Advertisements
Advertisements
Question
Show that `[(5, -1),(6,7)][(2,1),(3,4)] != [(2,1),(3,4)][(5,-1),(6,7)]`
Solution
right side`[(2,1),(3,4)][(5,-1),(6,7)]`
`[(10 + 6, -2 + 7),(15 + 24, -3 + 28)] `
`= [(16, 5),(39, 25)]`
left side `[(5,-1),(6,7)][(2, 1),(3, 4)]`
`= [(10 - 3, 5 - 4),(12 + 21, 6 + 28)]`
` =[(7 , 1),(33, 34)]`
left side `ne` right side
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.
Simplify, `cos theta[(cos theta, sintheta),(-sin theta, cos theta)] + sin theta [(sin theta, -cos theta), (cos theta, sin theta)]`
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 `A^2 - 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 ______.