The angles of a quadrilateral ABCD are `x^(@),(x+1)^(@),(x+2)^(@)` and `(x+3)^(@)`, taken in the same order. Then the quadrilateral ABCD is necessarily a
If `|A| ne 0`, then
`AB=AC`
or `A^(-1) AB=A^(-1) AC`
or `B=C`
Also, if A is orthogonal matrix, then `A A^(T)=I`
`implies |A A^(T)|=1 implies |A|^(2)=1implies` is invertible
`Let vecc=xhati+yhatj+zhatk`. Then `|veca|=|vecb|=|vecc|Rightarrowx^(2)+y^(2)+z^(2)=2`
it is given that the angles between the vectors taken in pairs are equal, say `theta` . Therefore,
`cos theta=(veca.vecb)/(|veca||vecb|)= (0+1+1)/(sqrt2sqrt2)=1/2`
`Rightarrow (veca.vecc)/(|veca||vecc|)=1/2and (vecb.vecc)/(|vecb||vecc|)=1/2`
`(x+y)/(sqrt2sqrt2)=1/2 and(y+z)/(sqrt2sqrt2)=1/2`
`Rightarrow...
Correct Answer - A::C
`(a,c)` `A+B=AB`
`impliesI-(A+B-AB)=I`
`implies(I-A)(I-B)=I`
`implies|I-A||I-B|=I`
`implies |I-A|`, `|I-B|` are non zero
Also `(I-B)(I-A)=I`
`impliesI-B-A+BA=I`
`impliesA+B=B+A`
`impliesAB=BA`
`implies(a)` and `(c )` are correct