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If A is 4x4 matrix and  det(A) = -4. Then  det(8A^-1) = ???,

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gomezzzzzzzzz... | Student | (Level 1) Honors

Posted April 28, 2013 at 1:23 AM via web

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If A is 4x4 matrix and  det(A) = -4. Then

 det(8A^-1) = ???,

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mvcdc | Student, Graduate | (Level 1) Associate Educator

Posted April 28, 2013 at 5:19 AM (Answer #1)

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We are given the following:

det(A) = -4

det(B) = 3

To solve for det(8A^-1), we first note the following:

i. The determinant of an inverse of a matrix, is just the reciprocal of the determinant of the original matrix.

ii. Scalar multiplication of a row by a constant c, also multiplies the determinant by c.

Hence, det(8A^-1) can be calculated as follows:

`det(8A^-1) = 8^4 det(A^-1) = 8^4 (1/det(A)) = 4096/-4 = -1024`

 

Note that we multiplied by 8^4 since A is a 4 x 4 matrix (i.e. there are 4 rows, each of which is multiplied by the scalar 8, thus, the determinant of A^-1 is multiplied to 8 for times or 8^4).

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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted April 28, 2013 at 8:04 AM (Answer #2)

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We know

`det(P^(-1))=1/det(P)`

`det(xP)=x^ndet(P)`    , x is scalar and  n is order of P.

Thus

`det(8A^(-1))=8^4det(A^(-1))=8^4/det(A)`

`=8^4/(det(A))`

`=8^4/(-4)=-1024`

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