Question

Show how to solve each of the following systems of linear equations by elimination or substitution, and using matrix inverses

a) 3x-y=1

-2+y=1

b) 3x+2y=5

-2x+3y+14

Answer 1

• a).

Substitution method :

The system of equations are

3x - y = 1 → ( 1 )

- 2 + y= 1 → ( 2 )

Solve eq (2), since the y has a coefficient of 1.

- 2 + y = 1

Add 2 to each side.

y = 1 + 2

⇒ y = 3.

Substitute the value of y = 3 in eq (1) to find the valueof x.

3x - 3 = 1

3x = 1 + 3

3x = 4

⇒ x = 4/3.

The solution is x = 4/3 and y= 3.

Inverse matrix method :

The systemof equations are

3x - y = 1 → ( 1 )

- 2 + y= 1 → ( 2 )

Rewrite the equations as 3x - y = 1 and (0)x + y = 3.

Write the equations in matrices form AX  = B .

A  is coefficeint mattrix , X  is variable matrix and B  is constant matrix.

and .

The solution is x = 4/3 and y = 3.

Answer 2

• b).

Substitution method :

Consider the system as

3x + 2y = 5     → ( 1 )

- 2x + 3y = 14 → ( 2 )

Solve eq ( 2 ) for x.

2x = 3y - 14

⇒ x = (3y - 14)/2.

Substitute the value of x = (3y - 14)/2 in eq ( 1 ), and solve for y.

3[ (3y - 14)/2] + 2y = 5

9y - 42 + 4y = 10

13y = 10 + 42 = 52

⇒ y = 52/13 = 4.

Substitute the value of y = 4 in x = (3y - 14)/2, tofind the value of x.

x = (3 * 4 - 14)/2

= (12 - 14)/2

= - 2/2

⇒ x = - 1.

The solution is x = - 1 and y = 4.

Inverse matrix method :

The systemof equations are

3x + 2y = 5    → ( 1 )

- 2x + 3y= 14 → ( 2 )

Write the equations in matrices form AX  = B .

A  is coefficeint mattrix , X  is variable matrix and B  is constant matrix.

and .

.

The solution is x = - 1 and y = 4.