Question

Let Aequals=left bracket Start 2 By 2 Matrix 1st Row 1st Column 1 2nd Column 2 2nd Row 1st Column 8 2nd Column 18 EndMatrix right bracket

1 | 2 |

8 | 18 |

, Bold b 1**b**1equals=left bracket Start 2 By 1
Matrix 1st Row 1st Column negative 5 2nd Row 1st Column negative 36
EndMatrix right bracket

−5 |

−36 |

, Bold b 2**b**2equals=left bracket Start 2 By 1
Matrix 1st Row 1st Column 3 2nd Row 1st Column 16 EndMatrix right
bracket

3 |

16 |

, Bold b 3**b**3equals=left bracket Start 2 By 1
Matrix 1st Row 1st Column 3 2nd Row 1st Column 22 EndMatrix right
bracket

3 |

22 |

, and Bold b 4**b**4equals=left bracket Start 2 By
1 Matrix 1st Row 1st Column 4 2nd Row 1st Column 24 EndMatrix right
bracket

4 |

24 |

.(a) Find

Upper A Superscript negative 1A−1

and use it solve the four equations

Axequals=Bold b 1**b**1,

Axequals=Bold b 2**b**2

,Axequals=Bold b 3**b**3,

and

Axequals=Bold b 4**b**4.

(b) The four equations in part (a) can be solved by the same set of operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix [A

Bold b 1**b**1

Bold b 2**b**2

Bold b 3**b**3

Bold b 4**b**4].

Find

Upper A Superscript negative 1A−1.

Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice

A.Upper A Superscript negative 1A−1equals=left bracket Start 2 By 2 Matrix 1st Row 1st Column nothing 2nd Column nothing 2nd Row 1st Column nothing 2nd Column nothing EndMatrix right bracket

9 | −1 |

−4 | 12 |

(Simplify your answers.)

B.

The matrix is not invertible.

Solve

Axequals=Bold b 1**b**1.

xequals=left bracket Start 2 By 1 Matrix 1st Row 1st Column nothing 2nd Row 1st Column nothing EndMatrix right bracket

−9 |

2 |

(Simplify your answers.)Solve

Axequals=Bold b 2**b**2.

xequals=left bracket Start 2 By 1 Matrix 1st Row 1st Column nothing 2nd Row 1st Column nothing EndMatrix right bracket

11 |

−4 |

(Simplify your answers.)Solve

Axequals=Bold b 3**b**3.

xequals=left bracket Start 2 By 1 Matrix 1st Row 1st Column nothing 2nd Row 1st Column nothing EndMatrix right bracket

5 |

−1 |

(Simplify your answers.)Solve

Axequals=Bold b 4**b**4.

12 |

−4 |

(Simplify your answers.)(b) Solve the four equations by row reducing the augmented matrix [A

Bold b 1**b**1

Bold b 2**b**2

Bold b 3**b**3

Bold b 4**b**4].

Write the augmented matrix [A

Bold b 1**b**1

Bold b 2**b**2

Bold b 3**b**3

Bold b 4**b**4]

in reduced echelon form.left bracket Start 2 By 6 Matrix 1st Row 1st Column nothing 2nd Column nothing 3rd Column nothing 4st Column nothing 5st Column nothing 6st Column nothing 2nd Row 1st Column nothing 2nd Column nothing 3rd Column nothing 4st Column nothing 5st Column nothing 6st Column nothing EndMatrix right bracket nbsp

(Simplify your answers.)

Are the solutions the same in (a) and (b)?

Answer #1

Find the equilibrium vector for the given transition matrix. P
equals left bracket Start 2 By 2 Matrix 1st Row 1st Column 0.42 2nd
Column 0.58 2nd Row 1st Column 0.32 2nd Column 0.68 EndMatrix right
bracket The equilibrium vector is nothing.

Find the equilibrium vector for the transition matrix below.
left bracket Start 2 By 2 Matrix 1st Row 1st Column one eighth 2nd
Column seven eighths 2nd Row 1st Column one ninth 2nd Column eight
ninths EndMatrix right bracket.

Find the equilibrium vector for the transition matrix below.
left bracket Start 3 By 3 Matrix 1st Row 1st Column 0.75 2nd Column
0.10 3rd Column 0.15 2nd Row 1st Column 0.10 2nd Column 0.60 3rd
Column 0.30 3rd Row 1st Column 0.10 2nd Column 0.35 3rd Column 0.55
EndMatrix right bracket The equilibrium vector is.

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A= 1 3 8 2
7
1 3 8 2 7
2 7
20 6 20
--- 0 1 4 2 6
-3 -12
-36 -7
-19
0 0 0 1 4
3 13 40
9
25
0 0 0 0 0
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