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

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