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April 15th, 2017, 07:53 AM  #1 
Newbie Joined: Apr 2017 From: India Posts: 20 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: 2,640 Thanks: 1319  
April 15th, 2017, 10:01 PM  #3 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,511 Thanks: 761 
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 $Det(A)=27$ 
April 16th, 2017, 12:21 PM  #4 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,511 Thanks: 761 
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 $b=\begin{pmatrix}1 &2 \\2 &1 \end{pmatrix}$ $Det[b]=14 = 3$ The determinant of the block diagonal matrix will be the product of the determinants of the blocks $Det[B] = (3)^3 = 27$ Last edited by romsek; April 16th, 2017 at 12:41 PM. 

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