Order 7 matrix with odd entries has determinant a multiple of 64? [closed]

if $A$ is $7\times 7$ matrix with all $49$ entries being odd numbers. Show that $|A|$ is a multiple of $64$. You can use the fact that an $n\times n$ matrix with integer entries has an integer determinant.

• Do you know the effect of row operations on the determinant? Can you think of a row operation that will get you lots of even entries? – Gerry Myerson Nov 5 '12 at 22:44
• This question matches this Yahoo! Answers question. – Douglas S. Stones Nov 6 '12 at 2:13

Hint: Put Gerry's comment and EuYu's hint together,

and note that: $$2^6 = 2^{7-1} = 64.$$

In case there's a language translation issue: I'll give an example - using a simple $5\times 5$ matrix of odd integer entries - of what is meant by "Add one row to the others:"

Suppose $\quad A=% \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1% \end{bmatrix}%$

Adding, say, the entries of row $5$ to each of the other rows would result as follows

$\quad\quad\quad\quad\;\; A^\prime=% \begin{bmatrix} 2 & 2 & 2 & 2 & 2 \\ 2 & 2 & 2 & 2 & 2 \\ 2 & 2 & 2 & 2 & 2 \\ 2 & 2 & 2 & 2 & 2 \\ 1 & 1 & 1 & 1 & 1% \end{bmatrix}%$

Hint: $|A^\prime| = |A|$

What must be the case when adding odd integers to odd integers?

In your (any) $7 \times 7$ matrix of with only odd integers and entries, if you add the odd integer entries of one row of entries to the corresponding entries in other six rows of odd integers, you'll end up with only one row of odd integers.

Hint: What elementary row operation can you apply to matrix $A^{\prime}$ four times? (See Gerry's answer.) And what effect does using that operation have on the determinant of $A^{\prime}$, and hence, on $A$? And what matrix results when applying that row operation on $A^{\prime}$?

Note: you don't need to know what, exactly, the values of the entries are (or the exact value of the determinant) to solve the problem; it is enough to know that the entries are all odd integers. You need only show that the $determinant$ of a $7\times 7$ matrix - whose entries are all/only odd integers - must be a multiple of $64$...

Hint: Add one row to the others.

• what do you mean by add one row to the others? – user48331 Nov 5 '12 at 23:13
• Do you know the effects of elementary row operations on the determinant? – EuYu Nov 5 '12 at 23:33
• @Micheal - Are you familiar with elementary row operations? – amWhy Nov 5 '12 at 23:42
• Yes, i'm familiar with elementary row operations, but i don't know where to start with this problem? – user48331 Nov 6 '12 at 0:58

It appears OP needs a bit more than a hint, so....

Let $$A=\pmatrix{3&5&7\cr11&13&17\cr19&23&29\cr}$$ Note that all the entries in $A$ are odd integers. $A$ has some determinant --- I don't know what it is --- let's call it, $Q$.

Now let's make a matrix $B$ by adding the top row of $A$ to the second row, thus: $$B=\pmatrix{3&5&7\cr14&18&24\cr19&23&29\cr}$$ Do you know how the determinant of $A$ relates to the determinant of $B$? Good. Now, notice that every entry in the second row of $B$ is even. Let $C$ be the matrix you get by taking out the factor of $2$ from the second row of $B$, so $$C=\pmatrix{3&5&7\cr7&9&12\cr19&23&29\cr}$$ Do you know how the determinant of $B$ relates to that of $C$? And, thus, how $\det A$ relates to $\det C$?

Does this give you the idea of what to do in your problem?

• Sorry, Gerry, I didn't see your post...I was wrapped up in figuring out how to format matrices to help provide a nudge(s) in the right direction! – amWhy Nov 6 '12 at 1:58
• @amWhy, no worries, maybe two nudges are better than one. – Gerry Myerson Nov 6 '12 at 2:03