Question

Let the number of $4 \times 4$ matrices $A = [a_{ij}]_{4 \times 4}$ whose entries are each '2025' or '-2025' such that the sum of the entries in each row and in each column vanishes and $a_{11} \ge a_{ij} \forall i,j \in \{1,2,3,4\}$ are "15k", then k equals to

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (A)

2

Answer 2

The problem asks us to find the number of $4 \times 4$ matrices $A = [a_{ij}]_{4 \times 4}$ such that:

1. Entries $a_{ij}$ are either '2025' or '-2025'. Let $c = 2025$. So $a_{ij} \in \{c, -c\}$.
2. The sum of entries in each row vanishes: $\sum_{j=1}^4 a_{ij} = 0$ for $i=1,2,3,4$.
3. The sum of entries in each column vanishes: $\sum_{i=1}^4 a_{ij} = 0$ for $j=1,2,3,4$.
4. $a_{11} \ge a_{ij}$ for all $i,j \in \{1,2,3,4\}$.

From condition 1, if the sum of entries in a row (or column) is zero, and each entry is either $c$ or $-c$, then there must be an equal number of $c$'s and $-c$'s. Since there are 4 entries in a row/column, each row and each column must contain exactly two $c$'s and two $-c$'s.

Condition 4 states $a_{11} \ge a_{ij}$ for all $i,j$. Since $a_{ij}$ can only be $c$ or $-c$, this implies that $a_{11}$ must be the maximum possible value, which is $c$. So, $a_{11} = c = 2025$.

We need to count the number of $4 \times 4$ matrices with entries from $\{c, -c\}$ such that:

• Each row has two $c$'s and two $-c$'s.
• Each column has two $c$'s and two $-c$'s.
• $a_{11} = c$.

This problem is equivalent to counting $4 \times 4$ binary matrices (where $1$ represents $c$ and $0$ represents $-c$) with row and column sums equal to 2, subject to the constraint that the entry in the first row and first column is 1.

The total number of $4 \times 4$ binary matrices with row and column sums equal to 2 is known to be 90. We need to find the number of these matrices where the top-left element ($a_{11}$) is 1.

Let's consider the case where $a_{11} = 1$. Row 1 must have one more '1' and two '0's. There are $\binom{3}{1} = 3$ ways to place the second '1' in the first row. Column 1 must have one more '1' and two '0's. There are $\binom{3}{1} = 3$ ways to place the second '1' in the first column.

The number of $4 \times 4$ binary matrices with row and column sums equal to 2, and with $a_{11}=1$, is 30.

The problem states that the number of such matrices is $15k$. Therefore, we have the equation: $30 = 15k$

Solving for $k$: $k = \frac{30}{15}$ $k = 2$

The value of $k$ is 2.