Question

hus, D satisfying ​(ABC)DequalsI exists. Why does the expression for D found above also satisfy ​D(ABC)equalsI​, thereby showing that ABC is​ invertible? Select the correct choice below​ and, if​ necessary, fill in the answer box within your choice. A. After substituting the expression for​ D, the product DABC simplifies to I by repeated application of the associative property and the definition of inverse matrices. B. After substituting the expression for​ D, left multiplying the product by nothing results in the equation IequalsABCD. C. After substituting the expression for​ D, taking the inverse of both sides of the equation results in the equation IequalsABCD. D. After substituting the expression for​ D, right multiplying the pr

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

First Question

Option A is correct

Second  Question

Option C is correct

Third   Question

     $D = A^{-1} * B^{-1} * C^{-1}$

Fourth   Question

  So substituting for D in  (ABC) D =  I

                 $(ABC) * A^{-1} * B^{-1} * C^{-1} = I$

                 $I = I$

This proof that  ABC is invertible

Step-by-step explanation:

From the question we are told that

   A , B and  C are invertible which means that $A^{-1} , B^{-1}, C^{-1}$ exist

Now

 From the question

          (ABC) D =  I

Where I is an identity matrix

   Now when we multiply both sides by  $A^{-1}$  we have

          $A^{-1} A BCD = A^{-1} * I$

          $IBCD = A^{-1}$

Now when we multiply both sides by  $B^{-1}$  we have  

         $B^{-1 } *I BCD = A^{-1} * B^{-1}$

         $I CD = A^{-1} * B^{-1}$

Now when we multiply both sides by  $C^{-1}$  we have  

          $C^{-1} * I CD = A^{-1} * B^{-1} * C^{-1}$

              $I D = A^{-1} * B^{-1} * C^{-1}$

                 $D = A^{-1} * B^{-1} * C^{-1}$

So substituting for D in the above equation

                 $(ABC) * A^{-1} * B^{-1} * C^{-1} = I$

                 $I = I$

This proof that  ABC is invertible