# Determinant simplification

• May 22nd 2010, 02:13 AM
aceband
Determinant simplification
Hi,

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

A
$

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

B
$
2*
\begin{bmatrix}
1 & 0 & z \\
z & 1 & 0 \\
0 & z & 1
\end{bmatrix}
$

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

Thanks guys
• May 22nd 2010, 02:37 AM
kompik
Quote:

Originally Posted by aceband
Hi,

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

A
$

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

B
$
2*
\begin{bmatrix}
1 & 0 & z \\
z & 1 & 0 \\
0 & z & 1
\end{bmatrix}
$

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

Thanks guys

$
\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}
=
$

$
\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}
$

$
=
\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}
$

$
=
\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}
$

$
=
\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.
• May 22nd 2010, 02:42 AM
kompik
Perhaps here Determinant as Sum of Determinants - ProofWiki you can find a better explanation of the result I have used in the first step.
• May 22nd 2010, 04:19 AM
Soroban
Hello, aceband!

Quote:

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

$A \;=\;\begin{bmatrix} 1 & z & z+1 \\ z+1 & 1 & z \\ z & z+1 & 1 \end{bmatrix}$

$B \;=\;2\cdot \begin{bmatrix} 1 & 0 & z \\ z & 1 & 0 \\ 0 & z & 1 \end{bmatrix}$

I used standard row operations . . .

$\text{Given: }\;\begin{bmatrix} 1 & z & z+1 \\ z+1 & 1 & z \\ z & z+1 & 1 \end{bmatrix}$

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

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

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

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

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

• May 22nd 2010, 05:11 AM
aceband
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.
• May 22nd 2010, 05:25 AM
kompik
Quote:

Originally Posted by aceband
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.

If you mean the result about the sum of determinants, it is often used in the proofs of results about effect of elementary row operations on the value of determinant, like here Multiple of Row Added to Row of Determinant - ProofWiki