Question

11. The inverse of the matrix

1 1 1

A 1 0 2

3 1 1

 

 



 

   

is

(a)

2 0 2

1

5 1 2

4

1 1 2

  

 



 

     

(b)

2 0 2

1

5 2 1

4

1 2 1

  

 

   

    

(c)

2 0 2

1

2 5 1

4

2 1 1

  

 



 

     

(d)

2 0 2

1

5 1 1

4

1 2 1

  

 



 

     ​

Answer 1

Answer:

101 you so much for your time and consideration

Answer 2

Answer:

5 Inverse Matrices

Suppose A is a square matrix. We look for an “inverse matrix” A-

1 of the same size, such

that A-

1 times A equals I . Whatever A does, A-

1 undoes. Their product is the identity

matrix—which does nothing to a vector, so A-

1Ax D x. But A-

1 might not exist.

What a matrix mostly does is to multiply a vector x. Multiplying Ax D b by A-

1

gives A-

1Ax D A-

1b. This is x D A-

1b. The product A-

1A is like multiplying by

a number and then dividing by that number. A number has an inverse if it is not zero—

matrices are more complicated and more interesting. The matrix A-

1 is called “A inverse.”

DEFINITION The matrix A is invertible if there exists a matrix A-

1 such that

A-

1A D I and AA-

1 D I: (1)

Not all matrices have inverses. This is the first question we ask about a square matrix:

Is A invertible? We don’t mean that we immediately calculate A-

1. In most problems

we never compute it! Here are six “notes” about A-

1.

Note 1 The inverse exists if and only if elimination produces n pivots (row exchanges

are allowed). Elimination solves Ax D b without explicitly using the matrix A-

1.

Note 2 The matrix A cannot have two different inverses. Suppose BA D I and also

AC D I . Then B D C, according to this “proof by parentheses”:

B.AC / D .BA/C gives BI D IC or B D C: (2)

This shows that a left-inverse B (multiplying from the left) and a right-inverse C (multi-

plying A from the right to give AC D I ) must be the same matrix.

Note 3 If A is invertible, the one and only solution to Ax D b is x D A-

1b:

Multiply Ax D b by A-

1: Then x D A-

1Ax D A-

1b:

Note 4 (Important) Suppose there is a nonzero vector x such that Ax D 0. Then A

cannot have an inverse. No matrix can bring 0 back to x.

If A is invertible, then Ax D 0 can only have the zero solution x D A-

10 D 0.

Note 5 A 2 by 2 matrix is invertible if and only if ad -

bc is not zero:

2 by 2 Inverse: -

a b

c d -

1

D 1

ad -

bc -

d -

b

-

c a

: (3)

This number ad -

bc is the determinant of A. A matrix is invertible if its determinant is not

zero (Chapter 5). The test for n pivots is usually decided before the determinant appears.