Answer:
Option C
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|=
![\left[\begin{array}{ccc}4&2&6\\2&1&3\\-2&3&-4\end{array}\right] =0](https://img.qammunity.org/2020/formulas/mathematics/high-school/ipmixxwj5wswsk8nzvkdxkridhd8yw27wz.png)
since II row is a multiple of I row
Hence no solution or infinite
OPtion B
|B|=
![\left[\begin{array}{ccc}2&0&-2\\-7&1&5\\4&-2&0\end{array}\right] \\=2(10)-2(10)=0](https://img.qammunity.org/2020/formulas/mathematics/high-school/yzso4vd2amy2nl6pz9nmub40dq52i26m3c.png)
Hence no solution or infinite
Option C
![\left[\begin{array}{ccc}6&0&-2\\-2&0&6\\1&-2&0\end{array}\right] \\=2(36-2)=68](https://img.qammunity.org/2020/formulas/mathematics/high-school/p39lbu6vlqigc0xpq4y2zjytr2ramgbfaj.png)
Hence there will be a unique solution
Option D
=0
(since I row is -5 times III row)
Hence there will be no or infinite solution
Option C is the correct answer