+0  
 
0
6
5
avatar+13 

For a certain value of k, the system

 

x + y + 3z &= 10, 
-4x + 2y + 5z &= 7, 
kx + z &= 3


has no solutions. What is this value of k?

 Mar 19, 2024
 #2
avatar+624 
0

The system

 

x + y + 3z = 10,
-4x + 2y + 5z = 7,
kx + z = 3

 

will have no solution if the corresponding matrix

 

[1 1 3]
[-4 2 5]
[k 0 1]

 

has a determinant of 0.  Taking determinant, we find that it is 2k - 14.

 

Therefore, the value of k such that the system has no solution is k = 7.

 Mar 19, 2024
 #5
avatar+128794 
+1

1  1  3  1   1
-4 2  5  -4  2
k  0  1  k   0

 

Determinant   =     0

 

2 + 5k + 0  - [ 6k  + 0 - 4] =   0

 

2 + 5k  - 6k + 4  = 0

 

6 - k   =   0 

 

k = 6

 

cool cool cool

CPhill  Mar 20, 2024
 #3
avatar+13 
0

That is not correct.

 Mar 20, 2024
 #4
avatar+128794 
0

https://web2.0calc.com/questions/for-a-certain-value-of-nbsp-k-nbsp-the-system_1

 

 

cool cool cool

 Mar 20, 2024

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