Determinants

Prepared by:

Joey F. Valdriz

Recall:

Inverse of a Matrix

SUB-TOPICS

Determinant of a Square Matrix Minors and Cofactors

Properties of Determinants Applications of Determinants Area of a Triangle

Condition of Collinearity of Three Points Solution of System of Linear Equations

(Cramer’s Rule)

DETERMINANT

Every square matrix has associated with it a scalar called its determinant.

Given a matrix A, we use det(A) or |A| to designate its

determinant.

We can also designate the determinant of matrix A by replacing the brackets by vertical straight lines. For

example,

⎣ ⎦

₃⎥

⎢₀

1_⎤

A = ^⎡2

2 1

0 3

det(A) =

Definition 1: The determinant of a 1×1 matrix [a] is the scalar

is the

a.

Definition 2: The determinant of a 2×2 matrix

scalar ad-bc.

For higher order matrices, we will use a recursive procedure to compute determinants.

⎣ ⎦

⎢_{c d} ⎥

_⎡a b _⎤

Example

Evaluate the determinant : ⁴

- 3

2 5

Solution : ⁴

^{- 3} = 4 × 5 - 2 × ₍-3₎ = 20 + 6 = 26

2 5

Solution

The determinant of a 3 × 3 matrix A,

where

⎢

21 22 23 _⎥

13

12

11

^⎣a31 ^a32 ^a33 ^⎦

_a ⎥

_⎡a a a _⎤

A = ^⎢a a

is a real number defined as

det A = a₁₁a₂₂a₃₃

^{+ a}12^a23^a31 ^{+ a}13^a21^a32

^−(a31^a22^a13 ^{+ a}32^a23^a11 ^{+ a}33^a21^a12 ^{) .}

Solution

If A =

is a square matrix of order 3, then

a

^⎡a11

⎢

^a23 ^⎥

⎢

^⎢⎣^a31

^a12 ^a13 ^⎤

21 ^a22 _⎥

^a32 ^a33 ^⎥⎦

22 23 21 23 21 22

a a

a a a a

| A |= a₂₁

^a11 ^a12 ^a13

^a22 ^a23

31 32 33

= a₁₁ - a₁₂ + a₁₃

^a32 ^a33 ^a31 ^a33 ^a31 ^a32

a

a a

[Expanding along first row]

^{= a}11 ^(a22^a33 ^{- a}32^a23 ^{) - a}12 ^(a21^a33 ^{- a}31^a23 ^{) + a}13 ^(a21^a32 ^{- a}31^a22 ⁾

^{= (a}11^a22^a33 ^{+ a}12^a31^a23 ^{+ a}13^a21^a32 ^{) − (a}11^a23^a32 ^{+ a}12^a21^a33 ^{+ a}13^a31^a22 ⁾

Example

= 2 ₄

1 - 2 _{- 3} 7

1 -3

- 2 _{+ (-5)} 7 1

1 -3 4

[Expanding along first row]

​

= 2 ₍1 + 8₎ - 3 ₍7 - 6₎ - 5 ₍28 + 3₎

= 18 - 3 - 155

= -140

Properties of Determinants

1. The value of a determinant remains unchanged, if its rows and columns are interchanged.

a₁ b₁ c₁ a₁ a₂ a₃ a₂ b₂ c₂ = b₁ b₂ b₃ a₃ b₃ c₃ c₁ c₂ c₃

i.e. A = A '

2. If any two rows (or columns) of a determinant are interchanged, then the value of the determinant is changed by minus sign.

a₁ b₁ c₁

a₂ b₂ c₂

a₃ b₃ c₃

a₂ b₂ c₂

= - a₁ b₁ c₁

a₃ b₃ c₃

_[Applying R₂ ↔ R_{1 ]}

Properties

3. If all the elements of a row (or column) is multiplied by a non-zero number k, then the value of the new determinant is k times the value of the original determinant.

​

ka₁ kb₁ kc₁ a₁ b₁ c₁ a₂ b₂ c₂ = k a₂ b₂ c₂ a₃ b₃ c₃ a₃ b₃ c₃

​

which also implies

​

a₁ b₁ c_{1 1} ma₁ mb₁ mc₁ a₂ b₂ c₂ = _m a₂ b₂ c₂ a₃ b₃ c₃ a₃ b₃ c₃

Properties

4. If each element of any row (or column) consists of two or more terms, then the determinant can be expressed as the sum of two or more determinants.

a₁ + x a₂ + y a₃ + z

b₁ c₁ a₁ b₂ c₂ = a₂ b₃ c₃ a₃

b₁ c₁ x b₁ c₁ b₂ c₂ + y b₂ c₂ b₃ c₃ z b₃ c₃

5. The value of a determinant is unchanged, if any row (or column) is multiplied by a number and then added to any other row (or column).

= a₂ + mb₂ - nc₂

a₁ b₁ c₁ a₁ + mb₁ - nc₁ a₂ b₂ c₂

a₃ b₃ c₃

a₃ + mb₃ - nc₃

b₁ c₁

b₂ c₂

b₃ c₃

_[Applying C₁ → C₁ + mC₂ - nC_{3 ]}

Properties

6. If any two rows (or columns) of a determinant are identical, then its value is zero.

7. If each element of a row (or column) of a determinant is zero, then its value is zero.

​

0 0 0

a₂ b₂ c₂ = 0 a₃ b₃ c₃

c₂ = 0

a₁ b₁ c₁ a₂ b₂

a₁ b₁ c₁

Properties

(⁸)

⎡^a

⁰⎤

Let A = ^⎢0

0

b 0^⎥ be a diagonal matrix, then 0

⎢

⎢_⎣0

⎥

_c⎥_⎦

a 0 0

A = 0 b 0 = abc

0 0 c

The Minor of an Element

Copyright © 2011 Pearson Education, Inc.

Slide 7.5-14

• The determinant of each 3 × 3 matrix is called a minor of the associated element.
• The symbol M_ij represents the minor when the

ith row and jth column are eliminated.

The Cofactor of an Element

Copyright © 2011 Pearson Education, Inc.

Slide 7.5-15

• To find the determinant of a 3 × 3 or larger square matrix:
1. Choose any row or column,
2. Multiply the minor of each element in that row or column by a +1 or –1, depending on whether the sum of i + j is even or odd,
3. Then, multiply each cofactor by its corresponding element in the matrix and find the sum of these products. This sum is the determinant of the matrix.

Let M_ij be the minor for element a_ij in an n × n matrix.

The cofactor of a_ij, written A_ij, is

ij

ij

A = ( )

i + j

−1 ⋅ M .

Finding the Determinant

Copyright © 2011 Pearson Education, Inc.

Slide 7.5-16

Example

− 3 − 2_⎤

• _{4 − 3⎥}, expanding

0 2_⎥⎦

by the second column.

Solution First find the minors of each element in the second column.

_⎢⎣−1

_⎡ 2

Evaluate det _⎢−1

⎢_⎣ ^{1 − 3}_⎦⎥

^{− 2⎤} = 2(−3) − (−1)(−2) = −8

= det^⎡₋²

⎢_⎣ ^{1 2}⎥_⎦

^{− 2⎤} = 2(2) − (−1)(−2) = 2

= det^⎡₋²

^M 32

^M 22

12

⎢_⎣^{−1 2}⎥_⎦

^{− 3⎤} = −1(2) − (−1)(−3) = −5

M = det^⎡−1

Finding the Determinant

Copyright © 2011 Pearson Education, Inc.

Slide 7.5-17

Now, find the cofactor.

The determinant is found by multiplying each cofactor by its corresponding element in the matrix and finding the sum of these products.

32 32

22

22

12 12

A = (−1)³⁺² ⋅ M = (−1)⁵ ⋅ (−8) = 8

A = (−1)²⁺² ⋅ M = (−1)⁴ ⋅ (2) = 2

A = (−1)¹⁺² ⋅ M = (−1)³ ⋅ (−5) = 5

− 3 − 2_⎤

− 4 − 3_⎥ = a₁₂ ⋅ A₁₂ + a₂₂ ⋅ A₂₂ + a₃₂ ⋅ A₃₂

0 2_⎥⎦

= −3(5) + (−4)(2) + (0)(8)

= −23

_⎡ 2

det_⎢−1

_⎢⎣−1

VALUE OF DETERMINANT IN TERMS OF MINORS AND COFACTORS

^⎡a11

If A = ^⎢a₂₁

a₂₃ ^⎥, then

^a12 ^a13 ^⎤

_⎢ ^a22 _⎥

^⎢⎣^a31 ^a32 ^a33 ^⎥⎦

3 3

^{A =} ∑^{(−1)i+ j aijMij =} ∑^aijCij j=1 j=1

​

​

= a_i1C_i1 + a_i2C_i2 + a_i3C_i3 , for i =1 or i = 2 or i = 3

ROW (COLUMN) OPERATIONS

Following are the notations to evaluate a determinant:

​

(i) R_i to denote ith row

3. R_i↔R_i + λR_j to denote the addition of λ times the elements of jth row to the corresponding elements of ith row.
4. λR_i to denote the multiplication of all elements of ith row by λ.

​

Similar notations can be used to denote column operations by replacing R with C.

(ii) R_i ^↔R_j

rows.

to denote the interchange of ith and jth

EVALUATION OF DETERMINANTS

If a determinant becomes zero on putting

_{x = α, then (x - α)}is the factor of the determinant.

For example, if Δ = x²

_x3

x 5 2

9 4 , then at x = 2

16 8

Δ = 0^{, because C}₁ ^{and C}₂ ^{are identical at x = 2}

Hence, (x – 2) is a factor of determinant .

Δ

SIGN SYSTEM FOR EXPANSION OF DETERMINANT

Sign System for order 2 and order 3 are given by

+ – +

+ –

₊ , – + –

–

+ – +

EXAMPLE - 1

6 1 6

= 7 4 7 4 _[Taking out 7 common from C_{1 ]}

2 3 2

6 -3 2

2 -1 2

-10 5 2

42 1 6

28 7 4

14 3 2

Find the value of the following determinants

(i)

(ii)

Solution :

= 7 × 0

=0

⎡_⎣ ^C₁ ^{and C}₃ ^{are identical}⎤_⎦

EXAMPLE –1 (II)

6 -3 2

2 -1 2

-10 5 2

(ii)

−3 × ₍−2₎ −3 2

= −1× ₍−2₎ −1 2 5 × ₍−2₎ 5 2

⎡_⎣^{Taking out − 2 common from C}₁ ⎤_⎦

−3 −3 2

= (−2) −1 −1 2

5 5 2

⎡_⎣ ^C₁ ^{and C}₂ ^{are identical}⎤_⎦

= (−2) × 0

= 0

EXAMPLE - 2

Evaluate the determinant

1 a b+c

1 b c+a

1 c a+b

Solution :

_[ Applying c₃ → c₂ +c_{3 ]}

1 a b+c 1 1 b c+a = 1

1 c a+b 1

a a+b+c b a+b+c c a+b+c

1 a 1

= ₍a+b+c₎ 1 b 1

1 c 1

⎡_⎣Taking ₍a+b+ c₎ common from C₃ ⎤_⎦

= ₍a + b + c₎ × 0

=0

⎡_⎣ ^C₁ ^{and C}₃ ^{are identical}_⎦⎤

EXAMPLE - 3

We have a²

a b c

_b2 _c2

bc ca ab

_[Applying C₁ → C₁ -C₂ and C₂ → C₂ - C_{3 ]}

b- c c

(b- c)(b+c) c²

-a(b- c) ab

(a-b)

= (a-b)(a+b)

-c(a-b)

1

1 c

=(a-b)(b- c) a+b b+ c c²

-c -a ab

⎡Taking ₍a-b₎ and ₍b- c₎ common ⎤

⎢

_⎣from C₁ and C₂ respectively

⎥

⎦

a b c

_a2 _b2 _c2

bc ca ab

Evaluate the determinant:

Solution:

SOLUTION CONT.

0

=(a-b)(b- c) -(c- a)

-(c- a)

=-(a-b)(b- c)(c- a) 1

0

1 c

b+ c c² _[Applying c → c - c _]

1 1 2

-a ab

1. 1 c b+ c c²
2. -a ab

1

= -(a-b)(b- c)(c- a) 0

c a+b+ c c² - ab

1 -a ab

_[Applying R₂ → R₂ -R_{3 ]}

Now expanding along C₁ , we get

(a-b) (b-c) (c-a) [- (c² – ab – ac – bc – c²)]

= (a-b) (b-c) (c-a) (ab + bc + ac)

EXAMPLE - 4

Without expanding the determinant,

prove that

3x = x³

3x+y 2x x 4x+3y 3x 5x+6y 4x 6x

3x+y

2x x 3x 3x 3x = 4x 4x 6x 5x

2x x y 2x x 3x 3x + 3y 3x 3x 4x 6x 6y 4x 6x

L.H.S= 4x+3y

5x+6y

3 2 1 1 2 1

= x³ 4 3 3 + x²y 3 3 3

5 4 6 6 4 6

Solution :

[

C₁ and C₂

are identical in II determinant_]

3 2 1

= x³ 4 3 3 + x²y×0

5 4 6

SOLUTION CONT.

1

1

2

= x³ 0

1 2 1

1 2 _[Applying R₂ →R₂ -R₁ and R₃ →R₃ -R_{2 ]}

0 1 3

1

_[Expanding along C _]

= x³ ×(3-2)

=x³ = R.H.S.

3 2 1

= x³ 4 3 3

5 4 6

EXAMPLE - 5

= 0 , where ω is cube root of unity.

_ω3 _ω5

1 ω⁴

ω⁵ 1

1

Prove that : _ω³

_ω5

L.H.S= ω³

_ω5

1 ω³ ω⁵ 1 ω³ ω³.ω²

1 ω⁴ = ω³ 1 ω³.ω

ω⁵ 1 ω³.ω² ω³.ω² 1

[

_ω2

1 1 ω²

= 1 1 ω

ω² 1

= 0 = R.H.S.

C₁ and C₂ are identical_]

⎡_⎣

ω³ =1^⎤_⎦

Solution :

EXAMPLE - 6

x+a b c

a x+b c = x+a+b+c a b x+C

L.H.S=

x+a+b+c

x+a+b+c b c

x+b c b x+c

_[Applying C₁ → C₁ +C₂ +C_{3 ]}

​

1 b c

= ₍x+a+b+c₎ 1 x+b c

1 b x+c

⎡_⎣Taking ₍x+a+b+c₎ commonfrom C₁ ⎤_⎦

SOLUTION CONT.

1 b c

=(x+a+b+c) 0 x 0

0 0 x

_[Applying R₂ → R₂ -R₁ and R₃ →R₃ -R_{1 ]}

Expanding along C₁ , we get

(x + a + b + c) [1(x²)] = x² (x + a + b + c)

= R.H.S

EXAMPLE - 7

2(a+b+c)

= c+a a+b

2(a+b+c) a+b b+c

2(a+b+c) b+c c+a

_[Applying R₁ →R₁ +R₂ +R_{3 ]}

1

=2(a+b+c) c+a

1 1

a+b b+c a+b b+c c+a

Solution :

Using properties of determinants, prove that

b+ c =2(a+b+ c)(ab +bc + ca- a² - b² - c² ).

b+ c c+ a a+b c+ a a+b

a+b b+ c c+ a

L.H.S= c+a

b+c c+a a+b

a+b b+c a+b b+c c+a

SOLUTION CONT.

_[Applying C₁ → C₁ - C₂ and C₂ → C₂ - C_{3 ]}

0

0 1

(a- c) b+ c

(b- a) c+ a

= 2(a+b+ c) (c-b)

(a- c)

Now expanding along R₁ , we get

2(a+b+c)^⎡_⎣(c -b)(b- a)-(a- c)²

⎤_⎦

=2(a+b+c)^⎡_⎣bc -b² - ac+ab-(a² +c² - 2ac)^⎤_⎦

​

=2(a+b+c)^⎡_⎣ab+bc+ac- a² -b² - c^{2 ⎤}_⎦

=R.H.S

EXAMPLE - 8

Using properties of determinants prove that

1 2x 2x

=(5x+ 4) 1 x+ 4 2x

1 2x x+ 4

Solution :

_[Applying C₁ → C₁ +C₂ +C_{3 ]}

L.H.S=

= 5x+ 4

5x+ 4

x+ 4 2x 2x 5x+ 4 2x 2x 2x x+ 4 2x x+ 4 2x

2x 2x x+ 4 2x x+ 4

SOLUTION CONT.

_[Applying R₂ → R₂ -R₁

and R₃ → R₃ -R_{2 ]}

=(5x+ 4) 0

1 2x 2x

-(x - 4) 0

0 x - 4

-(x - 4)

Now expanding along C₁ , we get

​

(5x+4)^⎡_⎣1(x - 4)² - 0^⎤_⎦

​

=(5x+4)(4- x)²

​

=R.H.S

EXAMPLE - 9

Using properties of determinants, prove that

x+9 x x x x+9 x x x x+9

L.H.S=

= 3x+9

3x+9

3x+9 x x x+9 x

x x+9

_[Applying C₁ → C₁ +C₂ +C_{3 ]}

Solution :

SOLUTION CONT.

_[Expanding along C_{1 ]}

= 3(x+3)×81

=243(x+3)

= R.H.S.

1 x x

=(3x+9) 1 x+9 x

1 x x+9

1 x x

= 3₍x+3₎ 0 9 0

0 -9 9

^⎡_⎣Applying R₂ →R₂ -R₁

and R₃ → R₃ -R₂

⎤_⎦

SOLUTION CONT.

_[Applying R₂ →R₂ -R₁ and R₃ →R₃ -R_{2 ]}

1

=(a² +b² +c²) 0

_a2

(b- a)(b+a)

0 (c-b)(c+b)

bc c(a-b)

a(b- c)

1

=(a² +b² +c² )(a-b)(b- c)(-ab- a² +bc+c² ) _[Expanding along C _]

=(a² +b² +c² )(a-b)(b- c)⎡_⎣b₍c - a₎+₍c- a₎₍c+a₎⎤_⎦

​

=(a² +b² +c² )(a-b)(b- c)(c- a)(a+b+c)=R.H.S.

EXAMPLE - 10

Solution :

(b+ c)²

_a2 _b2 _c2

bc b² + c² ca = c² + a² ab a² +b²

_a2 _b2 _c2

L.H.S.= (c+ a)²

bc

ca ^⎡_⎣Applying C₁ → C₁ - 2C₃ ^⎤_⎦ ab

(a+b)²

a² +b² + c²

a² +b² + c²

_a2 _b2 _c2

= a² +b² + c²

bc

ca _[Applying C₁ →C₁ +C_{2 ]} ab

1

=(a² +b² +c² ) 1

1

_a2 _b2 _c2

bc ca ab

(b + c)²

_a2 _b2 _c2

Show that _{(c + a)}²

ca = (a² +b² + c² )(a- b)(b - c)(c - a)(a+b + c)

(a+b)²

bc

ab

Applications of Determinants (Area of a Triangle)

The area of a triangle whose vertices are

​

^(x₁^{, y}₁^{), (x}₂^{, y}₂^{) and (x}₃^{, y}₃⁾ is given by the expression

​

x₁ y₁ 1

Δ = ¹ x y 1

₂ 2 2

x₃ y₃ 1

1

= ₂[x₁ (y₂ - y₃ ) + x₂ (y₃ - y₁ ) + x₃ (y₁ - y₂ )]

Example

Find the area of a triangle whose vertices are (-1, 8), (-2, -3)

and (3, 2).

Solution :

Area of triangle = ₂ x₂

1 = ₂ -2

x₁ y₁ 1 -1 8 1

¹ y ¹ -3 1

2

x₃ y₃ 1 3 2 1

= ¹ _[-1(-3- 2)- 8(-2 - 3)+1(-4+9)_]

2

= ¹ _[5+ 40+5_]= 25 sq.units

2

Condition of Collinearity of Three Points

If ^{A (x}₁^{, y}₁ ^{), B (x}₂ ^{, y}₂ ^{) and C (x}₃^{, y}₃ ⁾ are three points,

​

then A, B, C are collinear

​

⇔ Area of triangle ABC = 0

⇔ ₂ x₂

x₁ y₁ 1 x₁ y₁ 1

¹ y₂ 1 = 0 ⇔ x₂ y₂ 1 = 0

x₃ y₃ 1 x₃ y₃ 1

Example

If the points (x, -2) , (5, 2), (8, 8) are collinear, find x , using determinants.

Solution :

Since the given points are collinear.

​

x -2 1

∴ 5 2 1 = 0

8 8 1

​

⇒ x₍2- 8₎- ₍-2₎₍5- 8₎+1₍40-16₎= 0

​

⇒ -6x- 6+24=0

​

⇒ 6x=18 ⇒ x=3

Solution of System of 2 Linear

Equations (Cramer’s Rule)

Let the system of linear equations be

a₂x +b₂y = c₂

...₍ii₎

a₁x +b₁y = c₁

...₍i₎

Then x = ^D1 , y = ^D2

D D

provided D ≠ 0,

1

1

2

where D = ^a1

, D =

and D

b₁

a₂ b₂

c b₁

c₂ b₂

₌ a₁ c₁

a₂ c₂

Cramer’s Rule

then the system is inconsistent and has no solution.

Note :

₍1₎ If D ≠ 0,

then the system is consistent and has unique solution.

​

₍2₎ If D = 0 and D₁ = D₂ = 0,

​

then the system is consistent and has infinitely many solutions.

(³)

If D = 0 and one of D₁, D₂ ≠ 0,

Example

1

D = ^{7 -3} =7+15=22

5 1

2

D = ^{2 7} =10-21=-11

3 5

D ≠ 0

∴By Cramer's Rule x = ^D1 = ²² = 2 and y = ^D2 = ^-11 =-1

D 11 D 11

Using Cramer's rule , solve the following system of equations 2x-3y=7, 3x+y=5

Solution :

​

D= ^{2 -3} = 2+9 =11 ≠ 0

3 1

Solution of System of 3 Linear

Equations (Cramer’s Rule)

Let the system of linear equations be

a₁x +b₁y + c₁z = d₁

a₂x +b₂y + c₂z = d₂

a₃x +b₃y + c₃z = d₃

...₍i₎

...₍ii₎

...₍iii₎

z = ^D3

D

Then x = ^D1 , y = ^D2 ,

D D

provided D ≠ 0,

where D = a₂

d₁ b₁ c₁

b₂ c₂ , D₂ = a₂

d₃ b₃ c₃

a₁ d₁ c₁

d₂ c₂

a₃ d₃ c₃

and D₃ = a₂

a₁ b₁ c₁

b₂ c₂ , D₁ = d₂

a₃ b₃ c₃

​

a₁ b₁ d₁

b₂ d₂

a₃ b₃ d₃

Cramer’s Rule

Note:

​

1. If D ≠ 0, then the system is consistent and has a unique

solution.

​

2. If D=0 and D₁ = D₂ = D₃ = 0, then the system has infinite solutions or no solution.

​

• If D = 0 and one of D₁, D₂, D₃ ≠ 0, then the system is inconsistent and has no solution.

​

• If d₁ = d₂ = d₃ = 0, then the system is called the system of homogeneous linear equations.

​

1. If D ≠ 0, then the system has only trivial solution x = y = z = 0.

​

• If D = 0, then the system has infinite solutions.

Example

Using Cramer's rule , solve the following system of equations

5x - y+ 4z = 5 2x + 3y+ 5z = 2 5x - 2y + 6z = -1

Solution :

5 -1 4

D= 2 3 5

5 -2 6

5 -1 4

D₁ = 2 3 5

-1 -2 6

= 5(18+10)+1(12+5)+4(-4 +3)

= 140 +17 –4

= 153

= 5(18+10) + 1(12-25)+4(-4 -15)

= 140 –13 –76 =140 - 89

= 51 ^{≠ 0}

Solution

5 -1 5

D₃ = 2 3 2

5 -2 -1

= 5(-3 +4)+1(-2 - 10)+5(-4-15)

= 5 – 12 – 95 = 5 - 107

= - 102

D ≠ 0

∴By Cramer's Rule x = ^D1 = ¹⁵³ = 3, y = ^D2 = ¹⁰² = 2

D 51 D 51

and z = ^D3 = ^-102 =-2

D 51

5 5 4

D₂ = 2 2 5

5 -1 6

= 5(12 +5)+5(12 - 25)+ 4(-2 - 10)

= 85 + 65 – 48 = 150 - 48

= 102

Example

Solve the following system of homogeneous linear equations: x + y – z = 0, x – 2y + z = 0, 3x + 6y + -5z = 0

Solution:

⎡¹

^{- 1}⎤

We have D = ^⎢1

⎢

⎢_⎣3

1

- 2 1 ^⎥ = 1 ₍10 - 6₎ - 1 ₍-5 - 3₎ - 1 ₍6 + 6₎

⎥

6 - 5 ^⎥_⎦

= 4 + 8 - 12 = 0

​

∴ The system has infinitely many solutions.

​

Putting z = k, in first two equations, we get

​

x + y = k, x – 2y = -k

k

1

_{x =} D_{1 =} -k

^{- 2} ₌ -2k + k ₌ k ^D 1 1 ^{-2 - 1 3}

1 - 2

∴ By Cramer's rule

1

k

y = ^D2

₌ 1

D

^{- k} ₌ -k - k ₌ 2k

1 1 ^{-2 - 1 3}

1 - 2

∴ x = ^k , y = ^2k , z = k , 3 3

where k ∈ R

These values of x, y and z = k satisfy (iii) equation.

Seatwork

Find the determinant of each matrix.

−1_⎤

₄ ⎥

⎥

₆ ⎥_⎦

⎢

⎢₃

11^⎥

_⎡8 2 −1 −4 _⎤

5 −3

0 4

2 7

⎢

⎢₀

⎥

₀⎥

⎢₂

−1^⎥

⎣

⎦

Thank you