CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Consider A=a210b003c where a,b,c are the roots of the equation x33x2+2x1=0. If matrix B is such that AB=BA,|A+B2I|0 and A2B2=4I4B, where I is the 3×3 identity matrix, then the value of det(B) is 


Solution

|A|=∣ ∣a210b003c∣ ∣=abc=1
where x33x2+2x1=(xa)(xb)(xc)     (1)

Given, A2B2=4I4B
A2=B24B+4IA2(B2I)2=O(A+B2I)(AB+2I)=O     (AB=BA)
Since |A+B2I|0,
AB+2I=0B=A+2I=∣ ∣a+2210b+2003c+2∣ ∣

|B|=(a+2)(b+2)(c+2) 
Putting x=2 in equation (1), we get
81241(2a)(2b)(2c)25=(a+2)(b+2)(c+2)|B|=25|B|=5

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More


People also searched for
View More



footer-image