## Weird failure of Cramers Rule

This is very weird, but I found an inconsistency in the application of Cramer's Rule for a 3x3 simple linear matrix.

1x + 1y + 0z = 3
-1x + 3y + 4z = -3
0x + 4y + 3z = 2

Dz =
1 1 3
-1 3 -3
0 4 2

If you take the determinant across the first row To find Dz, I constantly get -16

If you take the determinant across any other rows or columns, you get the correct Dz = 8

What is going on?????????

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 Not really going to be able to help without seeing a step by step calculation. I get 8 no matter which minor I choose to expand by.
 Omg! Sorry... I calculated incorrectly by forgetting a negative. My mistake was 1(6-12) It should have been 1(6-(-12)) Thanks

## Weird failure of Cramers Rule

 Quote by epsilonOri This is very weird, but I found an inconsistency in the application of Cramer's Rule for a 3x3 simple linear matrix. 1x + 1y + 0z = 3 -1x + 3y + 4z = -3 0x + 4y + 3z = 2 Dz = 1 1 3 -1 3 -3 0 4 2 If you take the determinant across the first row To find Dz, I constantly get -16 If you take the determinant across any other rows or columns, you get the correct Dz = 8 What is going on????????? Help please.

What do you mean by "to take the determinant across a row? Do you mean to calculate it wrt the
minors determined by that row? Let's see:
$$\left|\begin{array}{rrr}1&1&3\\-1&3&-3\\0&4&2\end{array}\right|= 1\cdot\left|\begin{array}{rr}\,3&-3\\\,4&2\end{array}\right|+(-1)\cdot\left|\begin{array}{rr}-1&-3\\0&2\end{array}\right|+3\cdot\left|\begin{array}{rr}-1&3\\0&4\end{array}\right|=(6+12)-(-2)+3(-4)=18+2-12=8$$

If you meant the above then the result is 8, which is hardly surprising as this is the matrix's determinant ; if you

meant something else then I can't say.

DonAntonio

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