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+1 vote
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Let A be a 4 × 4 matrix with real entries such that -1, 1, 2, -2 are its eigen values. If B = A4 - 5A2 + 5I where I denotes 4 × 4 identity matrix, then which of the following is correct? (det(X) represents determinant of X)

(A) det(A + B) = 0
(B) det(B) = 1
(C) trace of A + B is 4
(D) all of these

asked in Linear Algebra by Active (5.1k points) | 256 views
0
All of these are correct :
0
how are a and b correct ?
0
Eigen value of B=1,1,1,1
A+B will have one zero thus det=0
and Det(B)=1
0
how Eigen value of B=1,1,1,1 ?
0

saxena0612 how you det ( a + b) =0 ?

I think b is correct option.

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What values are you getting for B?
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B=1,1,1,1

1 Answer

+3 votes
$(\lambda +1) (\lambda -1)(\lambda +2)(\lambda -2)=0$

$=>(\lambda ^{2}-1) (\lambda ^{2}-4)=0$

$=>\lambda ^{4} -5\lambda ^{2} +4=0$

$=> A^{4}-5A^{2}+4=0$

So, B= I   So eigen values of  identity matrix B are 1,1,1,1 and det(B)=1. option B is correct.

option A is also correct because on adding A and B and then after finding product of diagonals of A+B we will get 0.

option C is also true.

Answer : D
answered by Loyal (8.2k points)
edited by
0
How det(A+B) =0.Please explain??
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Let C=A+B. Therefore, C=A+I since B=I. Substituting the eigen values of A in C, we get the eigen values of C which is -1,0,2,3. and det(C)=det(A+B)= (-1)(0)(2)(3)=0. Hence det(A+B)=0.

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