Thread: Determinant of a 5 by 5 matrix

1. Determinant of a 5 by 5 matrix

I've been asked to: use determinant properties to evaluate the 5 x 5 determinant given below.

$\begin{bmatrix}1 & 3 & 1 & -6 & 8 \\ -1 & 2 & -1 & 6 & -8 \\ 0 & 0 & 4 & 8 & -4 \\ 2 & 6 & 0 & -16 & 18 \\ 0 & 0 & 1 & 2 & 2\end{bmatrix}$

I've only just started learning matrix algebra, but I was aiming to reduce this matrix to a triangle matrix so I could use the main diagonal to calculate the determinant. I have a couple of questions though:

1 - Does this method count as "using determinant properties"? Or should I be doing it differently in order to answer the question correctly?
2 - Is my working below (and resulting determinant) correct?

WORKING STEPS:

1 - Add the first row to the second.

$\begin{bmatrix}1 & 3 & 1 & -6 & 8 \\ 0 & 5 & 0 & 0 & 0 \\ 0 & 0 & 4 & 8 & -4 \\ 2 & 6 & 0 & -16 & 18 \\ 0 & 0 & 1 & 2 & 2\end{bmatrix}$

2 - Subtract 2x the first row from the fourth.

$\begin{bmatrix}1 & 3 & 1 & -6 & 8 \\ 0 & 5 & 0 & 0 & 0 \\ 0 & 0 & 4 & 8 & -4 \\ 0 & 0 & -2 & -28 & 2 \\ 0 & 0 & 1 & 2 & 2\end{bmatrix}$

3 - Add 2x the fifth row to the fourth.

$\begin{bmatrix}1 & 3 & 1 & -6 & 8 \\ 0 & 5 & 0 & 0 & 0 \\ 0 & 0 & 4 & 8 & -4 \\ 0 & 0 & 0 & -24 & 6 \\ 0 & 0 & 1 & 2 & 2\end{bmatrix}$

4 - Subtract 1/4x the third row from the fifth.

$\begin{bmatrix}1 & 3 & 1 & -6 & 8 \\ 0 & 5 & 0 & 0 & 0 \\ 0 & 0 & 4 & 8 & -4 \\ 0 & 0 & 0 & -24 & 6 \\ 0 & 0 & 0 & 0 & 3\end{bmatrix}$

Because this is now a triangle matrix, the determinant should just be equal to the multiple of the main diagonal right? i.e. 1 x 5 x 4 x -24 x 3 = -1440

I have a strong feeling that this is wrong, and possibly that the 0 5 0 0 0 row is more significant than I realise. Any comments/help would be greatly appreciated.

2. You are doing it correctly, you have just made an algebra mistake in step 2. That -28 should be a -4 you accidentally subtracted 12 when you should have added 12.

This is definitely the method you should use by adding scalar multiples of one row to another whenever possible, but you should note that it is also possible to swap rows (you must keep track of these steps because when you interchange two rows it changes the sign of the determinant) and you can multiply a row by a scalar (this multiplies the determinant by that scalar).

3. Ah, yes, you're right. Thanks. So the resulting matrix would then be:

$\begin{bmatrix}1 & 3 & 1 & -6 & 8 \\ 0 & 5 & 0 & 0 & 0 \\ 0 & 0 & 4 & 8 & -4 \\ 0 & 0 & 0 & 0 & 6 \\ 0 & 0 & 0 & 0 & 3\end{bmatrix}$

Does this mean the determinant of the original matrix is 0 (because the main diagonal is 1 x 5 x 4 x 0 x 3)? I attempted some alternate reductions and I can't see any that would result in a non-zero value for that row vector.

4. I think actually after that step where the mistake was made you can see that row 3 and row 4 are off by a multiple of -2, so you should get one of those two rows to 0 out completely showing you the determinant is 0.

I just put it into the calculator and it verifies that the determinant is indeed 0. As long as you have correctly entered the original matrix 0 should be your answer.

The determinant is unique to a matrix, so as long as you are following the correct rules, it doesn't matter which steps you took to row reduce it, you will get the same answer.

5. $\left| \begin{matrix}
1 & 3 & 1 & -6 & 8 \\
-1 & 2 & -1 & 6 & -8 \\
0 & 0 & 4 & 8 & -4 \\
2 & 6 & 0 & -16 & 18 \\
0 & 0 & 1 & 2 & 2
\end{matrix} \right|=\left| \begin{matrix}
1 & 3 & 1 & -8 & 6 \\
-1 & 2 & -1 & 8 & -6 \\
0 & 0 & 4 & 0 & -8 \\
2 & 6 & 0 & -16 & 18 \\
0 & 0 & 1 & 0 & 0
\end{matrix} \right|,$

thus your determinant equals $\left| \begin{matrix}
1 & 3 & -8 & 6 \\
-1 & 2 & 8 & -6 \\
0 & 0 & 0 & -8 \\
2 & 6 & -16 & 18
\end{matrix} \right|=16\left| \begin{matrix}
1 & 3 & -8 \\
-1 & 2 & 8 \\
1 & 3 & -8
\end{matrix} \right|=0.$

6. Try 'Easy Matrix Calculator'

It's a very useful matrix calculator.
See more in Easy Matrix Calculator

7. In that case I'd use Wolfram|Alpha.

But this doesn't make any sense.

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how to solve 5×5 determinant?

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