# Let A be a square matrix of order $3$ with integer entries such that $\det(A)=1$.

• What is the maximum possible number of entries of A that are even? What if A is a matrix of order n?

• What is the maximum possible number of entries of A that are PRIMES? What if A is a matrix of order n?

This question was asked during my class test and I was unable to do this.

• What set are the entries from? – joriki Jun 3 '16 at 5:49
• @joriki:Integers – Styles Aug 28 '16 at 14:39

For the first question, you can show that $I_3$ is an example of such a matrix with $6$ even entries. If there are $7$ or more event entries, then necessarily there is a row with only even entries, calculate your determinant by expanding this line, it is a sum of even terms, thus the determinant is even and can't be $1$.
Thus $6$ is the maximum possible number of entries of A that are even and such as $\det(A)=1$.

The generalisation for matrices of order $n$, for the first question is $n^2-n$, you can prove this with same reasoning that in the case $3 \times 3$ : $I_n$ has $n^2-n$ even entries and its determinant is $1$. And if your matrix has more that $n^2-n$ even entries then there is necessary a whole row with even entries and you can expand your determinant from this row and thus it will be even and so different of $1$.

• :But,its determinant is p,we've to consider those matrices whose determinant is 1 – Styles Jun 3 '16 at 6:12
• Try $\begin{matrix} 1 & 2 & 2 \\ 2& 1&2 \\ 0 &2 &1 \end{matrix}$. – Hetebrij Jun 3 '16 at 6:22

For first part answer is $6$. You cab take $A$ as $$A=\begin{bmatrix} 1&0&0\\b&a&ac-1\\d&1&c\end{bmatrix}$$ where $a,b,c,d$ are any even number. Now consider $$e_3=6,$$ $$e_4=e_3+2(4-1)$$ $$e_n=e_{n-1}+2(n-1)$$ where $e_n$ is no. of even entries when $A$ is of order $n\ge3$.

For second part take $A$ as $$A=\begin{bmatrix} 1&0&0\\a&p_1&p_2\\b&p_3&p_4\end{bmatrix}$$ where $a,b$ are any prime number and $p_1p_4-p_3p_2=1$ it is possible to find such primes e.g take $p_1=3,p_4=5,p_3=2,p_2=7$. Now consider $$p_3=6,$$ $$e_4=e_3+(4-1)$$ $$e_n=e_{n-1}+(n-1)$$ where $p_n$ is no. of prime entries when $A$ is of order $n\ge3$.

• For the first part, $\frac{1}{ab}$ is not an integer. – Hetebrij Jun 3 '16 at 6:29
• @Sry:How can we prove these recursion formulas. – Styles Jun 3 '16 at 7:00
• Say we are going for matrix of order $4$, it will be of same format i.e first row has $a_{11}=1$ entry fixed so rest three are left for even choices again first column has 3 entries left for even choices. Apart from first row and first column, leftover matrix is of $3\times3$ whose even choice we know. Similarly if we go for $5\times5$ matrix First row and column has $4$ choices each for even entries and leftover $4\times4$ matrix has $e_4$ choices we know already. So.... – Sry Jun 3 '16 at 7:34
• So, what is the maximum possible number of entries of a $3 \times 3$ matrix $A$ that are primes? – SARTHAK GUPTA Dec 6 '19 at 18:40

To a matrix of order 3, maximum possibility of even entries is 6 ( take identity matrix of order 3) but as you put even entries on more than 6 places then you can find a row consisting of even entries along which you can take determinant.We can also generalize this for order n, may be possibility for maximum even entries is n^{2}-n( which is Identity matrix)

• see identity matrix is always a case – Sunny Jun 3 '16 at 6:10
• There is no conviction in the argument – Shailesh Jun 3 '16 at 6:18