Exercise 2.1.55 ( Congruences and determinant)

Question

Let $A = [a_{ij}]$ and $B = [b_{ij}]$ be two $n	imes n $ matrices with integral entries. Show that if $a_{ij} \equiv b_{ij} \pmod m$ for all $i,j$, then $\det(A) \equiv \det(B) \pmod m)$. Show that
$$
\det
\begin{pmatrix}
4771 & 1452 & 8404 & 3275 & 9163\
6573 & 8056 & 7312 & 2265 & 3639\
9712 & 2574 & 4612 & 4321 & 7196\
8154 & 2701 & 6007 & 2147 & 7465\
2158 & 7602 & 5995 & 2327 & 8882
\end{pmatrix}

e 0.
$$
Hint:  Find a small modulus $m$ for which the given determinant is $
ot \equiv 0 \pmod m$.

Richard Ganaye · Accepted Answer

ewcommand{\F}{\mathbb{F}}

\begin{proof} If $a_{ij} \equiv b_{ij} \pmod m$ for all $i,j$, then $a_{\sigma(i),i} \equiv b_{\sigma(i),i}$ for all $i \in [\![1,n]\!]$ and all permutations $\sigma \in S_n$. Therefore 
$$\det(A) = \sum_{\sigma \in S_n} \varepsilon(\sigma) \prod_{i=1}^n a_{\sigma(i),i} \equiv \sum_{\sigma \in S_n} \varepsilon(\sigma) \prod_{i=1}^n b_{\sigma(i),i} = \det(B) \pmod m.$$
Take $$A = \begin{pmatrix}
4771 & 1452 & 8404 & 3275 & 9163\
6573 & 8056 & 7312 & 2265 & 3639\
9712 & 2574 & 4612 & 4321 & 7196\
8154 & 2701 & 6007 & 2147 & 7465\
2158 & 7602 & 5995 & 2327 & 8882
\end{pmatrix}$$
This matrix is not invertible modulo $2,3,5$, thus we reduce $A$ modulo $7$. The class of $\det(A)$ in the field $\F_7$ is, using Gauss' reduction,
\begin{align*}
\det\overline{A} = 
\left|\begin{array}{rrrrr}
4 & 3 & 4 & 6 & 0 \
0 & 6 & 4 & 4 & 6 \
3 & 5 & 6 & 2 & 0 \
6 & 6 & 1 & 5 & 3 \
2 & 0 & 3 & 3 & 6
\end{array}ight|
&=\phantom{-} +
\left|\begin{array}{rrrrr}
4 & 0 & 0 & 0 & 0 \
0 & 6 & 4 & 4 & 6 \
3 & 1 & 3 & 1 & 0 \
6 & 5 & 2 & 3 & 3 \
2 & 2 & 1 & 0 & 6
\end{array}ight| \quad (C_2\leftarrow C_2 + C_1, C_3 \leftarrow C_3 - C_1, C_4\leftarrow C_4+2C_1)\
&=\phantom{-} +
\left|\begin{array}{rrrrr}
4 & 0 & 0 & 0 & 0 \
0 & 6 & 0 & 0 & 0 \
3 & 1 & 0 & 5 & 6 \
6 & 5 & 1 & 2 & 5 \
2 & 2 & 2 & 1 & 4
\end{array}ight|\quad (C_3 \leftarrow C_3 +4C_2, C_4 \leftarrow C_4+4C_2,C_5 \leftarrow C_5 - C_2)\
&=-\left|\begin{array}{rrrrr}
4 & 0 & 0 & 0 & 0 \
0 & 6 & 0 & 0 & 0 \
3 & 1 & 5 & 0 & 6 \
6 & 5 & 2 & 1 & 5 \
2 & 2 & 1 & 2 & 4
\end{array}ight| \quad (C_3 \leftrightarrow C_4)\
&= 
-\left|\begin{array}{rrrrr}
4 & 0 & 0 & 0 & 0 \
0 & 6 & 0 & 0 & 0 \
3 & 1 & 5 & 0 & 0 \
6 & 5 & 2 & 1 & 4 \
2 & 2 & 1 & 2 & 0
\end{array}ight| \quad (C_5\leftarrow C_5 + 3 C_3)\
&=
-\left|\begin{array}{rrrrr}
4 & 0 & 0 & 0 & 0 \
0 & 6 & 0 & 0 & 0 \
3 & 1 & 5 & 0 & 0 \
6 & 5 & 2 & 1 & 0 \
2 & 2 & 1 & 2 & 6
\end{array}ight|  \quad (C_5\leftarrow C_5 + 3 C_4)\
&= - 4 \cdot 6 \cdot 5 \cdot 1 \cdot 6 = 1.\
\end{align*}
Therefore $\det(A) \equiv 1 \pmod 7$, so $\det(A) 
e 0$.
\end{proof} 
Note: With Sage, $\det(A) = 5354628584068202790$.

Exercise 2.1.55 ( Congruences and determinant)

Answers

Comments

Exercise 2.1.55 ( Congruences and determinant)

Answers

Comments

Add answer