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$$D=\begin{bmatrix} 246 & 427 & 327 \\ 1014 & 543 & 443 \\ -342 & 721 & 621 \\ \end{bmatrix}$$

What's the trick?

Hints?

Of course I know calculate by definition...

Please don't do that by cheating.

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3  
What do you mean by trick? What is wrong with using the definition? –  Amzoti Aug 29 '13 at 12:52
    
@Amzoti This is an exercise of the text, I wonder whether I should do that by definition. because they're not small numbers. –  User19912312 Aug 29 '13 at 12:53
    
But the dimension is small; just use a calculator to multiply. –  rfauffar Aug 29 '13 at 12:53
2  
Subtract column 3 from column 2 –  Gerry Myerson Aug 29 '13 at 12:54
1  
One simplification would be to subtract the 3rd column (@ZevChonoles, did you mean column?) from the 2nd and pull out $100$ common giving you a smaller determinant. –  Parth Thakkar Aug 29 '13 at 13:02

4 Answers 4

up vote 5 down vote accepted

First subtract the third column from the second as suggested by Gerry Myerson; this has no effect on the determinant.

$$\left[\begin{array}{ccc} 246 & 100 & 327\\ 1014 & 100 & 443\\ -342 & 100 & 621 \end{array}\right]$$

Now divide the first column by $2$ and the second by $100$. These operations multiply the determinant by $\frac{1}{2}$ and $\frac{1}{100}$ respectively.

$$\left[\begin{array}{ccc} 123 & 1 & 327\\ 507 & 1 & 443\\ -171 & 1 & 621 \end{array}\right]$$

Now add the first column to the third. This has no effect on the determinant.

$$\left[\begin{array}{ccc} 123 & 1 & 500\\ 507 & 1 & 1000\\ -171 & 1 & 450 \end{array}\right]$$

Now divide the third column by $50$. This multiplies the determinant by $\frac{1}{50}$.

$$\left[\begin{array}{ccc} 123 & 1 & 10\\ 507 & 1 & 20\\ -171 & 1 & 9 \end{array}\right]$$

Now substract the third row from the first and second. These have no effect on the determinant.

$$\left[\begin{array}{ccc} 294 & 0 & 1\\ 681 & 0 & 11\\ -171 & 1 & 9 \end{array}\right]$$

Now divide the first column by $3$. This multiplies the determinant by $\frac{1}{3}$.

$$\left[\begin{array}{ccc} 98 & 0 & 1\\ 227 & 0 & 11\\ -57 & 1 & 9 \end{array}\right]$$

If you expand along the second column, you should find that the computations are simple enough. To obtain the determinant of the original matrix, undo all of the effects of the column operations.

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Thanks, after check the solution of the book. I found that there is still a trick after the collect of 100. I calculate 3 times to come across the simple answer. –  User19912312 Aug 29 '13 at 13:45

To simplify the calculation, you can start by subtracting column three from column 2. That will not change the value of the determinant, but will simplify its calculation. Then subtract the first row from the other two, and you can then, essentially, compute the value of the determinant of a $2\times 2$ matrix by expansion of the minor using the appropriate column.

In problems like this (particularly when looking to simplify the calculation of determinants of even larger matrices, we can use elementary row/column operations can to simplify the calculation of the determinant, but recall if and how each operation impacts the value of the original determinant.

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Thanks, there is still one trick to go further, see my answer. –  User19912312 Aug 29 '13 at 13:53

Thanks, everyone, after check the solution of the book(have a glance). I found that there is still a trick after the collect of 100.

I calculate 3 times to come across this simple answer.

\begin{align*} d &= \left|\begin{array}{ccc} 246 & 427 & 327 \\ 1014 & 543 & 443 \\ -342 & 721 & 621 \\\end{array}\right| \\ &= \left|\begin{array}{ccc} 246 & 100 & 327 \\ 1014 & 100 & 443 \\ -342 & 100 & 621 \\\end{array}\right| \\ &= \left|\begin{array}{ccc} 246 & 100 & 327 \\ 768 & 0 & 116 \\ -588 & 0 & 294 \\\end{array}\right| \\ &=200 \left|\begin{array}{ccc} 123 & 1 & 327 \\ 384 & 0 & 116 \\ -294 & 0 & 294 \\\end{array}\right| \\ &=200 \left|\begin{array}{ccc} 123 & 1 & 327 \\ 384 & 0 & 500 \\ -294 & 0 & 0 \\\end{array}\right| \\ &=200 \times -294\times 500 \\ &=-29400000 \end{align*}

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I neatened this answer up a bit, and also corrected the arithmetic. –  Douglas S. Stones Aug 29 '13 at 14:58
    
There's still one mistake (although it doesn't affect the answer). I assume we get to the last matrix by adding column 1 to column 3, but if that's the case then the 327 should become 450. –  Gerry Myerson Aug 29 '13 at 23:26

Sarrus rule:

$$\det(A) = \begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix} = a_{11}a_{22}a_{33} + a_{21}a_{32}a_{13} + a_{31}a_{12}a_{23} -a_{11}a_{32}a_{23} - a_{21}a_{12}a_{33} - a_{31}a_{22}a_{13}$$

Look for example here: http://www.math.utah.edu/~gustafso/determinants.pdf

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8  
I doubt that's what OP wants. –  Gerry Myerson Aug 29 '13 at 12:57
2  
Please don't post link-only answers (see this discussion). Take screenshots or copy down the information yourself into the post, so that your answer will still make sense if external content is removed. –  Zev Chonoles Aug 29 '13 at 12:58
    
at G.M.: The answer was before the comments. at Z.C.: Okay, I wasn't aware of this. Thank you. –  plusepsilon.de Aug 29 '13 at 13:08
    
@ZevChonoles Thanks, there is still one trick to go further, see my answer. –  User19912312 Aug 29 '13 at 13:53

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