Question

Show that if A is​ invertible, then det Upper A Superscript negative 1det A−1equals=StartFraction 1 Over det Upper A EndFraction 1 det A. What​ theorem(s) should be used to examine the quantity det Upper A Superscript negative 1det A−1​? Select all that apply. A. If A and B are ntimes×n ​matrices, then det ABequals=(det Upper A )(det A)(det Upper B )(det B). Your answer is correct.B. If one row of a square matrix A is multiplied by k to produce​ B, then det Upper Bdet Bequals=ktimes•(det Upper A )(det A). C. A square matrix A is invertible if and only if det Upper Adet Anot equals≠0. Your answer is correct.D. If A is an ntimes×n ​matrix, then det Upper A Superscript Upper Tdet ATequals=det Upper Adet A. Consider the quantity (det Upper A )(det Upper A Superscript negative 1 Baseline )(det A)det A−1. To what must this be​ equal? A. det Upper I

Answer 1

Answer:

Therefore  

 $det(A)*det(A^{-1}) = 1$      and         $det(A^{-1}) = 1/det(A)$

Step-by-step explanation:

Remember that  if  

$I$  =  Identity Matrix    

then    $det(I)=1$.

Also remember that for any invertible matrix  

$A*A^{-1} = I =$Identity Matrix.

Therefore    

$det(A*A^{-1}) = det(A)*det(A^{-1}) = det(I) = 1$

Therefore  

 $det(A)*det(A^{-1}) = 1$      and         $det(A^{-1}) = 1/det(A)$