Find the product of the matrices 6 - 3 2 - 7 .

Given to find the product of the matrices 6 - 3 2 - 7 . Here, the column size of first matrix and row size of the second matrix is same. So, multiplication is possible. The multiplied matrix has the dimension 2 × 2 . Therefore, 6 - 3 2 × 1 2 - 7 1 × 2 = 6 × 2 - 7 × 6 - 3 × 2 - 3 × - 7 2 × 2 = 12 - 42 - 6 21 Therefore, the required product is 12 - 42 - 6 21 .

Solve x 1 1 0 - 2 - 3 x 5 = 0 .

Given, x 1 1 0 - 2 - 3 x 5 = 0 ⇒ x × 1 + 1 × - 2 x × 0 + 1 × - 3 x 5 = 0 ⇒ x - 2 - 3 x 5 = 0 ⇒ x x - 2 - 3 × 5 = 0 ⇒ x 2 - 2 x - 15 = 0 ⇒ x 2 - 5 x + 3 x - 15 = 0 ⇒ x x - 5 + 3 x - 5 = 0 ⇒ x + 3 x - 5 = 0 ⇒ x + 3 = 0 or x - 5 = 0 ⇒ x = - 3 or x = 5 Therefore, x = - 3 or 5

If A = 3 2 4 0 and B = 3 0 3 2 , then find A B and B A .

Given, A = 3 2 4 0 and B = 3 0 3 2 Now, A B is, 3 × 3 + 2 × 3 3 × 0 + 2 × 2 4 × 3 + 0 × 3 4 × 0 + 0 × 2 = 9 + 6 0 + 4 12 + 0 0 + 0 A B = 15 4 12 0 Now, B A is, 3 × 3 + 0 × 4 3 × 2 + 0 × 0 3 × 3 + 2 × 4 3 × 2 + 2 × 0 = 9 + 0 6 + 0 9 + 8 6 + 0 B A = 9 6 17 6

Solve the equation x 2 - 3 + 1 x + 3 = 0 by completing the square method

Given, x 2 - 3 + 1 x + 3 = 0 As we know that a - b 2 = a 2 - 2 a b + b 2 Now, x 2 - 3 + 1 x + 3 = 0 ⇒ x 2 - 2 3 + 1 x 2 + 3 = 0 Add and subtract 3 + 1 2 2 ⇒ x 2 - 2 3 + 1 x 2 + 3 + 1 2 2 - 3 + 1 2 2 + 3 = 0 ⇒ x - 3 + 1 2 2 = x 2 - 2 × x × 3 + 1 2 + 3 + 1 2 2 Now, ⇒ x - 3 + 1 2 2 - 3 + 1 2 2 + 3 = 0 ⇒ x - 3 + 1 2 2 = 3 + 1 2 2 - 3 ⇒ x - 3 + 1 2 2 = 3 + 1 + 2 3 4 - 3 ⇒ x - 3 + 1 2 2 = 3 + 1 + 2 3 - 4 3 4 ⇒ x - 3 + 1 2 2 = 3 + 1 - 2 3 4 ⇒ x - 3 + 1 2 2 = 3 2 + 1 2 - 2 3 4 ⇒ x - 3 + 1 2 2 = 3 - 1 2 2 And x - 3 + 1 2 = ± 3 - 1 2 Now, x = 3 - 1 2 + 3 + 1 2 x = 2 3 2 = 3 or x = - 3 + 1 2 + 3 + 1 2 x = 2 2 = 1 Thus, x = 3 & 1

A matrix consists of 30 elements. What are the possible orders it can have?

Since there are 30 elements, we can make the multiples of 30 , which are: 1 × 30 ; 30 × 1 ; 2 × 15 ; 15 × 2 ; 3 × 10 ; 10 × 3 ; 5 × 6 and 6 × 5 So, the possible orders are ( 1 , 30 ) , ( 2 , 15 ) , ( 3 , 10 ) , ( 5 , 6 ) , ( 6 , 5 ) , ( 10 , 3 ) , ( 15 , 2 ) , ( 30 , 1 )

Draw the graph of y = x 2 + 3 x + 2 and use it to solve the equation x 2 + 2 x + 4 = 0 .

Draw the graph of y = x 2 + 3 x + 2 by preparing the table of values as below: x - 4 - 3 - 2 - 1 0 1 2 3 4 y 6 2 0 0 2 6 12 20 30 Draw the parabola by plotting the points: ( - 4 , 6 ) , ( - 3 , 2 ) , ( - 2 , 0 ) , ( - 1 , 0 ) , ( 0 , 2 ) , ( 1 , 6 ) , ( 2 , 12 ) , ( 3 , 20 ) , ( 4 , 30 ) To solve x 2 + 2 x + 4 = 0 , subtract x 2 + 2 x + 4 = 0 from y = x 2 + 3 x + 2 x 2 + 3 x + 2 = y x 2 + 2 x + 4 = 0 x - 2 = y The equation y = x - 2 represents the straight line. Draw the graph of y = x - 2 by forming a table of values as below: x - 2 0 1 2 y - 4 - 2 - 1 0 Plotting the points ( - 2 , - 4 ) , ( 0 , - 2 ) , ( 1 , - 1 ) , 2 , 0 we get a straight line. The straight line does not intersect the curve y = x 2 + 3 x + 2 . Thus x 2 + 2 x + 4 = 0 has not real roots.