unit#1 matrice and derterminants (exercise 1.6)

Mudassar Nazar Notes

(Exercise 1.6)

Question # 2

The length of a rectangle is 4 times its width. The perimeter of the rectangle is 150

cm. Find the dimensions of the rectangle.

Solution

Let the width of the rectangle be x m and length be y m

According to first condition

Y = 4x

0 = 4x – y

4x – y = 0 (i)

According to second condition

Perimeter = 150 cm

2(L +W) = 150

2 ( y + x) = 150

x + y =

x + y = 75

Matrix Inversion Method

4x – y = 0

x + y = 75

=


Let

M =

=

= 4 – (-1 )

= 4 + 1

= 5 0

Adj M =

M-1 = Adj M

M-1 =

= M-1

=

=

=

Width = 15 cm and Length = 60 cm

Cramer’s Rule

4x – y = 0

x + y = 75


Let A = , Ax = , Ay =

=

= 4 – ( -1 )

= 4 + 1

= 5

=

= 0 – ( -75)

= 0 + 75

= 75

= 300

x = and y =

x = and y =

x = 15 and y = 60

Width = 15 cm and Length= 60 cm

Question # 3

Two sides of a rectangle differ by 3.5 cm. Find the dimensions of the rectangle if its perimeter

is 67 cm.

Solution Let the length of the rectangle be x m and width be y m

According to first condition

x – y = 3.5 cm


According to second condition

Perimeter = 67 cm

2(L +W)= 67

2 (y + x) = 67

x + y =

x + y = 33.5 cm


x – y = 3.5

x + y = 33.5

=

Let M =

=

= 1 – ( -1)

= 1 + 1

= 2

Adj M =

M-1 = Adj M

=

= M-1


=

=

Width = 18.5 cm and Length = 15 cm

Cramer’s Rule

x – y = 3.5

x + y = 33.5

Let A = , Ax = , Ay =

=

= 1 – ( -1 )

= 1 + 1

= 2

=

= 3.5 – ( -33.5)

= 3.5 + 33.5

= 37

=

= 33.5 – 3.5

= 30

x = and y =


x = and y =

x = 18.5 and y = 15

Length = 18.5 cm and Width= 15 cm

Question # 4

The third angle of an Isosceles is 16o less than the sum of two equal angles. Find three angles

of the triangle.

Solution

Let each equal angle be xo and third angle be yo

According to given condition

y = 2x – 16

16 = 2x – y

2x – y = 16

and

x + x + y = 1800

2x + y = 180


2x – y = 16

2x + y = 180

=


Let M =

=

= 2 – ( -2)

= 2 + 2

= 4

Adj M =

M-1 = Adj M

=

= M-1

=

=

So the angles are 49o, 49o and 82o

Cramer’s Rule

2x – y = 16

2x + y = 180

Let A = , Ax = , Ay =

=


= 2 – ( -2 )

= 2 + 2

= 4

=

= 16 – ( -180)

= 16 + 180

= 196

=

= 360 – 32

= 328

x = and y =

x = and y =

x = 49 and y = 82

So the angles are 49o, 49o and 82o

Question # 5 One acute angle of a right triangle is 12o more than twice the other acute angle. Find the

acute angles of the right triangle.

Solution

Let the two acute angles be xo and yo

According to the given condition

x = 2y + 12


x – 2y = 12

and

x + y = 90


x – 2y = 12

x + 2y = 90

=

Let M =

=

= 1 – ( -2)

= 1 + 2

= 3

Adj M =

M-1 = Adj M

=

= M-1

=

=


So two acute angles are 64o and 26o

Cramer’s Rule

x – 2y = 12

x + 2y = 90

Let A = , Ax = , Ay =

=

= 1 – ( -2 )

= 1 + 2

= 3

=

= 12 – ( -180)

= 12 + 180

= 192

=

= 90 – 12

= 78

x = and y =

x = and y =

x = 64 and y = 26

So two acute angles are 64o and 26o


Question # 6 Two cars that are 600km apart are moving towards each other. Their speeds differ by 6 km

per hour and the cars are 123 km apart after4 hours. Find the speed of each car.

Solution Let the speed of two cars be x kmh-1 and y kmh-1

According to the first condition

x – y = 6

According to the first condition

x + y = 477

(x + y) = 477

x + y = 477

x + y = 477

x + y = 106


x – y = 6

x + y = 106

=

Let M =

=

= 1 – ( -1)


= 1 + 1

= 2

Adj M =

M-1 = Adj M

=

= M-1

=

=

The speed of two cars are 56 kmh-1 and 50 kmh-1

Cramer’s Rule

x – y = 6

x + y = 106

Let A = , Ax = , Ay =

=

= 1 – ( -1)

= 1 + 1

= 2


=

= 6 – ( -106)

= 6 + 106

= 112

=

= 106 – 6

= 100

x = and y =

x = and y =

x = 56 and y = 50

The speed of two cars are 56 kmh-1 and 50 kmh-1

unit#1 matrice and derterminants (exercise 1.6)

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