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

[karush]

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..

 

[topsquark]

$\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

 

[karush]

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

 

[topsquark]

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:
\(\displaystyle det \left [ \begin{matrix} a & b \\ c & d \end{matrix} \right ] = \left | \begin{matrix} a & b \\ c & d \end{matrix} \right |\)

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
\(\displaystyle det \left [ \begin{matrix} 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{matrix} \right ] = - det \left [ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right ] \)
And continue from here.

-Dan

 

[karush]

Do we really need to do all these co factor if the original matrix can become rref

 

[HOI]

In fact, the last three columns and rows are just the identity matrix so this determinant can quickly be determined as $$\left|\begin{array}{cc}0 & 1 \\ 1 & 0 \end{array}\right|= 0(0)- 1(1)= -1$$.