Advertisements
Advertisements
Question
If A = `[(1, 2),(-3, -1)], "B" = [(-1, 0),(1, 5)]`, then AB =
Options
`[(1, -10),(1, 20)]`
`[(1, 10),(-1, 20)]`
`[(1, 10),(2,- 5)]`
`[(1, 10),(-1, -20)]`
Solution
If A = `[(1, 2),(-3, -1)], "B" = [(-1, 0),(1, 5)]`, then AB = `[(1, 10),(2,- 5)]`.
Notes
Option (C) has been modified.
APPEARS IN
RELATED QUESTIONS
Solve the following equations by the inversion method :
2x + 3y = - 5 and 3x + y = 3.
Find the adjoint of the following matrix.
`[(1, -1, 2),(-2, 3, 5),(-2, 0, -1)]`
If A = `[(1,-1,2),(3,0,-2),(1,0,3)]` verify that A (adj A) = (adj A) A = | A | I
Find the inverse of the following matrix.
`[(1,2),(2,-1)]`
Find the inverse of the following matrix (if they exist):
`((1,3),(2,7))`
Choose the correct answer from the given alternatives in the following question:
If A = `[(2,-4),(3,1)]`, then the adjoint of matrix A is
Find the inverse of the following matrices by transformation method: `[(1, 2),(2, -1)]`
Fill in the blank :
If A = `[(3, -5),(2, 5)]`, then co-factor of a12 is _______
State whether the following is True or False :
A(adj. A) = |A| I, where I is the unit matrix.
Solve the following :
If A = `[(2, -3),(3, -2),(-1, 4)],"B" = [(-3, 4, 1),(2, -1, -3)]`, verify (3A – 5BT)T = 3AT – 5B.
Check whether the following matrices are invertible or not:
`[(1, 0),(0, 1)]`
If A = `[(cos alpha, sin alpha),(-sin alpha, cos alpha)]`, then A10 = ______
If A = `[(1, -1, 1),(2, 1, -3),(1, 1, 1)]`, 10B = `[(4, 2,2),(-5, 0, ∞),(1, -2, 3)]` and B is the inverse of matrix A, then α = ______
If A(α) = `[(cos alpha, sin alpha),(-sin alpha, cos alpha)]` then prove that A2(α) = A(2α)
If A = `[(-1),(2),(3)]`, B = `[(3, 1, -2)]`, find B'A'
If A = `[(6, 5),(5, 6)]` and B = `[(11, 0),(0, 11)]` then find A'B'
Choose the correct alternative:
If A is a non singular matrix of order 3, then |adj (A)| = ______
Complete the following activity to find inverse of matrix using elementary column transformations and hence verify.
`[(2, 0, -1),(5, 1, 0),(0, 1, 3)]` B−1 = `[(1, 0, 0),(0, 1, 0),(0, 0, 1)]`
C1 → C1 + C3
`[("( )", 0, -1),("( )", 1, 0),("( )", 1, 3)]` B−1 = `[("( )", 0, 0),("( )", 1, 0),("( )", 0, 1)]`
C3 → C3 + C1
`[(1, 0, 0),("( )", 1, "( )"),(3, 1, "( )")]` B−1 = `[(1, 0, "( )"),(0, 1, 0),("( )", 0, "( )")]`
C1 → C1 – 5C2, C3 → C3 – 5C2
`[(1, "( )", 0),(0, 1, 0),("( )", 1, "( )")]` B−1 = `[(1, 0, "( )"),("( )", 1, -5),(1, "( )", 2)]`
C1 → C1 – 2C3, C2 → C2 – C3
`[(1, 0, 0),(0, 1, 0),(0, 0, 1)]` B−1 = `[(3, -1, "( )"),("( )", 6, -5),(5, "( )", "( )")]`
B−1 = `[("( )", "( )", "( )"),("( )", "( )", "( )"),("( )", "( )", "( )")]`
`[(2, "( )", -1),("( )", 1, 0),(0, 1, "( )")] [(3, "( )", "( )"),("( )", 6, "( )"),("( )", -2, "( )")] = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`
Find the inverse of the following matrix:
`[(3,1),(-1,3)]`
Find the inverse of the following matrix:
`[(1,2,3),(0,2,4),(0,0,5)]`
Find the inverse of the following matrix:
`[(-3,-5,4),(-2,3,-1),(1,-4,-6)]`
Solve by matrix inversion method:
2x + 3y – 5 = 0; x – 2y + 1 = 0.
The sum of the cofactors of the elements of second row of the matrix `[(1, 3, 2), (-2, 0, 1), (5, 2, 1)]` is ____________.
If A is non-singular matrix such that (A - 2l)(A - 4l) = 0 then A + 8A-1 = ______.
If A and Bare square matrices of order 3 such that |A| = 2, |B| = 4, then |A(adj B)| = ______.
If A = `[(0, 0, 1), (0, 1, 0), (1, 0, 0)]`, then A-1 = ______
If A = `[(5, -4), (7, -5)]`, then 3A-1 = ______
If AB = I and B = AT, then _______.
If A is a solution of x2 - 4x + 3 = 0 and `A=[[2,-1],[-1,2]],` then A-1 equals ______.
If matrix A = `[(1, -1),(2, 3)]` such that AX = I, then X is equal to ______.
Choose the correct option:
If X, Y, Z are non zero real numbers, then the inverse of matrix A = `[(x, 0, 0),(0, y, 0),(0, 0, z)]`
If A = `[(1, 2, -1),(-1, 1, 2),(2, -1, 1)]`, then det (adj (adj A)) is ______.
The inverse of the matrix `[(1, 0, 0),(3, 3, 0),(5, 2, -1)]` is ______.
If A = `[(1, 1, 0),(2, 1, 5),(1, 2, 1)]`, then a11A21 + a12A22 + a13A23 is equal to ______.
If A = `[(2, 2),(-3, 2)]`, B = `[(0, -1),(1, 0)]`, then (B–1 A–1)–1 is equal to ______.
If matrix A = `[(1, 2),(4, 3)]`, such that AX = I, then X is equal to ______.
If A = `[(4, 3, 2),(-1, 2, 0)]`, B = `[(1, 2),(-1, 0),(1, -2)]`
Find (AB)–1 by adjoint method.
Solution:
AB = `[(4, 3, 2),(-1, 2, 0)] [(1, 2),(-1, 0),(1, -2)]`
AB = [ ]
|AB| = `square`
M11 = –2 ∴ A11 = (–1)1+1 . (–2) = –2
M12 = –3 A12 = (–1)1+2 . (–3) = 3
M21 = 4 A21 = (–1)2+1 . (4) = –4
M22 = 3 A22 = (–1)2+2 . (3) = 3
Cofactor Matrix [Aij] = `[(-2, 3),(-4, 3)]`
adj (A) = [ ]
A–1 = `1/|A| . adj(A)`
A–1 = `square`
If A = `[(2, 3),(4, 5)]`, show that A2 – 7A – 2I = 0