A and B are square matrices such that det. `(A)=1, B B^(T)=I`, det `(B) gt 0`, and `A( adj. A + adj. B)=B`. The value of det `(A+B)` is

Question

A and B are square matrices such that det. `(A)=1, B B^(T)=I`, det `(B) gt 0`, and `A( adj. A + adj. B)=B`. The value of det `(A+B)` is

A and B are square matrices such that det. `(A)=1, B B^(T)=I`, det `(B) gt 0`, and `A( adj. A + adj. B)=B`.
The value of det `(A+B)` is
A. `-2`
B. `-1`
C. 0
D. 1

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

Related Questions

Let `A` be a square matrix of order 3 such that adj. (adj. (adj. A)) `=[(16,0,-24),(0,4,0),(0,12,4)]`. Then find (i) `|A|` (ii) adj. A

2 Answers 7 Views

If A is order 3 square matrix such that `|A|=2`, then `|"adj (adj (adj A))"|` is

2 Answers 4 Views

If `A=[a_("ij")]_(4xx4)`, such that `a_("ij")={(2",","when "i=j),(0",","when "i ne j):}`, then `{("det (adj (adj A))")/(7)}` is (where `{*}` represent

2 Answers 5 Views

Let A be a square matrix of order 3 such that det. `(A)=1/3`, then the value of det. `("adj. "A^(-1))` is

2 Answers 4 Views

If A and B are two non-singular matrices of order 3 such that `A A^(T)=2I` and `A^(-1)=A^(T)-A`. Adj. `(2B^(-1))`, then det. (B) is equal to

2 Answers 5 Views

If A is a square matric of order 5 and `2A^(-1)=A^(T)`, then the remainder when `|"adj. (adj. (adj. A))"|` is divided by 7 is

2 Answers 5 Views

Let A and B are two square matrices of order 3 such that det. `(A)=3` and det. `(B)=2`, then the value of det. `(("adj. "(B^(-1) A^(-1)))^(-1))` is eq

2 Answers 7 Views

A, B and C are three square matrices of order 3 such that A= diag. (x, y, x), det. `(B)=4` and det. `(C)=2`, where `x, y, z in I^(+)`. If det. (adj. (

2 Answers 4 Views

Let matrix `A=[{:(x,y,-z),(1,2,3),(1,1,2):}]` , where `x,y,z in N`. If `|adj(adj(adj(adjA)))|=4^(8)*5^(16)`, then the number of such `(x,y,z)` are

2 Answers 5 Views

A be a square matrix of order `2` with `|A| ne 0` such that `|A+|A|adj(A)|=0`, where `adj(A)` is a adjoint of matrix `A`, then the value of `|A-|A|adj

2 Answers 5 Views

Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - D
`B B^(T)=I`
`:. |B B^(T)|=1`
`implies |B||B^(T)|=1`
`implies |B||B^(T)|=1`
`implies |B|=1`
A(adj. A + adj. B)=B
`implies |A|I+Axx"adj "B=B`
`implies I+Axx"adj "B=B`
`implies IxxB+Axx("adj B")xxB=B^(2)`
`implies B+A|B|I=B^(2)`
`implies B+A=B^(2)` (1)
`implies |A+B|=|B|^(2)=1`

4 views Answered 2023-02-05 15:29