Advertisements
Advertisements
Question
If A = `[(1,-1),(2,3)]` show that A2 - 4A + 5I2 = 0 and also find A-1.
Solution
Given A = `[(1,-1),(2,3)] |"A"| = |(1,-1),(2,3)|` = 3 + 2 = 5 ≠ 0
∴ A-1 exists.
LHS = A2 - 4A + 5I
Premultiply by A-1 we get
LHS = `"A"^-1. "A"^2 - 4"A"^-1."A" + 5"A"^-1."I"`
`= ("A"^-1 "A")."A" - 4"I" + 5"A"^-1` ...[∵ A-1A = I]
= IA - 4I + 5A-1
= A - 4I + 5A-1
Now `"A"^-1 = 1/|"A"| "adj A"`
`= 1/5 [(3,1),(-2,1)]`
∴ LHS = `[(1,-1),(2,3)] - 4[(+1,0),(0,1)] + 5 . 1/5 [(3,1),(-2,1)]`
`= [(1,-1),(2,3)] + [(-4,0),(0,-4)] + [(3,1),(-2,1)]`
`= [(1-4+3,-1+0+1),(2+0-2,3-4+1)] = [(0,0),(0,0)]`
= RHS
APPEARS IN
RELATED QUESTIONS
Find the inverse of the following matrix by the adjoint method.
`[(2,-2),(4,3)]`
Find the inverses of the following matrices by the adjoint method:
`[(1,2,3),(0,2,4),(0,0,5)]`
Find matrix X, if AX = B, where A = `[(1, 2, 3),(-1, 1, 2),(1, 2, 4)] "and B" = [(1),(2),(3)]`.
Fill in the blank :
If A = `[(2, 1),(1, 1)] "and" "A"^-1 = [(1, 1),(x, 2)]`, then x = _______
A = `[(cos alpha, - sin alpha, 0),(sin alpha, cos alpha, 0),(0, 0, 1)]`, then A−1 is
Find A–1 using adjoint method, where A = `[(cos theta, sin theta),(-sin theta, cos theta)]`
The inverse matrix of `((3,1),(5,2))` is
If A = `[(1 + 2"i", "i"),(- "i", 1 - 2"i")]`, where i = `sqrt-1`, then A(adj A) = ______.
If a 3 × 3 matrix A has its inverse equal to A, then A2 = ______
Find the inverse of the matrix `[(1, 1, 1),(1, 2, 3),(3, 2, 2)]` by elementary column transformation.
