Respuesta :
Answer:
If k = −1 then the system has no solutions.
If k = 2 then the system has infinitely many solutions.
The system cannot have unique solution.
Step-by-step explanation:
We have the following system of equations
[tex]x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2[/tex]
The augmented matrix is
[tex]\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right][/tex]
The reduction of this matrix to row-echelon form is outlined below.
[tex]R_2\rightarrow R_2-R_1[/tex]
[tex]\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right][/tex]
[tex]R_3\rightarrow R_3-2R_1[/tex]
[tex]\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right][/tex]
[tex]R_3\rightarrow R_3-R_2[/tex]
[tex]\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right][/tex]
The last row determines, if there are solutions or not. To be consistent, we must have k such that
[tex]k^2-k-2=0[/tex]
[tex]\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2[/tex]
Case k = −1:
[tex]\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right][/tex]
If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.
Case k = 2:
[tex]\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right][/tex]
This gives the infinite many solution.
