My Math Forum Determinant of 6*6 matrix

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 April 15th, 2017, 07:53 AM #1 Member   Joined: Apr 2017 From: India Posts: 73 Thanks: 0 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.
 April 15th, 2017, 08:10 AM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1588
 April 15th, 2017, 10:01 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 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 $Det(A)=-27$ Thanks from Country Boy and shashank dwivedi
 April 16th, 2017, 12:21 PM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 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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how to find the determinant of a 6 by 6 matrix

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