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I have to calculate the determinant of this matrice. I want to use the rule of sarrus, but this does only work with a $3\times3$ matrice: $$ A= \begin{bmatrix} 1 & -2 & -6 & u \\ -3 & 1 & 2 & -5 \\ 4 & 0 & -4 & 3 \\ 6 & 0 & 1 & 8 \\ \end{bmatrix} $$ $|A|=35$

Any idea how to solve this?

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4 Answers 4

up vote 2 down vote accepted

$$A=\begin{pmatrix}1 & -2 & -6 & u \\-3 & 1 & 2 & -5 \\4 & 0 & -4 & 3 \\6 & 0 & 1 & 8 \\\end{pmatrix}\stackrel{ 3R_1+R_2}{\stackrel{-4R_1+R_3}{\stackrel{-6R_1+R_4}\longrightarrow}}\begin{pmatrix}1 & -2 & \;\;-6 & \;\;\;u \\0 & -5 & -16 & \;\;\,3u-5 \\0 & \;\;\;8 & \;\;\;20 &-4u+ 3 \\0 & \;\,12 & \;\;\;37 & -6u+8 \\\end{pmatrix}$$

Now develop the above wrt the first column and you get a $\,3\times 3\,$ determinant. Compute now directly or repeat the above process.

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what do you mean "develope by first colume"? Would you be so kind to explain that further? –  Le Chifre Jan 16 '13 at 12:33
    
By minors, @maximus. Read it here mathworld.wolfram.com/Determinant.html –  DonAntonio Jan 16 '13 at 13:19

(1) Expand the determinant with respect to the second column and you will end up with two $3\times3$ determinants.

Or first add two times the second row to the first and then do (1).

Then use the rule of Sarrus.

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You can write

$$\det{A} = 2 \left| \begin{array}{ccc} -3 & 2 & -5 \\ 4 & -4 & 3 \\ 6 & 1 & 8 \end{array} \right| + 1 \left| \begin{array}{ccc} 1 & -6 & u \\ 4 & -4 & 3 \\ 6 & 1 & 8 \end{array} \right| $$

which you should easily be able to expand to get a simple linear equation for $u$, which comes out to a nice whole number.

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Use Leibniz formula or the Laplace expansion. For more an Laplace expansion see youtube video .

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