Advertisements
Advertisements
Question
Evaluate without using tables:
`[(2cos 60°, -2sin 30°),(-tan45°, cos 0°)] [(cos 45°, cosec 30°),(sec 60°, sin 90°)]`
Solution
We know that,
cos 60° = `1/2`, sin 30° = `1/2`, tan 45° = 1, cos 0° = 1
tan 45° = 1, cosec 30° = 2, sec 60° = 2, sin 90° = 1
Now `[(2cos 60°, -2sin 30°),(-tan45°, cos 0°)] [(cos 45°, cosec 30°),(sec 60°, sin 90°)]`
= `[(2 xx 1/2, -2 xx 1/2),(-1, 1)][(1, 2),(2,1)]`
= `[(1, -1),(-1, 1)][(1, 2),(2, 1)]`
= `[(1 xx 1 + (-1) xx 2, 1 xx 2 + (-1) xx 1),(-1 xx 1 + 1 xx 2, -1 xx 2 + 1 xx 1)]`
= `[(1 - 2, 2 - 1),(-1 + 2, -2 + 1)]`
= `[(-1, 1),(1, -1)]`
APPEARS IN
RELATED QUESTIONS
If M = `[(1, 2),(2, 1)]` and I is a unit matrix of the same order as that of M; show that: M2 = 2M + 3I.
If A = `[(2, 1, -1),(0, 1, -2)]`, Find A . At where At is the transpose of matrix A.
If M = `[(4,1),(-1,2)]`, show that 6M – M2 = 9I; where I is a 2 × 2 unit matrix.
If A =` [(1,-2 ,1),(2,1,3)]and B=[(2,1),(3,2),(1,1)]`
would it be possible to form the product matrix BA? If so, compute BA; if not give a reason
why it is not possible.
If A = `[(3, 5),(4,- 2)]` and B = `[(2),(4)]`, is the product AB possible ? Given a reason. If yes, find AB.
`[(2sin 30° ,- 2 cos 60°),(- cot 45° , sin 90°)]`
`[(tan 45° , sec 60°),("cosec" 30° , cos 0°)]`
A = `[(1, 2),(3, 4)] and "B" = [(6, 1),(1, 1)], "C" = [(-2, -3),(0, 1)]` find each of the following and state if they are equal.CA + B
Find x and y if `[(x + y, y),(2x, x - y)] [(2),(-1)] = [(3),(2)]`
If `[(3, 4),(5, 5)] = [(a, b),(c, d)] [(1, 0),(0, 1)]`write down the values of a,b,c and d
Choose the correct answer from the given four options :
If A = `[(0, 1),(1, 0)]`, then A2 =
