How to find the determinant of this 5×5 matrix? I can't put it in Lower or Upper Triangular form so I'm confused. I dont really know how to use laplace expansion
$\begin{bmatrix}3&0&0&3&0\\-3&0&-2&0&0\\0&-1&0&0&-3\\0&0&0&3&3\\0&-1&2&0&0\end{bmatrix}$
I tried to get all 0's in the diagonal and i then deleted row 1 column 1 so that i now have a 4×4 matrix so i did
$R_2 = R_1 + R_2$ then I did $R_4 = R_2 – R_4$ then I got $3 \begin{bmatrix}0&-2&3&0\\-1&0&0&-3\\0&-2&0&-3\\-1&2&0&0\end{bmatrix}$ = $3(-1)(-2)(2)(-1) + 3(-2)(3)(-3)(-3) = -498 $ but i did $R_2 – R_4$ so i divided by a factor of -1 and got $\det(A) = -498/-1 = 498$ which was still incorrect
Best Answer
ERO Method:
$$\begin{align}\begin{vmatrix}3&0&0&3&0\\-3&0&-2&0&0\\0&-1&0&0&-3\\0&0&0&3&3\\0&-1&2&0&0\end{vmatrix} &=\begin{vmatrix}3&0&0&3&0\\0&0&-2&3&0\\0&-1&0&0&-3\\0&0&0&3&3\\0&-1&2&0&0\end{vmatrix} \tag{$R_2\to R_2+R_1$}\\ &=\color{red}{-}\begin{vmatrix}3&0&0&3&0\\0&-1&0&0&-3\\0&0&-2&3&0\\0&0&0&3&3\\0&-1&2&0&0\end{vmatrix} \tag{$R_3\leftrightarrow R_2$} \\ &=-\begin{vmatrix}3&0&0&3&0\\0&-1&0&0&-3\\0&0&-2&3&0\\0&0&0&3&3\\0&0&2&0&3\end{vmatrix} \tag{$R_5\to R_5-R_2$} \\ &=-\begin{vmatrix}3&0&0&3&0\\0&-1&0&0&-3\\0&0&-2&3&0\\0&0&0&3&3\\0&0&0&3&3\end{vmatrix} \tag{$R_5\to R_5+R_3$} \\ &=-\begin{vmatrix}3&0&0&3&0\\0&-1&0&0&-3\\0&0&-2&3&0\\0&0&0&3&3\\0&0&0&0&0\end{vmatrix} \tag{$R_5\to R_5-R_4$} \\ &=0\end{align}$$
Laplace Expansion Method:
Each row and column has $2$ entries so I don't see any strategic way to choose where to expand this. So I'll just go across the first row first: $$\begin{align}\begin{vmatrix}3&0&0&3&0\\-3&0&-2&0&0\\0&-1&0&0&-3\\0&0&0&3&3\\0&-1&2&0&0\end{vmatrix} &= 3\begin{vmatrix}0&-2&0&0\\-1&0&0&-3\\0&0&3&3\\-1&2&0&0\end{vmatrix} + -3\begin{vmatrix}-3&0&-2&0\\0&-1&0&-3\\0&0&0&3\\0&-1&2&0\end{vmatrix}\end{align}$$
Now I'll expand along the first row of the first and first column of the second:
$$=3\left[-(-2)\begin{vmatrix}-1 & 0 & -3 \\ 0 & 3 & 3 \\ -1 & 0 & 0\end{vmatrix}\right]+-3\left[-3\begin{vmatrix}-1 & 0 & -3 \\ 0 & 0 & 3 \\ -1 & 2 & 0\end{vmatrix}\right]$$
From here expand along the third and second rows and evaluate the $2\times 2$ determinant in the normal way to get $$=6[-(0--9)]+9[-3(-2-0)] = 0$$
Exterior Product Method:
$$\begin{align}&(3e_1-3e_2)\wedge(-e_3-e_5)\wedge(-2e_2+2e_5)\wedge(3e_1+3e_4)\wedge (-3e_3+3e_4) \\ &= \big[3e_1\wedge(-e_3-e_5)\wedge(-2e_2+2e_5)\wedge(3e_1+3e_4)\wedge (-3e_3+3e_4)\big] \\ &\quad -\big[3e_2\wedge(-e_3-e_5)\wedge(-2e_2+2e_5)\wedge(3e_1+3e_4)\wedge (-3e_3+3e_4)\big] \\ &= \big[3e_1\wedge(-e_3-e_5)\wedge(-2e_2+2e_5)\wedge(3e_4)\wedge (-3e_3+3e_4)\big] \\&\quad -\big[3e_2\wedge(-e_3-e_5)\wedge(2e_5)\wedge(3e_1+3e_4)\wedge (-3e_3+3e_4)\big] \\ &= -9\big[e_1\wedge e_3\wedge(-2e_2+2e_5)\wedge e_4\wedge (-3e_3+3e_4)\big] \\ &\quad -9\big[e_1\wedge e_5\wedge(-2e_2+2e_5)\wedge e_4\wedge (-3e_3+3e_4)\big] \\ &\quad +6\big[e_2\wedge e_3\wedge e_5\wedge(3e_1+3e_4)\wedge (-3e_3+3e_4)\big]\\ &\quad +6\big[e_1\wedge e_5\wedge e_5\wedge(3e_1+3e_4)\wedge (-3e_3+3e_4)\big] \\ &= -9\big[0\big]-9\big[e_1\wedge e_5\wedge(-2e_2)\wedge e_4\wedge (-3e_3)\big] +6\big[e_2\wedge e_3\wedge e_5\wedge(3e_1+3e_4)\wedge (3e_4)\big]+6\big[0\big] \\ &= (-9\cdot 6)I+(6\cdot9)I \\ &=0I\end{align}$$
Happen-to-Notice-Linear-Dependence Method:
Notice that $R_1 = -R_2+R_3+R_4-R_5$. So because the rows are not linearly independent, the determinant must be $0$.