# Finding the determinant of a $3 \times 3$ matrix via Laplace Expansion

I have a matrix here where I need to calculate the determinant using Laplace expansion.

$$\begin{pmatrix} 4 & 0 & 1\\19 & 1 & -3\\7 & 1 & 0 \end{pmatrix}$$

So I did the simple diagonal & antidiagonal way to calculate and get

\begin{align*} |A| &= 0 + 0 + 19-7-0-(-12) \\&=24 \end{align*}

Right now, I have to show that, by Laplace expansion, and am having a mistake:

\begin{align*} |A| &= 4\begin{vmatrix}1 & -3\\1 & 0\end{vmatrix}+ 0\begin{vmatrix}19 & -3\\7 & 0\end{vmatrix}+ 1\begin{vmatrix}19 & 1\\7 & 1\end{vmatrix}\\ &= 4(-3) + 0 + 1(12)\\ &= 0 \end{align*}

Could someone help and spot my errors here?

UPDATE: The 2nd last row should be $3$, not $-3$, as shown in the answer.

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it should be $4 \cdot (-(-3))$ since $$\begin{vmatrix} a & b\\c&d\\\end{vmatrix}=ad-bc$$ –  Dominic Michaelis Feb 17 '13 at 16:37
On the second last line, you computed the determinant as -3 instead of 3. –  copper.hat Feb 17 '13 at 16:39
And (though it doesn't matter here), the middle term should be $\color {red}- 0\cdot\Bigl|{19\atop 7}\ {-3\atop 0}\Bigr|$. –  David Mitra Feb 17 '13 at 16:42
@DavidMitra can I say that the $+/-$ depends on the relative position of the particular element (scalar outside)? From top-left, take positions down & right, add up all the positions and take $-1$ raised to that value? –  bryansis2010 Feb 17 '13 at 16:45
$$\left|\begin{array}{cc}1 & -3\\1 & 0\end{array}\right|=(1\cdot 0)-(1\cdot-3)=0-(-3)=3$$