expert :Norman Wagnerplease see attachment Note- please make sure that you give 100% right answer because last time you did i got only 80% right answer thanks 1. The matrix A = —4 [1 —2 2 —7 —6 —1 3 5...

Sol: (1) (a) the matrix A is invertible, . .,i e
 
17 4 1
14 3 1
10 2 1
Adj A
   
 

 
  

And
1A 
So, the inverse of a matrix is
 1
1
17 4 1
1
14 3 1
1
10 2 1
17 4 1
14 3 1
10 2 1
Adj A
A
A
A



   
 

 
  
   
 

 
  

(b) The matrix5 5 ,
1 2 1 0 0
4 7 3 0 0
2 6 5 0 0
0 0 0 1 0
0 0 0 5 6
P Q
B
R S
 
 
 
   
     
   
 
  

Here we have
1 2 1 0 0
0 0 0 1 0
4 7 3 , 0 0 , ,
0 0 0 5 6
2 6 5 0 0
P Q R S
   
      
               
       

We now have
 
1
1
17 4 1
14 3 1
10 2 1
t S RP Q

 
   
 

 
 
 
1
0 0
0 0
0 0
Y P Qt 
 
 

 
  
1
0 0 0
0 0 0
Z RP t 
 
  
 
 1
1 0
0.834 0.167
X P I QZ 
 
  
 

The final inverse is given by

1
17 4 1 0 0
14 3 1 0 0
10 2 1 0 0
0 0 0 1 0
0 0 0 0.834 0.167
t Y
B
Z X

   
 
  
   
  
 
  

Sol: (2)
2 3 4
4 8 7
6 5 14
6 9 12
8 6 19
A
 
 
 
 
  
 
  
  

Row reduced of matrix A is
2 1 2 3 1 3 4 1 4 5 1 5
3 2 3 5 2 5
2 3 4
0 2 1
2 , 3 , 3 , 4 ,0 4 2
0 0 0
0 6 3
2 3 4
0 2 1
2 , 3 , 0 0 0
0 0 0
0 0 0
R R R R R R R R R R R R
R R R R R R
 
 
 
           
 
 
  
 
 
 
       
 
 
  

So,
2 3 4
0 2 1
0 0 0
0 0 0
0 0 0
U
 
 
 
 
 
 
  

This has only two non-zero rows. Thus L is given by replacing entries below
diagonal pivotal...