Question

Find the determinant of the following matrix.

Answer 1

One of the methods you can use is co-factor expansion, or Gaussian elimination.

Answer 2

Well, I would think #2 is faster.
Have you ever solved for a determinant this way (using Gaussian elimination)?

Answer 3

nope

Answer 4

Just as a timeout, you do know what Gaussian elimination is, though, am I right?

Answer 5

I will assume you do, unless you say otherwise ... so I'll proceed.
(Sorry if this appeared to be any-what insulting)

Answer 6

Suppose I want to find the determinant of the following matrix.
((Notation for a determinant of a matrix \(M\) is \(\left|M\right|\).))
\(\color{blue}{\displaystyle M=\left[\begin{matrix}1 & 2 & 2\\ 4 & 2 & 1\\ 3 & 5 & 2\end{matrix}\right]~}\)

This is what you want to find:
\(\color{blue}{\displaystyle \left|M\right|=\left|\begin{matrix}1 & 2 & 2\\ 4 & 2 & 1\\ 3 & 5 & 2\end{matrix}\right|~}\)

Rules:
1) When you switch any two rows, you multiply the determinant by -1.
2) Subtracting/Adding a rows does NOT have any effect on the determinant.
3) When you multiply a row by \(c\), divide the determinant by \(c\) (multiply by \(1/c\)).

Answer 7

And, you start from setting the determinant equal to 1, and then preform your steps.

Answer 8

Step 1:

So far, \(\color{blue}{\displaystyle \left|M\right|=1}\)

Multiply the first row by -4 (thus, also multiply the determinant by -1/4).
\(\color{blue}{\displaystyle \left|\begin{matrix}1 & 2 & 2\\ 4 & 2 & 1\\ 3 & 5 & 2\end{matrix}\right|= \left|\begin{matrix}-4 & -8 & -8\\ 4 & 2 & 1\\ 3 & 5 & 2\end{matrix}\right|~}\)

Then, your determinant is: \(\color{blue}{\displaystyle \left|M\right|=1\times (-1/4)}\).

Step 2: \(\rm Row_2-Row_1=\rm {new}~Row_2\)

Subtract \(\rm Row_1\) from \(\rm Row_2\), (this doesn't change the determinant).

\(\color{blue}{\displaystyle \left|\begin{matrix}-4 & -8 & -8\\ 4 & 2 & 1\\ 3 & 5 & 2\end{matrix}\right|= \left|\begin{matrix}-4 & -8 & -8\\ 0 & -6 & -7\\ 3 & 5 & 2\end{matrix}\right|~}\)

Your determinant is still: \(\color{blue}{\displaystyle \left|M\right|=1\times (-1/4)}\). (unchanged).

Answer 9

there we go that last one you posted made sense

Answer 10

Step 3: \(\rm Row_1\times {3/4}\) (Then, \(\rm Determinant\times {1/(3/4)=Determinant\times4/3}\))

\(\color{blue}{\displaystyle \left|\begin{matrix}-4 & -8 & -8\\ 0 & -6 & -7\\ 3 & 5 & 2\end{matrix}\right|= ~\left|\begin{matrix}-3 & -6 & -6\\ 0 & -6 & -7\\ 3 & 5 & 2\end{matrix}\right|}\)

Your determinant is still: \(\color{blue}{\displaystyle \left|M\right|=1\times (-1/4)\times (4/3)}\).

Answer 11

I am copy pasting often.
In the last post:

The determinant IS CHANGED, (and I said "is still" ... a copy typo, apologize)

Answer 12

Step 3: \(\rm Row_3-Row_1=~{new}~~Row_3\)
(determinant is unchanged)

\(\color{blue}{\displaystyle \left|\begin{matrix}-4 & -8 & -8\\ 0 & -6 & -7\\ 3 & 5 & 2\end{matrix}\right|= ~\left|\begin{matrix}-3 & -6 & -6\\ 0 & -6 & -7\\ 0 & -1 & -4\end{matrix}\right|}\)

Your determinant is still: \(\color{blue}{\displaystyle \left|M\right|=1\times (-1/4)\times (4/3)}\). (This time, it really does not change.)

Answer 13

Step 4: \(\rm Row_1 \times (-1)\)

\(\color{blue}{\displaystyle \left|\begin{matrix}-4 & -8 & -8\\ 0 & -6 & -7\\ 3 & 5 & 2\end{matrix}\right|= ~\left|\begin{matrix}-3 & -6 & -6\\ 0 & -6 & -7\\ 0 & 1 & 4\end{matrix}\right|}\)

Your determinant is then: \(\color{blue}{\displaystyle \left|M\right|=1\times (-1/4)\times (4/3)\times (-1)}\).

Answer 14

oh my

Answer 15

Step 4: \(\rm Row_1\times (-1/3)\)

\(\color{blue}{\displaystyle \left|\begin{matrix}1 & 2 & 2\\ 0 & 6 & 7\\ 0 & 1 & 4\end{matrix}\right|}\)

Your determinant is then: \(\color{blue}{\displaystyle \left|M\right|=1\times (-1/4)\times (4/3)\times (-1)\times (-3)}\).

Answer 16

do you see the technique?

Answer 17

I didn't finish the problem, I will do it now shorty ...

Answer 18

i suppose i get it

Answer 19

\(\color{blue}{\displaystyle \left|\begin{matrix}1 & 2 & 2\\ 0 & 1 & 4\\ 0 & 6 & 7\end{matrix}\right|}\)

Your determinant is then: \(\color{blue}{\displaystyle \left|M\right|=1\times (-1/4)\times (4/3)\times (-1)\times (-3)\times (-1)}\).
(This time I multiplied by -1, b/c I switched rows)

Note, (to simplify): \(\color{blue}{\displaystyle \left|M\right|=1}\) (So far).

Answer 20

\(\color{blue}{\displaystyle \left|\begin{matrix}1 & 2 & 2\\ 0 & -2 & -8\\ 0 & 6 & 7\end{matrix}\right|}\)

Your determinant is then: \(\color{blue}{\displaystyle \left|M\right|=1\times (-1/2)}\)
(Multiplied Row1 by -2, thus multiply determinant by -1/2.)

Answer 21

\(\color{blue}{\displaystyle \left|\begin{matrix}1 & 0 & -6\\ 0 & -2 & -8\\ 0 & 6 & 7\end{matrix}\right|}\)

Your determinant is still: \(\color{blue}{\displaystyle \left|M\right|=-1/2}\)
(adding/subtracting rows doesn't change the determinant)

Answer 22

\(\color{blue}{\displaystyle \left|\begin{matrix}1 & 0 & -6\\ 0 & -6 & -24\\ 0 & 6 & 7\end{matrix}\right|}\)

Your determinant is then: \(\color{blue}{\displaystyle \left|M\right|=-1/6}\)
( [Row1] × 3, thus determinant ÷3)

Answer 23

\(\color{blue}{\displaystyle \left|\begin{matrix}1 & 0 & -6\\ 0 & -6 & -24\\ 0 & 0 & -17\end{matrix}\right|}\)

Your determinant is still: \(\color{blue}{\displaystyle \left|M\right|=-1/6}\)
(adding/subtracting rows doesn't change the determinant)

Answer 24

Oh, two steps ago, I said [Row1]×3, SORRY, this is really [Row\(\color{red}{2}\)]×3.

Answer 25

\(\color{blue}{\displaystyle \left|\begin{matrix}1 & 0 & -6\\ 0 & 1 & 4\\ 0 & 0 & -4\end{matrix}\right|}\)

Your determinant is still: \(\color{blue}{\displaystyle \left|M\right|=17/4}\)
( [Row3] × 4/17, thus determinant × (17/4) )

Answer 26

\(\color{blue}{\displaystyle \left|\begin{matrix}1 & 0 & -6\\ 0 & 1 & 0\\ 0 & 0 & -4\end{matrix}\right|}\)

adding rows doesn't change the determinant so,
\(\color{blue}{\displaystyle \left|M\right|=17/4 }\) (still)

Answer 27

\(\color{blue}{\displaystyle \left|\begin{matrix}1 & 0 & -6\\ 0 & 1 & 0\\ 0 & 0 & 6\end{matrix}\right|}\)

\(\color{blue}{\displaystyle \left|M\right|=17/6 }\)
( [Row3] × -3/2, thus determinant × (-2/3) )

Answer 28

Oh, that should say |M|=-17/6

Answer 29

\(\color{blue}{\displaystyle \left|\begin{matrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 6\end{matrix}\right|}\)

\(\color{blue}{\displaystyle \left|M\right|=-17/6 }\)
(adding rows doesn't change the determinant)

Answer 30

\(\color{blue}{\displaystyle \left|\begin{matrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{matrix}\right|}\)

\(\color{blue}{\displaystyle \left|M\right|=-17 }\)
([Row3] × 1/6, thus determinant × 6)

Answer 31

Let me check my steps to make sure I am done
(Wow this is indeed long to write out)

Answer 32

yeah it did lol

Answer 33

Yes, there is an error.
Where I said, \(|M|=1\), and I simplified the result, it should be \(-1\), because I switched rows and didn't in fact account for this, despite that I explained that I should (account for switching rows by multiplying |M| times -1).

Answer 34

Everything else is right.

Answer 35

basically, it is 17.

Answer 36

Once you master it, you won't have to write too much out.

Answer 37

I will do another example in a faster fashion (and hopefully without mistakes or typos).
You will note that I am skipping steps (as opposed to an official layout), and that I am accounting for the RULES I mentioned in the beginning.

Answer 38

kay

Answer 39

Oh, just in case it comes across, do you know what a Transpose of a matrix is?

Answer 40

(Denoted as \(A^T\) for a matrix \(A\).
I am asking this, because there is a rule: \(|A^T|=|A|\).)

Answer 41

fraid not

Answer 42

\(\color{blue}{\displaystyle \left|\begin{matrix}-4 & 4 & 1\\ 7 & -2 & 1\\ 2 & 1 & 1\end{matrix}\right|~}\) \(\color{blue}{\displaystyle M=1}\)

\(\color{red}{\Huge \text{_____________________}}\)

\(\color{blue}{\displaystyle \left|\begin{matrix}-4 & 4 & 1\\ 7 & -2 & 1\\0 & 6 & 5\end{matrix}\right|~}\) \(\color{blue}{\displaystyle M=1/2}\)
(Multiplied ROW1 times 2, and added ROW1 to ROW3.)
\(\color{red}{\Huge \text{_____________________}}\)

\(\color{blue}{\displaystyle \left|\begin{matrix}-4 & 4 & 1\\ 0 & 20/7 & 11/7\\0 & 6 & 5\end{matrix}\right|~}\) \(\color{blue}{\displaystyle M=7/8}\)
(Multiplied ROW2 times 4/7, and added ROW1 to ROW2.)
\(\color{red}{\Huge \text{_____________________}}\)

\(\color{blue}{\displaystyle \left|\begin{matrix}-4 & 0 & -6/5\\ 0 & 20/7 & 11/7\\0 & 6 & 5\end{matrix}\right|~}\) \(\color{blue}{\displaystyle M=-5/8}\)
(Multiplied ROW2 times -7/5, and added ROW2 to ROW1.)
\(\color{red}{\Huge \text{_____________________}}\)

\(\color{blue}{\displaystyle \left|\begin{matrix}-4 & 0 & -6/5\\ 0 & -4 & -11/5\\0 & 6 & 5\end{matrix}\right|~}\) \(\color{blue}{\displaystyle M=-5/8}\)
(Multiplied ROW2 times -7/5, and added ROW2 to ROW1.)
\(\color{red}{\Huge \text{_____________________}}\)

\(\color{blue}{\displaystyle \left|\begin{matrix}-4 & 0 & -6/5\\ 0 & -6 & -33/10\\0 & 0 & 17/10\end{matrix}\right|~}\) \(\color{blue}{\displaystyle M=-5/12}\)
(Multiplied ROW2 times 3/2, and added ROW2 to ROW3.)
\(\color{red}{\Huge \text{_____________________}}\)

\(\color{blue}{\displaystyle \left|\begin{matrix}-4 & 0 & -6/5\\ 0 & -6 & -33/10\\0 & 0 & 1\end{matrix}\right|~}\) \(\color{blue}{\displaystyle M=-17/24}\)
(Multiplied ROW3 times 10/17.)
\(\color{red}{\Huge \text{_____________________}}\)

\(\color{blue}{\displaystyle \left|\begin{matrix}-4 & 0 & 0\\ 0 & -6 & 0\\0 & 0 & 1\end{matrix}\right|~}\) \(\color{blue}{\displaystyle M=-17/24}\)
(I multiply row 3 times 33/10, add ROW 3 to ROW 2, and then divide back by 33/10, so the determinant is not affected overall. Then I do the same for ROW1. I multiply ROW3 times 5/6, add ROW3 to ROW2, and then divide by 5/6. Then again my determinant is not affected. And the matrix is simplified.)
\(\color{red}{\Huge \text{_____________________}}\)

\(\color{blue}{\displaystyle \left|\begin{matrix}1 & 0 & 0\\ 0 & 1 & 0\\0 & 0 & 1\end{matrix}\right|~}\) \(\color{blue}{\displaystyle M=(-17/24)(-6)(-4)=-17.}\)
(Divided rows to make the identity matrixes, and adjusted for the determiantn with the rules.)

Answer 43

I would advise at first, though, to be careful.

Answer 44

Don't worry about the transpose.
For transpose you just "flip" the matrix, so that the rows are columns and the columns are rows. (That has its own use as regards to linear algebra.)

Answer 45

oh lord

Answer 46

You can use Cofactor expansion (if you don't like the previous method - Gaussian elimination, or if your teacher ever requires you to... (I might be out of time if I try to post this method as thoroughly as the previous one, but I believe it should be online on some websites that don't overwhelm with scholarliness)

Basically, with your example, Cofactor expansion is:
\(\small \color{blue}{\displaystyle \left|\begin{matrix}2 & -3 & 4\\ 5 & 1 & -2\\ 10 & 3 & -1\end{matrix}\right|=2\cdot\left|\begin{matrix} 1 & -2\\ 3 & -1\end{matrix}\right|+(\color{red}{-}5)\cdot\left|\begin{matrix} -3 & 4\\ 3 & -1\end{matrix}\right|+10\cdot\left|\begin{matrix} -3 & 4\\ 1 & -2\end{matrix}\right|}\)

Or, if you choose a column (as opposed to choosing a row)
\(\small \color{blue}{\displaystyle \left|\begin{matrix}2 & -3 & 4\\ 5 & 1 & -2\\ 10 & 3 & -1\end{matrix}\right|=2\cdot\left|\begin{matrix} 1 & -2\\ 3 & -1\end{matrix}\right|+(\color{red}{-}(-3))\cdot\left|\begin{matrix} 5 & -2\\ 10 & -1\end{matrix}\right|+4\cdot\left|\begin{matrix} 5 & -2\\ 10 & -1\end{matrix}\right|}\)

Answer 47

be carefull though ... if you for example choose ROW2, your entry is
Column\(\color{red}{1}\) \(\color{blue}{\times }\) Row\(\color{red}{2}\) = \(\color{red}{1}\color{blue}{\times }\color{red}{2}=\color{red}{2}\) <--- even entry,
so you would start from a negative, and then alternate.