# linear algebra question.the matrices A and B are defined by A={{3,2},{-2,-2}} respective B={{5,3},{2,1}}. calculate the determinant for the matrix  5(A^T)(B^7)((A^-1)^5)(B^T). (T is transpose).

magnesi | Certified Educator

I think this is what you are asking.

Find the Determinant of 5(A^t)(B^7)((a^-1)^5)(B^t)

You need to first find A^T = [[3,-2];[2,-2]] and B^T =[[5,2];[3,1]]

Multiply as given in the problem. To find the determinant you multiply and subtract. Example  [[a,b];[c,d]] the determinant is a*d-c*d.

The answer I got for the determinate of 5(A^t)(B^7)((a^-1)^5)(B^t) =25/16

it is difficult to show the work because there is no equation editor available.

[[15,10][-10,-10]]*[[275423,160059][106706,62011]]*[[3/8,11/16][-11/16][-43/32]]*[[52,2][3,1]]=[[-7202655/16,-54845935/32][-4621585/16,-35191905/32]]

det([[-7202655/16,-54845935/32][-4621585/16,-35191905/32]])=25/16

neela | Student

Given the 2x2 square matrices,A={{3,2},{-2,-2}},B={{5,3},{2,1}} and B={{5,3},{2,1}}.

To find, |5(A^T)(B^7)((A^-1)^5)(B^T)|, where A^T  or B^t stands for transpose of A and respectively and A^(-1) stands for the inverse matrix of A.

We use the following results to find the determinantal value of the expression,5(A^T)(B^7)((A^-1)^5)(B^T),(of course without going through the multiplplying and writing matrices ptoducts.)

For a square matrix,

|A*B| = |A|*|B| and so, |A^n| = {|A|}^n , n is an integer.

|kA| = ( k^2)*|A| for 2X2 matrix A , where k is a scalar ,(a constant). For a mXm matrix A, |kA| =  (k^m)*|A|.

|A^T| = |A| and similarly |B^T| =|B| and finally,

|[A^(-1)]^n| = {1/|A|}^n , where n is an integer.

|A^T|=|A|= |{{3,2},{-2,-2}}| = 3*(-2)-2(-2) = -2

|B^T|=|B|=|{{5,3},{2,1}}| =5*1-3*2= -1

With these we can rewrite, |5(A^T)(B^7)((A^-1)^5)(B^T)| like:

(5^2)*|A|*|B|^7 * {1/|A|}^5 * |B|

25*(-2)*(-1)^7 * {1/(-2)}5 * (-1)

=50/32,

=25/16,

=25/16.