Matrices and Determinants (2×2, 3×3)

 

FIRST TERM

SUBJECT: FURTHER MATHEMATICS

CLASS: SSS 3

WEEK 1

SUBJECT: Further Mathematics[mediator_tech]

CLASS: SS 3

TOPIC: Matrices and Determinants (2×2, 3×3)

CONTENT:

1. Matrices as linear transformations
2. Determinants
3. Solutions of 2 and 3 simultaneous equations

Subtopic 1: MATRICES

• Basic definitions

1. A matrix is a rectangular away of numbers arranged in rows and column. A matrix with m rows and n columns is called an ‘m by n matrix’ or a matrix of order m x n.

Examples:

1. Is a matrix of order 2 x 2
2. Is a matrix of order 2 x 3
3. Is a matrix of order 3 x 2
4. A matrix with only one row is called a row matrix e.g. (2 6 5) is a 1 x 3 matrix

Solution:

• MULTIPLICATION OF MATRICES

1. Scalar Multiplication

Let

Matrix kP is the calar multiplication of the matrix P by the scalar k and is denoted by kP, thus

For example, Let

Solution:

1. Matrix Matriculation

Two matrices X and Y are said to be conformable for multiplication if the numbers of columns in one equal the number of rows in the other

Then the product of x and y written as XY or X.Y. is

Z₁₁ = x₁₁y₁₁ + x₁₂y₂₁

Z₁₂ = x₁₁y₁₂ + x₁₂y₂₂

Z₂₁ = x₂₁y₁₁ + x₂₂y₂₁

Z₂₂ = x₂₁y₁₂ + x₂₁y₂₂

Example 2:

Given that and B =

Find: –

1. 3A
2. AB
3. BA

Solution

• PROPERTIES OF MATRICES

1. Community

If A and B are matrices, then A + B = B + A

but AB ≠ BA

thus addition of matrices is commutative but multiplication of matrices is not generally commutative.

1. Associativity

If A, B and C are matrices then (A + B) + C = A + (B + C) and

(AB) C = A (BC)

Thus the operation of addition and multiplication in matrices are Associative.

1. Distributivity

If A, B, and C are matrices then

A (B + C) = AB + BC also

(B + C) A = BA + CA

Thus in matrices, multiplication distributes over addition.

• SPECIAL MATRICES

1. Transpose of a matrix

Let A = A^T is called the transpose of the matrix A and is obtained by interchanging the columns and rows of A.

1. Symmetric Matrix

If A = A^T that is A is equal to its transpose then the matrix A is said to be symmetric.

e.g.

(3) Skew symmetric matrix

For a matrix A, if A^T = -A then A is said to be skew symmetric

e.g.[mediator_tech]

(4) Scalar matrix: –

A scalar matrix is a square matrix in which the elements in the principal diagonal are equal and non-zero.

e.g.

(5) triangular matrix

This is a scalar matrix in which the entries above or below the principal diagonal are zero.

Examples

(6) Identity Matrix

This is a scalar matrix in which all the elements in the principal diagonal are unity, it is also called a unity, it is also called a unit matrix.

Examples

(7) Null Matrix

This is a matrix with all its elements being zero it is also called zero matrix

EVALUATION

1. Given that

find

1. A + B
2. B – A
3. 2A – B
4. Given that ,

Y=

Find

1. 4B
2. AB
3. Let a X = ,

B =

Show that (A + b)^T

SUBTOPIC 2: DETERMINANTS

• Second order determinants of matrix A is denoted det A.

If A = then, det a is denoted by det A

Then, det A is defined as det A =

The determinant above is said to be of second order since it is obtained from a matrix of order 2 x 2.

Example 1: – Evaluate each of the following determinants

Solution

(a)

= (4 x 2) – (-6 x -3)

= 8 – 18

= -10

= (x-3) (x +2) – [x (x+1)] = 0

= -2x – 6

= 0

-2x = 6

−3

Let A be a third order determinant. That is,

The minor of a particular element of the determinant A is obtained by deleting the row and column of the particular element. For example

Let A_ij be the minor of aij

Thus,

=

And so on.

Example 1: – given the matrix

Determine the following minors

1. A₂₃
2. A₁₂

Solution

1. A₂₃ =

=

= (4 x -2) – (1 x 0)

= -8 – 1

= = -9

Similarly,

A₁₂ =

= (2 x 1) – (-5 x 1)

= 2 + 5

= 7

Cofactors

For each minor of an element of a determinant, a sign (positive or negative) is attached, this gives the cofactor. Usually if C_ij denotes the cofactor of A_ij then,

C_ij =A_ij if (i + j) is even

and C_ij = (i + j) is odd

thus the signs of the cofactor are as follows

Consider

Then,

C₁₁ = +A₁₁ = 7-25 = -18

C₁₂ = -A₁₂

=

= 1-(-35)

= 36

These leads us to calculating third order determinants if ∆ denotes the determinant of the matrix

then

= a₁₁c₁₁ + a₁₂c₁₂ + a₁₃c₁₃

Example 3: – Evaluate each of the following determinants

1. ∆ = 5

=

= –

= 5 (0-6) -2 (-7+8) – 4 (-3-0)

= -20

1. ∆= -1

= -8

= +1

= -1(5-0) -8(15-0) +1(-9+2)

= -5-120-7

= -132

• Some properties of determinants

1. If the row and column of a determinant are interchanged, the value of the determinant is unchanged.

That is,

If ∆ =

and

∆* =

Then ∆ = ∆*

1. If two adjacent columns or rows of a determinant are interchanged, the sign of the determinant changes but its numerical value is unchanged.

If ∆ =

and

∆* =

1. If two rows of columns of a determinant are identical then the value of the determinant is zero.

For example if

∆ =

That is, row 1 = row 3

Then ∆ = 0

Check

∆ = 2[mediator_tech]

= 2 x 18 – (12 – 12) + 2 x -18

= 36 – 0 – 36 = 0

1. If every element in a rwo or column of a determinant is multiplied by the same constant then the value of the determinant is multiplied by that constant for example, if ∆ =

Then = 2D

1. If by putting x =a , the value of a determinant becomes zero, then x – a is a factor of the determinant. For example by putting x = 1 in the determinant below

f(x) = = 0

therefore, x-1 is a factor of f(x).

EVALUATION

1. Evaluate the determinants below
2. Show that = 1
3. Evaluate the determinant of the matrix
4. Find the values of x for which

= 0

1. Find the value of x if
2. = 17

Sub-Topic 3:

Solutions of Simultaneous Equations

• Two equations in two unknowns

If we consider the system of two equations with two unknowns

a₁x + b_1y = C_{1 …} (i)

a₂x + b₂y = C_{2 …} (ii)

to eliminate the y variables, we

a₁b₂x + b₁b₂y = c₁b₂

• a₂b₁x + b₁b₂y = c₂b₁

(a₁b₂ – a₂b₁)x = c₁b₂ – c₂b₁

x =

Similarly,

that the denominators are the same and is the value of the determinant

which we denote as

Therefore for a system of two linear equations in two unknowns

a₁x + b₁y = c₁

a₂x + b₂y = c₂

and ∆=

Example 1: – use Cramer’s rule to solve the follwoing simultaneous equations.

1. 4x + 3y = 13

x – 5y = -14

1. 2x – 7y = 12

3x – y = -1

Solution

1. ∆ = = -20-3 = -23

∆₁ = = -65+42 = -23

∆₂ = = -56-13 = -69

= = 1

= = 3

1. 2x – 7y = 12

3x – y = -1

Solution

= -2 + 21 = 19

∆₁ = = -2 – 36 = -38

x = = = -1

and y = = -2

• Three Linear equations in three unknown

Applying the same rule to the following system of three equations in three unknowns.

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z=d₂

a₃x + b₃y + c₃z = d₃

Let ∆ =

∆₂ =

∆₁ =

∆₃ =

It can also be shown that

x = , y = , z =

This is the Cramer’s rule for a system of three equations in three unknowns.

Example 2: – use determinants to solve each of the following systems of equations

1. 2x – y – z = -10

x – 3y + z = 13

4x – y + 2z = 3

1. x – y + z = 12

2x – 3y – 2z = 7

X + y + z = 6

Solution

1. Let ∆ =

= 2 + – 1

= 2 (-6+1) )2-4) – (-1+12)

= -10 – 2 – 11

= -23

Let ∆₁ =

= – 2 (-6+1) (2-4) – (-1+12)

= 50 + 23 + 4

= 77

∆₂ =

= 2(26 – 3) + 10 (2-4) – 1(3-52)

= 46 – 20 + 49

= 75

∆₃ =

= 2(-9 + 13) + (3 – 52) – 10 (-1 + 12)

= 8 – 49 – 110

= – 151

x = = = –

y = = =

z = =

1. Let ∆ =

= (-3 + 2) + (2+2) + (2+3)

= – 1 + 4 + 5

= 8

Let ∆₁ =

= 12 (-3 + 2) + (7+12) + (7+18)

= -12 + 19 + 25

= 32

∆₂ =

= (7+12) – 12 (2+2) + (12-7)

= 19 – 48 + 5

= -24

∆₃ =

= (-18-7) + (12-7) + 12(2+3)

= -25 + 5+ 60

= 40

x = = = 4

= = -3

= = 5

The adjoint of a given square matrix is the transpose of th matrix formedc by taking the co-factors of the matrix. It is someties called the adjugate

Examples 3: – determine the adjoint of the matrix

A =

Solution

Let the matrix of the co-factors be

C =

Where C₁₁ = – = 54

C₁₂= – = +15

C₁₃= – = 33

C₂₁= – = -3

C₂₂ = – = 5

C₂₃ = – = 1

C₃₁ = – = -30

C₃₂ = – = -10

C₃₃ = – = 25

Hence,

C = , let the transpose of C

be C^T =

then the adjoint of A is given by

Adj A = C^T =

Singular matrix

A square matrix whose determinant is equal to zero is called a singular matrix.

• The inverse of a square maatrices of the same order. If AB = BA = I (i.e. the identity matrix)
• the inverse of a square matrix

let A and B be non singular square matrices of the same order. If AB = BA = I (i.e. the identity matrix) then B is called the inverse of A denoted as A^-1

that is A. A^-1 = I

the inverse of a matrix A is defined as

A^-1 =

for a 2 x 2 matrix A = the inverse

A^-1 = where =

• matrix method of solving simultaneous equations

example 4: use matrix method to solve each of the following systems of equations

1. 4x + y = 1

5x – 2y = 11

1. 2x + y + 3z = 16

X + 2y – z = -2

3x + y + 27 = 14[mediator_tech][mediator_tech]

Solution

The equations

4x + y = 1

5x – 2y = 11

Can be written in the matrix form as

=

Let A =

B =

C =

We can rewrite the matrix equations as

AB = C … … … *

If we pre-multiply both sides of * by A^-1, then we have

A^-1 A.B = A^-1 C

IB = A^-1C

B = A^-1C

Recall A^-1 =

A* =

= -8-5 = -13

A^-1 = –

= –

=

Thus, =

Hence x = 1, and y = -2

(b) The equations

2x + y – z = 16

X + 2y – z = -2

3x + y + 2z = 14

In matrix form is =

Let P =

Q =

R =

We write the matrix as

PQ = R … … … **

Pre-multiplying both sides of ** by P^-1, gives

P^-1.PQ = P^-1R

IQ = P^-1R

Recall:

P^-1 =

Let x be the matrix of cofactors of P

x =

C₁₁ = + = 5

C₁₂ = – = -5

C₁₃ = + = -5

C₂₁ = – = 1

C₂₂ = + = -5

C₂₃ = + = 3

C₃₁ = + = -7

C₃₂ = – = 5

C₃₃= + = 3

Therefore,

X = ,

X^T =

= Adj P

= 2 – 1 + 3

= (2 x 5) – 5+(3×5)

= 10 – 5-15[mediator_tech]

= -10

Thus,

P^-1 = Adj P = –

and

Q = –

= –

Hence x = z, y = 0, z = 4.

EVALUATION

1. Use determinants to solve each of the following systms of equations.
2. x + 3y = 5

x – y = -11

1. x + 2y – z = -10

3x – y + z = 13

2x + y + 2z = 3

1. find the matrix of the cofactors of the elements of the determinant
2. use matrix method to solve

2x + 3y = 12

X – y = 1

GENERAL EVALUATION

1. If A = , B = , C =

Find: –

1. A + b
2. 2A – C
3. Given A =
4. Evaluate
5. Find the values of x for which
6. Evaluate

Hence solve the following equations

3x + 4y + 2z = 4

X – 5y + 3z = -1

2x + 3y + z = 3