MHB What is the determinant of the identity matrix in a larger matrix?

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karush
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Suppose A is a $5\times5$ matrix such that det(A)=2. Let
$$E=\left[\begin{array}{c}0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array}\right]$$
then
$\left[\begin{array}{c}0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array}\right]=
\left[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array}\right]
=-\left[\begin{array}{c}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]
=-\det\left[\begin{array}{c}1 & 0 \\ 0 & 1 \\ \end{array}\right]=-(1\cdot1)-(0 \cdot 0)=-1$
so
$det(AE)=(2)(-1)=-2$

ok W|A retured $\det(E)=-1$

But I got confused on the signs and assummed things
saw no need to demonstrate the 0 co factors
suggestions are welcome here..
 
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karush said:
$\left[\begin{array}{c}0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array}\right]=
\left[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array}\right]
=-\left[\begin{array}{c}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]
=-\det\left[\begin{array}{c}1 & 0 \\ 0 & 1 \\ \end{array}\right] $
How can these possibly be equal?? The three matrices aren't the same size and the determinant is a number!

-Dan
 
topsquark said:
How can these possibly be equal?? The three matrices aren't the same size and the determinant is a number!

-Dan

They are co factors multipled by 1
 
karush said:
They are co factors multipled by 1
Cofactors are matrices but they are not the same as the matrix you started with! Please get your notations correct. We can write a determinant in two ways:
[math]det \left [ \begin{matrix} a & b \\ c & d \end{matrix} \right ] = \left | \begin{matrix} a & b \\ c & d \end{matrix} \right |[/math]

What you wrote was neither.

The determinant of the 5 x 5 matrix is -1. You are good right up to your last step. Expanding in cofactors as you did shows that the determinant is indeed -1. That is to say
[math]det \left [ \begin{matrix} 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{matrix} \right ] = - det \left [ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right ] [/math]
And continue from here.

-Dan
 
Do we really need to do all these co factor if the original matrix can become rref
 
In fact, the last three columns and rows are just the identity matrix so this determinant can quickly be determined as \left|\begin{array}{cc}0 & 1 \\ 1 & 0 \end{array}\right|= 0(0)- 1(1)= -1.
 

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