Exercise 3.2

Let 2 4 1 3 2 5A , B , C3 2 2 5 3 4 −     = = =      −      Find each of the following: (i) A + B (ii) A - B (iii) 3A - C (iv) AB (v) BA

Compute the following: (i) a b a b

    +   −    (ii) 2 2 2 2 / 2 2 2 2 / 2 2 / 2 2 a b b c ab bc

  + +  +   − −+ +     (iii) 1 4 6 12 7 6 / 8 5 16 8 0 5 / 2 8 5 3 2 4 − −        +           (iv) 2 2 2 2 / 2 2 2 2 / cos sin sin cos / sin cos cos sin

    +          

Compute the indicated products. (i) a b a b

−       −    (ii) 1 2/3 /   /        [2 3 4] (iii) 1 2 1 2 3 2 3 2 3 1 −           (iv) 2 3 4 1 3 5 / 3 4 5 0 2 4 / 4 5 6 3 0 5 −                   (v) 2 1 1 0 13 2 1 2 11 1         −  −  (vi) 2 33 1 3 1 01 0 2 3 1 − −      −    

If / 1 2 3 3 1 2 4 1 2 A 5 0 2 , B 4 2 5 and C 0 3 2 1 1 1 2 0 3 1 2 3 − −           = = =           − −      , then compute (A+B) and (B - C). Also, verify that A + (B - C) = (A + B) - C.

If / 2 5 2 31 13 3 5 5 1 2 4 1 2 4A and B3 3 3 5 5 5 7 2 7 6 223 3 5 5 5                = =                   , then compute 3A - 5B.

Simplify cos sin sin coscos + sinsin cos cos sin θ θ θ − θ   θ θ   − θ θ θ θ   

Find X and Y, if (i) 7 0 3 0X + Y and X - Y2 5 0 3    = =       (ii) 2 3 2 22X + 3Y and 3X 2Y4 0 1 5 −   = + =   −   

Find X, if Y = 3 2 1 4      and 2X + Y = 1 0 3 2    − 

y / x      + =          

x z / y t −     + =          

If / 2 1 10 3 1 5x y −     + =           , find the values of x and y.

+     = +     − +      , find the values of x, y, z and w.

If / cos sin 0 / F ( ) sin cos 0 0 0 1 x x

−   =      , show that F(x) F(y) = F(x + y).

Show that (i) 5 1 2 1 2 1 5 1 / 6 7 3 4 3 4 6 7 − −        ≠               (ii) 1 2 3 1 1 0 1 1 0 1 2 3 / 0 1 0 0 1 1 0 1 1 0 1 0 / 1 1 0 2 3 4 2 3 4 1 1 0 − −                − ≠ −                      

Find A2 - 5A + 6I, if 2 0 1

1 1 0 /    =     − 

If / 1 0 2

2 0 3 /    =      , prove that A3 - 6A2 + 7A + 2I = 0

If 3 2 1 0A and I=4 2 0 1 −   =    −    , find k so that A 2 = kA - 2I

If / 0 tan 2A tan 02 α −  =  α     and I is the identity matrix of order 2, show that I + A = (I - A) cos sin sin cos α − α   α α 

A trust fund has 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: (a) 1800 (b) 2000

The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are 80, 60 and ` 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra. Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Choose the correct answer in Exercises 21 and 22.

The restriction on n, k and p so that PY + WY will be defined are: (A) k = 3, p = n (B) k is arbitrary, p = 2 (C) p is arbitrary, k = 3 (D) k = 2, p = 3

If n = p, then the order of the matrix 7X - 5Z is: (A) p × 2 (B) 2 × n (C) n × 3 (D) p × n 3.5. Transpose of a Matrix In this section, we shall learn about transpose of a matrix and special types of matrices such as symmetric and skew symmetric matrices. Definition 3 If A = [aij] be an m × n matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A. Transpose of the matrix A is denoted by A′ or (AT). In other words, if A = [aij]m × n, then A′ = [aji]n × m. For example, if 2 3 3 2 3 5 3 3 0 A 3 1 , then A 15 10 1 5 × ×       = ′ =  −    −     3.5.1 Properties of transpose of the matrices We now state the following properties of transpose of matrices without proof. These may be verified by taking suitable examples. For any matrices A and B of suitable orders, we have (i) (A′)′ = A, (ii) (kA)′ = kA′ (where k is any constant) (iii) (A + B)′ = A′ + B′ (iv) (A B)′ = B′ A′ Example 20 If 2 1 23 3 2A and B 1 2 44 2 0   − = =      , verify that (i) (A′)′ = A, (ii) (A + B)′ = A′ + B′, (iii) (kB)′ = kB′, where k is any constant. Solution (i) We have A = ( ) 3 4 3 3 2 3 3 2A 3 2 A A 4 2 0 4 2 02 0        ′′ ′⇒ = ⇒ = =           Thus (A′)′ = A (ii) We have A = 3 3 2 , 4 2 0       B = 2 1 2 5 3 1 4A B1 2 4 5 4 4  −  −⇒ + =       Therefore (A + B)′ = 5 5 3 1 4 4 4     −     Now A′ = 3 4 2 1 3 2 , B 1 2 , 2 0 2 4       ′ = −          So A′ + B′ = 5 5 3 1 4 4 4     −     Thus (A + B)′ = A′ + B′ (iii) We have kB = k 2 1 2 2 2 1 2 4 2 4

− −   =       Then (kB)′ = 2 2 1 2 1 2 B 2 4 2 4 k k

k k         ′− = − =           Thus (kB)′ = kB′ Example 21 If [ ] A 4 , B 1 3 6 −   = = −     , verify that (AB) ′ = B′A′. Solution We have A = [ ] 4 , B 1 3 6 −    = −     then AB = [ ] 4 1 3 6 −    −     = 2 6 12 4 12 24 5 15 30 − −   −   −  Now A′ = [-2 4 5] , 1    ′ =    −  B′A′ = [ ] 1 2 4 5 3 2 4 5 6 12 15 (AB) 6 12 24 30 −        ′− = − =       − − −    Clearly (AB)′ = B′A′