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I tried finding the determinant of A using the co-factors technique, yet always get it wrong, This is my approach:

A = $$ \begin{matrix} 3 & 1 & 0 \\ -2 & -4 & 3 \\ 5 & 4 & -2 \\ \end{matrix} $$

B = $$ \begin{matrix} -4 & 3 \\ 4 & -2 \\ \end{matrix} $$

C = $$ \begin{matrix} -2 & 3 \\ 5 & -2 \\ \end{matrix} $$

D = $$ \begin{matrix} -2 & -4 \\ 5 & 4 \\ \end{matrix} $$

|A|= (3)* |A|+ (1)|B| + (0)|D| |A| = (3)(-4) + (-11) |A| = -21

Which is wrong, I looked up the solution, it is -1

sorry for my bad writing, I am new to the website.

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    $\begingroup$ You forgot that the signs alternate. $\endgroup$ – amd Nov 28 '19 at 19:11
  • $\begingroup$ det(B)=-4 det(C)= -11 det(D)= 12. 3*(-4) -1*(-11) + 0*12 = -1. Remeber the pattern goes like this: + - + - + etc. So you need a “minus-sign” in front of your “1”. $\endgroup$ – Carl Nov 28 '19 at 19:12
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The formula is somewhat wrong:

$${\mathrm{det}}(A) = 3 \cdot {\mathrm{det}}(B) + (-1) \cdot 1 \cdot {\mathrm{det}}(C) + 0 \cdot {\mathrm{det}}(D) = -1$$.

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  • $\begingroup$ My text book says the formula is: |A| = $$\sum_{k=1}^n (a_{kj})*(A_{kj}) =$$ $\endgroup$ – Anttarm Nov 28 '19 at 19:30
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    $\begingroup$ In that case $j$ is fixed. Which means you may expand through any column. Similar principle work for expansion through rows. Here you are expanding through first row: $$\sum_{k=1}^{n} (a_{1k}) \ast (A_{1k})$$. Also this means $A_{1k}$ include the sign. $\endgroup$ – Siddhartha Nov 28 '19 at 19:35
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    $\begingroup$ @Anttarm To put it another way, $A_{kj}$ in that formula isn’t simply the determinant of the submatrix with that row and column deleted. It’s the cofactor, which includes a factor of $(-1)^{k+j}$. $\endgroup$ – amd Nov 28 '19 at 19:49
  • $\begingroup$ This is the best ahaha moment that I had in a while $\endgroup$ – Anttarm Nov 28 '19 at 20:02

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