Calculating Determinants: Using Laplace Expansion or Echelon Form?

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In summary, the determinant of the matrix $b$ is: \begin{align*}\det b&=&\begin{vmatrix}2 & 1 & -2 & 1 & 7 & 3 \\ 3 & 4 & 1 & 9 & -1 & 2 \\ 0 & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 2 & 1 & 6 \\ 0 & 0 & 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 3\end{vmatrix}
  • #1
mathmari
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Hey! :eek:

I want to calculate the determinants of the matrices $a=\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{pmatrix}$ and $b=\begin{pmatrix}2 & 1 & -2 & 1 & 7 & 3 \\ 3 & 4 & 1 & 9 & -1 & 2 \\ 0 & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 2 & 1 & 6 \\ 0 & 0 & 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 3\end{pmatrix}$

For that do we use the Laplace expansion theorem or can we transform these matrices firstly in echelon form and calculate the determinant then? But is the determinant of the initial matrix equal to the determinant of the matrix in echelon form? (Wondering)
 
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  • #2
Hey mathmari!

We can do both. (Nod)

If we choose the echelon form, there are a couple of special rules for determinants though.
1. We can add a multiple of a row to another.
2. If we swap 2 rows, then the sign of the determinant swaps as well.
3. If we multiply a row by a factor, then the determinant is multiplied by that factor as well.
4. We can do the same with columns instead of rows.
(Thinking)
 
  • #3
Klaas van Aarsen said:
We can do both. (Nod)

If we choose the echelon form, there are a couple of special rules for determinants though.
1. We can add a multiple of a row to another.
2. If we swap 2 rows, then the sign of the determinant swaps as well.
3. If we multiply a row by a factor, then the determinant is multiplied by that factor as well.
4. We can do the same with columns instead of rows.
(Thinking)

Ok! Which of the two ways do you suggest to use in this case? (Wondering)
 
  • #4
mathmari said:
Ok! Which of the two ways do you suggest to use in this case? (Wondering)

A combination! (Happy)

Start with working towards echelon form or some such until Laplace expansion becomes easy enough.

If you reach echelon form, the Laplace expansion is just the product of the elements on the diagonal. (Thinking)
 
  • #5
Klaas van Aarsen said:
If we choose the echelon form, there are a couple of special rules for determinants though.
1. We can add a multiple of a row to another.

How does the determinant change in this case? Do we multiply also the determinant by this factor? Or doesn't the determinant change by using this rule? (Wondering)
 
  • #6
mathmari said:
How does the determinant change in this case? Do we multiply also the determinant by this factor? Or doesn't the determinant change by using this rule?

The determinant does not change if we add a multiple of a row to a different row. (Emo)
 
  • #7
Klaas van Aarsen said:
The determinant does not change if we add a multiple of a row to a different row. (Emo)

Ok! So, as for the matrix $b$ we have the following:
\begin{align*}\det b=&\begin{vmatrix}2 & 1 & -2 & 1 & 7 & 3 \\ 3 & 4 & 1 & 9 & -1 & 2 \\ 0 & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 2 & 1 & 6 \\ 0 & 0 & 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 3\end{vmatrix} \ \overset{R_2:R_2-\frac{3}{2}\cdot R_1}{=} \ \begin{vmatrix}2 & 1 & -2 & 1 & 7 & 3 \\ 0 & \frac{5}{2} & 4 & \frac{15}{2} & -\frac{23}{2} & -\frac{5}{2} \\ 0 & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 2 & 1 & 6 \\ 0 & 0 & 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 3\end{vmatrix} \\ & \ \overset{R_4:R_4-R_3}{=} \ \begin{vmatrix}2 & 1 & -2 & 1 & 7 & 3 \\ 0 & \frac{5}{2} & 4 & \frac{15}{2} & -\frac{23}{2} & -\frac{5}{2} \\ 0 & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 6 \\ 0 & 0 & 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 3\end{vmatrix} \ \overset{R_5:R_5-R_4}{=} \ \begin{vmatrix}2 & 1 & -2 & 1 & 7 & 3 \\ 0 & \frac{5}{2} & 4 & \frac{15}{2} & -\frac{23}{2} & -\frac{5}{2} \\ 0 & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 6 \\ 0 & 0 & 0 & 0 & 1 & -7 \\ 0 & 0 & 0 & 0 & 0 & 3\end{vmatrix} \\ & =2\cdot \frac{5}{2}\cdot 1\cdot 1\cdot 1\cdot 3=15
\end{align*}
Right? (Wondering)
 
  • #8
Yep. (Nod)
 
  • #9
Klaas van Aarsen said:
Yep. (Nod)

And as for matrix $a$ we have:
\begin{align*}\det a&=\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_2:R_2-2\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix}\\ & \ \overset{R_3:R_3-3\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & -2 & -4 & -11 & -13 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_4:R_4-4\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & -2 & -4 & -11 & -13 \\ 0 & -3 & -11 & -14 & -17 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \\ & \ \overset{R_5:R_5-5\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & -2 & -4 & -11 & -13 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \ \overset{R_3:R_3-2\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & 0 & 0 & -5 & -5 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \\ & \ \overset{R_4:R_4-3\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & 0 & 0 & -5 & -5 \\ 0 & 0 & -2 & -5 & -5 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \ \overset{R_5:R_5-9\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & 0 & 0 & -5 & -5 \\ 0 & 0 & -2 & -5 & -5 \\ 0 & 0 & 5 & 10 & 15\end{vmatrix} \\ & \ \overset{R_5:R_5+\frac{5}{2}\cdot R_4}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & 0 & 0 & -5 & -5 \\ 0 & 0 & -2 & -5 & -5 \\ 0 & 0 & 0 & -\frac{5}{2} & \frac{5}{2}\end{vmatrix} \ \overset{R_5:R_5-\frac{1}{2}\cdot R_3}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & 0 & 0 & -5 & -5 \\ 0 & 0 & -2 & -5 & -5 \\ 0 & 0 & 0 & 0 & 5\end{vmatrix} \\ & \ \overset{R_3\leftrightarrow R_4}{=} \ -\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & 0 & -2 & -5 & -5 \\ 0 & 0 & 0 & -5 & -5 \\ 0 & 0 & 0 & 0 & 5\end{vmatrix}=-1\cdot (-1)\cdot (-2)\cdot (-5)\cdot 5=50
\end{align*}

Is this correct? And is this the way you meant or is there a better/faster way if we use also the Laplace expansion? (Wondering)
 
  • #10
mathmari said:
And as for matrix $a$ we have:
\begin{align*}\det a&=\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_2:R_2-2\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix}
\end{align*}

Shouldn't that be
\begin{align*}\det a&=\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_2:R_2-2\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & {\color{red}\mathbf{-\enclose{circle}{9}}} \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix}
\end{align*}
(Worried)

mathmari said:
And is this the way you meant or is there a better/faster way if we use also the Laplace expansion?

This is the way yes. I think it is the fastest way to calculate this determinant. (Thinking)
 
  • #11
Klaas van Aarsen said:
Shouldn't that be
\begin{align*}\det a&=\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_2:R_2-2\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & {\color{red}\mathbf{-\enclose{circle}{9}}} \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix}
\end{align*}
(Worried)

Now I get
\begin{align*}\det a&=\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_2:R_2-2\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix}\\ & \ \overset{R_3:R_3-3\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & -2 & -4 & -11 & -13 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_4:R_4-4\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & -2 & -4 & -11 & -13 \\ 0 & -3 & -11 & -14 & -17 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \\ & \ \overset{R_5:R_5-5\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & -2 & -4 & -11 & -13 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \ \overset{R_3:R_3-2\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \\ & \ \overset{R_4:R_4-3\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -2 & -5 & 10 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \ \overset{R_5:R_5-9\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -2 & -5 & 10 \\ 0 & 0 & 5 & 10 & 60\end{vmatrix} \\ & \ \overset{R_5:R_5+\frac{5}{2}\cdot R_4}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -2 & -5 & 10 \\ 0 & 0 & 0 & -\frac{5}{2} & 85\end{vmatrix} \ \overset{R_5:R_5-\frac{1}{2}\cdot R_3}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -2 & -5 & 10 \\ 0 & 0 & 0 & 0 & \frac{165}{2}\end{vmatrix} \\ & \ \overset{R_3\leftrightarrow R_4}{=} \ -\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & -2 & -5 & 10 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & 0 & 0 & \frac{165}{2}\end{vmatrix}=-1\cdot (-1)\cdot (-2)\cdot (-5)\cdot \frac{165}{2}=825
\end{align*}

But according to Wolfram the result is $1875$ but I don't see where I could have done something wrong. Do you maybe have an idea? (Wondering)
 
  • #12
mathmari said:
\begin{align*}\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} & \ \overset{R_4:R_4-3\cdot R_2}{=} \
\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -2 & -5 & 10 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix}\end{align*}

But according to Wolfram the result is $1875$ but I don't see where I could have done something wrong. Do you maybe have an idea? (Wondering)

Shouldn't it be:
\begin{align*}\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} & \ \overset{R_4:R_4-3\cdot R_2}{=} \
\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & {\color{red}-\enclose{circle}{5}} & -5 & 10 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix}\end{align*}
(Worried)
 
  • #13
Klaas van Aarsen said:
Shouldn't it be:
\begin{align*}\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} & \ \overset{R_4:R_4-3\cdot R_2}{=} \
\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & {\color{red}-\enclose{circle}{5}} & -5 & 10 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix}\end{align*}
(Worried)

Oh yes! So it must be as follows:
\begin{align*}\det a&=\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_2:R_2-2\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix}\\ & \ \overset{R_3:R_3-3\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & -2 & -4 & -11 & -13 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_4:R_4-4\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & -2 & -4 & -11 & -13 \\ 0 & -3 & -11 & -14 & -17 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \\ & \ \overset{R_5:R_5-5\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & -2 & -4 & -11 & -13 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \ \overset{R_3:R_3-2\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \\ & \ \overset{R_4:R_4-3\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -5 & -5 & 10 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \ \overset{R_5:R_5-9\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -5 & -5 & 10 \\ 0 & 0 & 5 & 10 & 60\end{vmatrix} \\ & \ \overset{R_5:R_5+R_4}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -5 & -5 & 10 \\ 0 & 0 & 0 & 5 & 70\end{vmatrix} \ \overset{R_5:R_5+ R_3}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -5 & -5 & 10 \\ 0 & 0 & 0 & 0 & 75\end{vmatrix} \\ & \ \overset{R_3\leftrightarrow R_4}{=} \ -\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & -5 & -5 & 10 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & 0 & 0 & 75\end{vmatrix}=-1\cdot (-1)\cdot (-5)\cdot (-5)\cdot 75=1875
\end{align*}

Now the result is correct (Malthe)
 
  • #14
mathmari said:
Now the result is correct (Malthe)

Nice! (Cake)
 

FAQ: Calculating Determinants: Using Laplace Expansion or Echelon Form?

What is the purpose of calculating the determinant?

The determinant is a mathematical tool used to determine various properties of a matrix, such as whether it is invertible or singular. It is also used in solving systems of linear equations and finding the area or volume of a parallelogram or parallelepiped defined by the matrix.

How is the determinant calculated?

The determinant of a matrix can be calculated using various methods, such as cofactor expansion, row reduction, or using the properties of determinants. The method used depends on the size and type of the matrix.

What are the properties of determinants?

Some of the key properties of determinants include:

  • The determinant of a matrix and its transpose are equal.
  • Multiplying a row or column of a matrix by a scalar multiplies the determinant by the same scalar.
  • Swapping two rows or columns of a matrix changes the sign of the determinant.
  • The determinant of a diagonal matrix is the product of its diagonal elements.
  • If a matrix has a row or column of all zeros, the determinant is zero.

Can determinants be negative?

Yes, the determinant of a matrix can be positive, negative, or zero. The sign of the determinant depends on the number of row/column swaps needed to transform the matrix into its reduced row echelon form.

In what real-world applications are determinants used?

Determinants have various applications in fields such as physics, engineering, and economics. They are used in solving systems of linear equations, calculating the stability of a system, determining the area or volume of a shape, and in finding the eigenvalues and eigenvectors of a matrix.

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