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April 15th, 2017, 07:53 AM   #1
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Determinant of 6*6 matrix

I am looking for the determinant of the following matrix.

$$
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 2 \\
0 & 1 & 0 & 0 & 2 & 0 \\
0 & 0 & 1 & 2 & 0 & 0 \\
0 & 0 & 2 & 1 & 0 & 0 \\
0 & 2 & 0 & 0 & 1 & 0 \\
2 & 0 & 0 & 0 & 0 & 1 \\
\end{bmatrix}
$$


Is there any trick to find the determinant of this matrix?If yes, please help me out.
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April 15th, 2017, 08:10 AM   #2
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Determinant Calculator
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April 15th, 2017, 10:01 PM   #3
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let your matrix be $A$

$A x =\begin{pmatrix} x_1+2 x_6\\x_2+2 x_5\\x_3+2 x_4\\2 x_3+x_4\\2 x_2+x_5\\2 x_1+x_6 \end{pmatrix}$

we notice that rows 1 and 6, 2 and 5, and 3 and 4 share the same variables.

Now suppose we have $Ax = \lambda x$

so for example using rows 1 and 6 we see

$x_1 + 2 x_6 = \lambda x_1$
$2x_1 + x_6 = \lambda x_6$

this results in eigenvalues $\lambda = 3,~\lambda = -1$

The equations are identical for the other pairs of variables.

So we end up with a set of eigenvalues $(3,3,3,-1,-1,-1)$

and the determinant is just the product of all these which is

$Det(A)=-27$
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April 16th, 2017, 12:21 PM   #4
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it occurs to me there's a cleaner way to look at this.

Do a row swap of rows 2 and 6 and then a column swap of columns 2 and 6.

These 2 elementary operations cancel out each others effect on the determinant with zero net effect.

After the swaps we have a block diagonal matrix $B$ with 3 blocks

$b=\begin{pmatrix}1 &2 \\2 &1 \end{pmatrix}$

$Det[b]=1-4 = -3$

The determinant of the block diagonal matrix will be the product of the determinants of the blocks

$Det[B] = (-3)^3 = -27$
Thanks from shashank dwivedi

Last edited by romsek; April 16th, 2017 at 12:41 PM.
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