Answer:
Option C
Step-by-step explanation:
We are given a coefficient matrix along and not the solution matrix
Since solution matrix is not given we cannot check for infinity solutions.
But we can check whether coefficient matrix is 0 or not
If coefficient matrix is zero, the system is inconsistent and hence no solution.
Option A)
|A|=[tex]\left[\begin{array}{ccc}4&2&6\\2&1&3\\-2&3&-4\end{array}\right] =0[/tex]
since II row is a multiple of I row
Hence no solution or infinite
OPtion B
|B|=[tex]\left[\begin{array}{ccc}2&0&-2\\-7&1&5\\4&-2&0\end{array}\right] \\=2(10)-2(10)=0[/tex]
Hence no solution or infinite
Option C
[tex]\left[\begin{array}{ccc}6&0&-2\\-2&0&6\\1&-2&0\end{array}\right] \\=2(36-2)=68[/tex]
Hence there will be a unique solution
Option D
[tex]\left[\begin{array}{ccc}5&10&5\\4&1&4\\-1&-2&-1\end{array}\right] \\=2(10)-2(10)=0[/tex]=0
(since I row is -5 times III row)
Hence there will be no or infinite solution
Option C is the correct answer
