 + b 2β 3 + c 2 γ 3

a 3 α 1 + b 3 β 1 + c 1γ 1

a 3 α 2 − b 3β 2 + c 3 γ 2

a 3 α 3 + b 3β3 + c 3 γ 3

γ3

Differentiation of a Determinant :

(i)

st

Let ∆ (x) be a determinant of order two. If we write ∆ (x) = |C1 C2| where C1 and C2 denote the 1 and 2nd columns, then ∆ (x) = |C1' C2| + | C1 C2‘| where C'i denotes the column which contains the derivative of all the functions in the ith column. In a similar fashion, if we write

∆(x) = (ii)

/ R1 R R the ∆ (x) = 1 + /1 R2 R2 R2

Let ∆ (x) be of order three. If we write ∆ (x) = C1 C 2

'

C 3 then ∆ (x) = C ' 1 C 2

C3

+

C1 C 2 ' C 3 + C1 C 2 C 3 ' and similarly if we consider R1 ∆ (x) = R 2 R3

then (iii)

R1 ' R1 R1 ∆' (x) = R 2 + R 2 ' + R 2 R3 R3 R3 '

If only one row (or column) consists functions of x and other rows are constant, viz. f1 ( x )

∆ (x) = b 1

c1

f2 ( x )

f3 ( x )

b2

b3

c2

f1 ' ( x )

c3

f1n ( x )

n

(Say), then ∆’ (x) = b 1

and in general ∆ (x) = b 1

c1

f 2n ( x )

f3 ' ( x )

b2

b3

c2

c3

f 3n ( x )

b2

c1

f2 ' ( x )

c2

b3 c3

where n is any positive integer.

CHARACTERISTIC EQUATION: If A is any square matrix of order n, we can form the matrix A – λ I, where I is the nth order unit matrix. The determinant of this matrix equated to zero, i.e. a11 − λ a12

| A – λI | =

...

a1n

a 21

a 22 − λ ....

a2n

... a n1

.... an 2

... ... ... ann − λ

=0

is called the characteristic equation of A. On expanding the determinant, the characteristic equation takes the form (– 1)nλn + k1λn – 1 + k2λn – 2 + …..kn = 0,

_________________________________________________________________________________________________ Page : 6 www.TCYonline.com

Top Careers & You _________________________________________________________________________________ ®

where k’s are expressible in terms of the elements aij. The roots of this equation are called the characteristic roots or latent roots or eigen-values of the matrix A.

The process of finding the eigen values of a matrix: Let A = [aij]nxn be a square matrix of order n. First we should write the characteristic equation of the matrix A i.e. the equation [A – λ I] = 0. This equation will be of degree n in λ . So it will have n roots. These n roots will give us the eigen values of the matrix A.

Exam.

⎡5 4⎤ Find the eigen values of the matrix ⎢ ⎥. ⎣1 2 ⎦

The characteristic equation is [A – λI] = 0 i.e.

or

5−λ 1

4 2−λ

=0

(λ − 6) (λ − 1) = 0

or λ2 – 7λ + 6 = 0 ∴ λ = 6, 1.

Thus the eigen values are 6 and 1.

Properties of eigen values: 1.

The sum of the eigen values of A is equal to the trace of A i.e. sum of the diagonal elements.

2.

The product of the eigen values of A is equal to |A|

3.

The non-zero eigen values of A B are equal to the non-zero eigen values of B A. As a consequence, the traces of B A and A B are equal.

4.

The eigen values of a diagonal matrix are the diagonal elements.

5.

The non-zero eigen values of A B are equal to the non-zero eigen values of B A. As a consequence, the traces of B A and A B are equal.

6.

If λ be the eigen value of A, then eigen value of A–1 is λ–1 =

7.

If λ be the eigen value of A, then the eigen value of

1 . λ

akAk + ak – 1Ak – 1 + ……… + a1A + a0I is akλk + ak – 1λk – 1 + ……… + a1λ + a0, where ak ,ak −1,K,a1,a0 are real numbers and k is a positive integer.

Ex.

The eigen values of A are 2 and –1. Thus, the eigen values of A5 + 2A – 12I are (2)5 + 2(2) – 12 = 32 + 4 – 12 = 24 and (– 1)5 + 2(– 1) – 12 = – 1 – 2 – 12 = – 15 …

_________________________________________________________________________________________________ Page : 7 www.TCYonline.com