Advertisements
Advertisements
Question
If A = `[(3,1),(-1,2)]` show that `A^2 - 5A + 7I = 0`.
Solution
We have to proved that A2 - 5A + 7I = 0
`A^2 = [(3,1), (-1, 2)][(3,1), (-1,2)]`
`= [(9 - 1,3 + 2),(-3 -2,-1 + 4)] = [(8,5), (-5,3)]`
`5A = 5 = [(3,1), (-1,2)] = [(15,5), (-5, 10)]`
Now, substituting the values in A2 - 5A + 71, we have,
`A^2 - 5A + 7I = [(8,5), (-5, 3)] - [(15,5),(-5,10)] + [(7,0),(0,7)]`
= `[(-7,0), (0,-7)] + [(7,0), (0,7)]`
= `[(0,0), (0,0)] = 0`
APPEARS IN
RELATED QUESTIONS
if `A = [(1,0,2),(0,2,1),(2,0,3)]` , prove that `A^2 - 6A^2 + 7A + 2I = 0`
if A = `[(3, -2),(4,-2)] and l = Matric [(1,0),(0,1)]` find k so that `A^2 = kA - 2I`
Find x, if `[x, -5, -1][(1,0,2),(0,2,1),(2,0,3)][(x),(4),(1)] = O`
A `= [(1,1,2),(0,2,-3),(3,-3,4)] "and B"^-1 = [(1,2,0),(0,3,-1),(1,0,2)]`
The restriction on n, k and P so that PY + WY will be defined are
If X, Y, Z, Wand P are matrices of order 2 × n, 3 × k, 2 × P and P × K respectively. if n = P, then the order of the matrix 7x – 5z is
If A = `[(1, 0),(2, 1)]`, B = `[(x, 0),(1, 1)]` and A = B2, then x equals ______.
If A = `[(1, 2, 3),(3, -2, 1),(4, 2, 1)]`, then show that A3 – 23A – 40I = 0.
