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