# EQUALITY OF MATRICES

Two matrices are said to be equal if :

Both the matrices are of the same order i.e., they have the same number of rows and columns Am × n = Bm × n .

Problem 1 :

Solution :

Equation (2), multiplying (2) on each sides.

2r + 8s = -4 --- (3)

Subtracting the equation (1) and (3).

(2r - 3s - 2r - 8s) = 4 + 4

-11s = 8

s = -8/11

s = -8/11 substitute the equation (1).

2r - 3(-8/11) = 4

2r + 24/11 = 4

2r = 4 - (24/11)

2r = (44 - 24)/11

2r = 20/11

Dividing 2 on each sides.

r = 1/2 × 20/11

r = 10/11

So, the values of r and s is -8/11 and 10/11.

Problem 2 :

Solve each matrix equation or system of equations by using inverse matrices.

Solution :

So, the values of a and b is 1.5 and -4.

Problem 3 :

Solve each equation.

Solution :

Equating the equation :

2y - x = 3 ---- (1)

x = 4y - 1

x  - 4y = -1--- (2)

Solving equation (1) and (2)

(2y - x) + (x - 4y) = 3 - 1

2y - x + x - 4y = 2

-2y = 2

y = -2/2

y = -1

y = -1 substitute the equation (2)

x - 4(-1) = -1

x + 4 = -1

x = -1 - 4

x = -5

So, the values of x and y is -1 and -5.

Problem 4 :

Solution :

7x = 5 + 2y

7x - 2y = 5 --- (1)

x + y = 11 --- (2)

Solving equation (1) and (2)

(7x - 2y) + (x + y) = 5 + 11

8x - y = 16

8x = 16 + y

x = (16 + y)/8 --- (3)

x = (16 + y)/8 substitute the equation (2).

(16 + y)/8 + y = 11

Multiplying 8 on each sides.

16 + y + 8y = 88

16 + 9y = 88

9y = 88 - 16

9y = 72

y = 72/9

y = 8

x = 8 substitute the equation (2).

x + 8 = 11

x = 11 - 8

x = 3

So, the values of x and y is 3 and 8.

Problem 5 :

Solution :

Equating the corresponding elements, we get

2x = -16

x = -16/2

x = -8

y + 1 = -7

y = -7 - 1

y = -8

z - 8  = -2

z = -2 + 8

z = 6

So, the values of x, y and z is -8, -8 and 6.

Problem 6 :

Solution :

Problem 7 :

Solution :

4|(-3 × 3) - (2 × 1)| - x|(-x × 3) - (-6 × 1)| - 2|(-x × 2) - (-6 × (-3))| = -3

4(-9 - 2) - x(-3x + 6) - 2(-2x - 18) = -3

4(-11) + 3x2 - 6x + 4x + 36 = -3

-44 + 3x2 - 2x + 36  + 3 = 0

3x2 - 2x - 5 = 0

3x2 - 5x + 3x - 5 = 0

x(3x - 5) + 1(3x - 5) = 0

(x + 1) (3x - 5) = 0

x + 1 = 0 and 3x - 5 = 0

x = -1 and 3x = 5

x = 5/3

So, the value of x is -1 and 5/3.

Problem 8 :

Solution :

Equating the corresponding elements, we get

x2 + 1 = 5

x2 = 5 - 1

x2 = 4

x = ±2

 5 - y = xPut x = 25 - y = 2-y = 2 - 5-y = -3y = 3 Put x = -25 - y = -2-y = -2 - 5-y  = -7y = 7

So, the values of x and y is ±2 and 3 and 7.

Problem 9 :

Solution :

Equating the corresponding elements, we get

3x - 5 = 10

3x = 10 + 5

3x = 15

x = 15/3

x = 5

x + y = 8

Put x = 5

5 + y = 8

y = 8 - 5

y = 3

9z = 3x + y

Put x = 5 and y = 3

9z = 3(5) + 3

9z = 15 + 3

9z = 18

z = 18/9

z = 2

So, the values of x, y and z is 5, 3 and 2.

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