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$ det A= \left(\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array}\right) =12$

$A= \left(\begin{array}{ccc} 3a & 3b & 3c \\ d+2a & e+2b & f+2c \\ -\frac{1}{2}g & -\frac{1}{2}h& -\frac{1}{2}i \end{array}\right).$

so according to determinant property, multiply first row with 3, $-\frac{1}{2}$times row three, and add 2 times first row to second row,but it wont change the determinant. so the new determinant will be $3\cdot \left(-\frac{1}{2}\right)\cdot 12$. Is my assumption true?

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  • $\begingroup$ Looks okay to me. $\endgroup$ – Jacky Chong Nov 29 '17 at 6:47
  • $\begingroup$ You should label the new matrix by $B$ (not $A$). $\endgroup$ – farruhota Nov 29 '17 at 7:07
  • $\begingroup$ @Vixf If you are ok, you can set as solved. Thanks! $\endgroup$ – user Dec 3 '17 at 11:50
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yes it's a good reasoning way, it looks ok

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