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Let A be (3×3) real matrix. Suppose 1 & -1 are two of three Eigen values of A and 18 is one of the Eigen values of A2+3A. Then

 [ A ]     A is invertible as well as  A2  + 3A is invertible
[ B ]

A^2+3A is invertible but A is not invertible

 

[ C ]

A is invertible but A^2+3A is not invertible

 

[ D ]

Both A & A^2+3A are not invertible

 
 
asked in Linear Algebra by Junior (769 points)  
edited by | 96 views

1 Answer

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Best answer

Here we have to recall the property :

a)  If  a matrix A is modified  A = f(A) then the individual eigen values will also be updated in the same manner 

b) The product of  eigen values will give us the determinant of the matrix.

c) If the determinant of the matrix is non zero then it is said to be invertible matrix.

d) In an n * n matrix , at most n distinct eigen values possible.

In this context a) point  means ,

If say x is the eigen value of A , then the corresponding eigen value

for  A2  +  3A     will   be    :   x2  + 3x

So given that one of the eigen values of A+ 3A  is 18..So using a) property mentioned above , we have :

           x+ 3x = 18

==>     x  =  3

So 3 is the corresponding eigen value of original matrix..And two of the eigen values already given : 1 , -1

So determinant of matrix A  =  3 * 1 * -1

                                         =  -3  which is non zero  

Hence A is invertible..

Now for A2  + A  matrix we need to calculate other 2 eigen values which can be found using the eigen values 1 and -1 of A.

So eigen value of the new matrix corresponding to the eigen value 1  =  12  + 3(1)   = 4

eigen value of the new matrix corresponding to the eigen value  -1   =  (-1)2  + 3(-1)   =  1 - 3 = -2

So product of eigen values of new matrix =  18 * 4 * -2   =  -144

Hence determinant of new matrix   =   -144 which is non zero as well

So the new matrix is also invertible.

Hence A is invertible as well as  A2  + 3A is invertible.

Hence A) is the correct option.

 

answered by Veteran (65k points)  
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