Advertisements
Advertisements
Question
Simplify `cos theta[(cos theta, sintheta),(-sin theta, cos theta)] + sin theta[(sin theta, -cos theta), (cos theta, sin theta)]`
Advertisements
Solution
`cos theta[(cos theta, sin theta),(-sin theta, cos theta)] + sin theta[(sin theta, -cos theta),(cos theta, sin theta)]`
= `[(cos^2 theta, cos theta sin theta),(-sin theta cos theta, cos^2 theta)] + [(sin^2 theta, -sin theta cos theta),(sin theta cos theta, sin^2 theta)]`
= `[(cos^2 theta + sin^2 theta, cos theta sintheta - sin theta cos theta),(-sin theta cos theta + sin theta cos theta, cos^2 theta + sin^2 theta)]`
= `[(1, 0),(0, 1)]` ...(∵ cos2θ + sin2θ = 1)
APPEARS IN
RELATED QUESTIONS
If `A = [(1, 2, -3),(5, 0, 2),(1, -1, 1)], B = [(3, -1, 2),(4, 2, 5),(2, 0, 3)] and C = [(4, 1, 2),(0, 3, 2),(1, -2, 3)]` then compute (A + B) and (B – C). Also verify that A + (B – C) = (A + B) – C.
If ` A = [(2/3, 1, 5/3),(1/3, 2/3, 4/3),(7/3, 2, 2/3)]` and `B = [(2/5, 3/5, 1),(1/5, 2/5, 4/5),(7/5, 6/5, 2/5)]`, then compute 3A – 5B.
Show that `[(5, -1),(6, 7)][(2, 1),(3, 4)] ≠ [(2, 1),(3, 4)][(5, -1),(6, 7)]`
Show that `[(1, 2, 3),(0, 1, 0),(1, 1, 0)][(-1, 1, 0),(0, -1, 1),(2, 3, 4)] ≠ [(-1, 1, 0),(0, -1, 1),(2, 3, 4)][(1, 2, 3),(0, 1, 0),(1, 1, 0)]`
Find A2 – 5A + 6I, if A = `[(2, 0, 1),(2, 1, 3),(1, -1, 0)]`
The product of any matrix by the scalar ______ is the null matrix.
If A `= [(1,2),(2,1)]` and f(x) = (1 + x) (1 - x), then f(a) is ____________.
If A `= [(2"x", 0),("x","x")] "and A"^-1 = [(1,0),(-1,2)],` then x equals ____________.
If | A | = | kA |, where A is a square matrix of order 2, then sum of all possible values of k is ______.
