Finding the determinant of the 5x5 matrix but can't put it in lower triangular form

Question

How to find the determinant of this 5x5 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 4x4 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

Just do elementary row operations, using your knowledge of how each one changes the value of the determinant. — , Jul 04 '16 at 18:04
Your matrix is somewhat sparse, why not just calculate the determinant by definition? — Aweygan, Jul 04 '16 at 18:05
@Bye_World You're implying that i put it in either Lower Triang or Upper Triang form correct? — Shammy, Jul 04 '16 at 18:07
@Shammy Basically yes. I'm saying put it in REF form (keeping track of how the determinant changes) -- which just happens to be a triangular matrix. — , Jul 04 '16 at 18:08
@Bye_World I tried to get it in LTF but I can't because I get left with a number that I can't get rid of! — Shammy, Jul 04 '16 at 18:12
$\begin{bmatrix}3&0&0&3&0\0&0&-2&3&0\0&-1&0&0&-3\0&0&0&3&3\0&-1&2&0&0\end{bmatrix}$ — Shammy, Jul 04 '16 at 18:13
@shammy Are you still interested in applying laplace expansion ? — callculus42, Jul 04 '16 at 18:20
ill reward with a upvote but i gave byeworld the answer check mark@callculus — Shammy, Jul 04 '16 at 18:23
Remember that swapping rows (which changes the sign of the determinant) is one of the tools in your kit. See this question for another example of a $5\times 5$ determinant. — amd, Jul 04 '16 at 18:46
@Shammy I posted an answer. If you have any question feel free to ask. — callculus42, Jul 04 '16 at 18:55

score 2 · Accepted Answer · 2016-07-07T01:19:24.797

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 $0$.

god i can never figure out how u guys see things so fast i can never see. i never ever would of thought to swap them -_- — Shammy, Jul 04 '16 at 18:17
If you have a question about any of the above methods, just let me know. — , Jul 04 '16 at 18:58

score 1 · Answer 2 · answered Jul 04 '16 at 18:53

laplace expansion

I go on at your intermediate result.

$3\begin{bmatrix}0&\color{red}{-2}&\color{red}{3}&\color{red}{0}\\ \color{blue}{-1}&0&0&-3\\ 0&\color{red}{-2}&\color{red}{0}&\color{red}{-3}\\ \color{green}{-1}&\color{red}{2}&\color{red}{0}&\color{red}{0}\end{bmatrix}$

$\color{blue}{-1}$ is positioned in the second row and and in the first column. Thus $(-1)*(-1)^{2+1}=1$ The correspoonding submatrix consist the red numbers.

Similar calculation for the green $\color{green}{-1}$. The sign is positive as well, because $(-1)*(-1)^{4+1}=1$

Thus $det \ A= 3\cdot 1 \cdot \begin{bmatrix}\color{red}{-2}&\color{red}{3}&\color{red}{0}\\ \color{red}{-2}&\color{red}{0}&\color{red}{-3}\\ \color{red}{2}&\color{red}{0}&\color{red}{0}\end{bmatrix}+3\cdot 1\cdot \begin{bmatrix}-2&3&0\\ 0&0&-3\\ -2&0&-3\end{bmatrix}$

$=3\cdot 1\cdot (0-18+0-0-0-0)+3\cdot 1\cdot (0+18+0-0-0-0)=-54+54=0$

Finding the determinant of the 5x5 matrix but can't put it in lower triangular form

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