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For a certain value of k, the system has no solutions. What is the value of k x + y + 3z = 10, -4x + 2y + 5z = 7, kx + z = 3

 Jul 23, 2019
 #1
avatar+111393 
+2

kx + y + 3z  = 10      multiply through by -2 ⇒  -2kx - 2y - 6z  =  -20      (1)

-4x + 2y + 5z  = 7   (2)

kx + z  = 3     (3)

 

Add (1) and (2)   and we have that

 

-(2k + 4)x  - z  =  -13     multiply through by -1    ⇒    (2k + 4)x  + z =  13      (4)

 

Note that   if      (2k + 4)  =  k,  then the system will have no solutions

 

So

 

2k + 4  = k

k  = - 4

 

 

cool cool cool

 Jul 23, 2019
 #2
avatar+25237 
+1

For a certain value of k, the system has no solutions. What is the value of

\(k x + y + 3z = 10,\\ -4x + 2y + 5z = 7,\\ kx + z = 3 \)

There is no solution, if the determinant \(\begin{vmatrix} k&1&3\\-4&2&5\\k&0&1\end{vmatrix} = 0\)

 

\(\begin{array}{|rcll|} \hline \begin{vmatrix} k&1&3\\-4&2&5\\k&0&1\end{vmatrix} &=& k\begin{vmatrix} 1&3\\2&5\end{vmatrix}+\begin{vmatrix} k&1\\ -4&2\end{vmatrix} \\ &=& k(5-6)+3k+4 \\ &=& -k+2k+4\\ &=& k+4 \quad &| \quad \begin{vmatrix} k&1&3\\-4&2&5\\k&0&1\end{vmatrix} = 0 \\ 0 &=& k+4\\ \mathbf{k} &=& \mathbf{-4} \\ \hline \end{array} \)

 

 

laugh

 Jul 24, 2019

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