Determinant simplification

Nov 2009
58
0
Hi,

Can anyone see a way of getting from A to B?

A
\(\displaystyle

\begin{bmatrix}
1 & z & z+1 \\
z+1 & 1 & z \\
z & z+1 & 1
\end{bmatrix}
\)

B
\(\displaystyle
2*
\begin{bmatrix}
1 & 0 & z \\
z & 1 & 0 \\
0 & z & 1
\end{bmatrix}
\)



So far the best I can do is:
\(\displaystyle
2*
\begin{bmatrix}
1 & 0 & z \\
z & 1 & z-1 \\
0 & z & 1
\end{bmatrix}
\)

Thanks guys
 
Mar 2010
116
41
Bratislava
Hi,

Can anyone see a way of getting from A to B?

A
\(\displaystyle

\begin{bmatrix}
1 & z & z+1 \\
z+1 & 1 & z \\
z & z+1 & 1
\end{bmatrix}
\)

B
\(\displaystyle
2*
\begin{bmatrix}
1 & 0 & z \\
z & 1 & 0 \\
0 & z & 1
\end{bmatrix}
\)



So far the best I can do is:
\(\displaystyle
2*
\begin{bmatrix}
1 & 0 & z \\
z & 1 & z-1 \\
0 & z & 1
\end{bmatrix}
\)

Thanks guys

\(\displaystyle
\begin{bmatrix}
1 & z & z+1 \\
z+1 & 1 & z \\
z & z+1 & 1
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & z \\
z+1 & 1 & z \\
z & z+1 & 1
\end{bmatrix}
+
\begin{bmatrix}
0 & z & 1 \\
z+1 & 1 & z \\
z & z+1 & 1
\end{bmatrix}
=
\)
\(\displaystyle
\begin{bmatrix}
1 & 0 & z \\
z & 1 & 0 \\
z & z+1 & 1
\end{bmatrix}
+
\begin{bmatrix}
1 & 0 & z \\
1 & 0 & z \\
z & z+1 & 1
\end{bmatrix}
+
\begin{bmatrix}
0 & z & 1 \\
1 & 0 & z \\
z & z+1 & 1
\end{bmatrix}
+
\begin{bmatrix}
0 & z & 1 \\
z & 1 & 0 \\
z & z+1 & 1
\end{bmatrix}
\)
\(\displaystyle
=
\begin{bmatrix}
1 & 0 & z \\
z & 1 & 0 \\
z & z+1 & 1
\end{bmatrix}
+
\begin{bmatrix}
0 & z & 1 \\
1 & 0 & z \\
z & z+1 & 1
\end{bmatrix}
\)
\(\displaystyle
=
\begin{bmatrix}
1 & 0 & z \\
z & 1 & 0 \\
0 & z & 1
\end{bmatrix}
+
\begin{bmatrix}
0 & z & 1 \\
1 & 0 & z \\
z & 1 & 0
\end{bmatrix}
\)
\(\displaystyle
=
\begin{bmatrix}
1 & 0 & z \\
z & 1 & 0 \\
0 & z & 1
\end{bmatrix}
+
\begin{bmatrix}
1 & 0 & z \\
z & 1 & 0 \\
0 & z & 1
\end{bmatrix}
\)

In the first step I used the property, that if two matrices have all lines with the exception of one of them equal, then the sum of their determinants is the determinant of the matrix that has the sum of these two lines instead of this line (and the remaining lines are the same).

Except for this property, all the remaining things I have used seems to be standard - determinant doesn't change when I subtract one line from another, switching two lines change the sign, if one of the lines is linear combination of the remaining ones, then the value of determinant is zero.

I hope there is not a typo somewhere.
 

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, aceband!

Can anyone see a way of getting from \(\displaystyle A\) to \(\displaystyle B\)?

\(\displaystyle A \;=\;\begin{bmatrix} 1 & z & z+1 \\ z+1 & 1 & z \\ z & z+1 & 1 \end{bmatrix}\)

\(\displaystyle B \;=\;2\cdot \begin{bmatrix} 1 & 0 & z \\ z & 1 & 0 \\ 0 & z & 1 \end{bmatrix} \)
I used standard row operations . . .


\(\displaystyle \text{Given: }\;\begin{bmatrix} 1 & z & z+1 \\ z+1 & 1 & z \\ z & z+1 & 1 \end{bmatrix}\)


\(\displaystyle \begin{array}{c} \\ R_2-R_1 \\ \\ \end{array}\begin{bmatrix}1&z&z+1 \\ z&1-z&\text{-}1 \\ z&z+1& 1 \end{bmatrix}\)


\(\displaystyle \begin{array}{c}\\ \\ R_3-R_2\end{array}\begin{bmatrix}1 & z&z+1 \\ z&1-z&\text{-}1 \\ 0 & 2z & 2 \end{bmatrix} \)


\(\displaystyle \text{Factor: }\;2\!\cdot\!\begin{bmatrix}1&z&z+1 \\ z&1-z&\text{-}1 \\ 0&z&1\end{bmatrix}\)


\(\displaystyle \begin{array}{c}R_1-R_3 \\ \\ \\ \end{array}\;2\!\cdot\!\begin{bmatrix}1&0&z \\ z & 1-z & \text{-}1 \\ 0&z&1\end{bmatrix}\)


\(\displaystyle \begin{array}{c} \\ R_2+R_3 \\ \\ \end{array}\quad2\!\cdot\! \begin{bmatrix}1&0&z \\ z&1&0 \\ 0&z&1\end{bmatrix}\)

 
Nov 2009
58
0
Wow, did not know you could do that! I think that'll be one of those techniques i never forget now! Thank you so much.