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# ISI 2018 Subjective Problem 8 , A Problem from Matrix Try this beautiful Subjective Matrix Problem appeared in ISI Entrance - 2018.

## Problem

$a_{i j} \in\{1,-1\}$ for all $1 \leq i,j \leq n$.

Suppose that

$a_{k 1}=1 \text { for all } 1 \leq k \leq n$

and k=1nakiakj=0 for all ij

Show that $n$ is a multiple of $4$.

Algebra

Matrix

## Suggested Book | Source | Answer

Suggested Book: IIT mathematics by Asit Das Gupta

Source of the Problem: ISI UG Entrance - 2018 , Subjective problem number - 8

## Try with Hints...

Hint 1:
We have $a_{k 1}=1 \forall k=1,2, \ldots, n$.
Now,
$\sum_{k=1}^{n} a_{k 1} a_{k 2}=0$

$\Rightarrow \sum_{k=1}^{n} a_{k 2}=0$

Similarly,
$\sum_{k=1}^{n} a_{k 3}=0$
Hence proceed.

Hint 2:

As $a_{i j}=+1$ or $-1$
so number of $+1^{\prime} s$ and $-1^{\prime} s$ are same in every column.
Therefore $n$ must be even . $(n=2 m$ say $)$
Proceed to work with the following:

$\sum_{k=1}^{n} a_{k 2} a_{k 3}=0$

Hint 3
Observe in column 2,
$\prod_{k=1}^{n} a_{k 2}=(1)^{m}(-1)^{m}=(-1)^{m} \ldots \ldots(1)$

Similarly in column 3,
$\prod_{k=1}^{n} a_{k 3}=(-1)^{m} \ldots \ldots(2)$

And we also have

$\sum_{k=1}^{n} a_{k 2} a_{k 3}=0 \ldots \ldots$
So proceed.

Hint 4:
Obviously,
Hence, $m$ of the $a_{k 2} a_{k 3}$ are $+1^{\prime}$ s and $m$ of them are $-1$ 's.
Now
$\prod_{k=1}^{n} a_{k 2} a_{k 3}=(1)^{m}(-1)^{m}=(-1)^{m} \ldots(4)$

Hint 5:
But,
$\prod_{k=1}^{n} a_{k 2} a_{k 3}=\prod_{k=1}^{n} a_{k 2} \Pi_{k=1}^{n} a_{k 3}$
we get by (1) and (2)
$=(-1)^{m}(-1)^{m}=1 \ldots \ldots(5)$

Hint 6:
Comparing (4) and (5),
we get $(-1)^{m}=1$
Hence, $m$ is even .
Therefore, $n$ is obviously a multiple of 4 .

ISI Entrance Program at Cheenta

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Try this beautiful Subjective Matrix Problem appeared in ISI Entrance - 2018.

## Problem

$a_{i j} \in\{1,-1\}$ for all $1 \leq i,j \leq n$.

Suppose that

$a_{k 1}=1 \text { for all } 1 \leq k \leq n$

and k=1nakiakj=0 for all ij

Show that $n$ is a multiple of $4$.

Algebra

Matrix

## Suggested Book | Source | Answer

Suggested Book: IIT mathematics by Asit Das Gupta

Source of the Problem: ISI UG Entrance - 2018 , Subjective problem number - 8

## Try with Hints...

Hint 1:
We have $a_{k 1}=1 \forall k=1,2, \ldots, n$.
Now,
$\sum_{k=1}^{n} a_{k 1} a_{k 2}=0$

$\Rightarrow \sum_{k=1}^{n} a_{k 2}=0$

Similarly,
$\sum_{k=1}^{n} a_{k 3}=0$
Hence proceed.

Hint 2:

As $a_{i j}=+1$ or $-1$
so number of $+1^{\prime} s$ and $-1^{\prime} s$ are same in every column.
Therefore $n$ must be even . $(n=2 m$ say $)$
Proceed to work with the following:

$\sum_{k=1}^{n} a_{k 2} a_{k 3}=0$

Hint 3
Observe in column 2,
$\prod_{k=1}^{n} a_{k 2}=(1)^{m}(-1)^{m}=(-1)^{m} \ldots \ldots(1)$

Similarly in column 3,
$\prod_{k=1}^{n} a_{k 3}=(-1)^{m} \ldots \ldots(2)$

And we also have

$\sum_{k=1}^{n} a_{k 2} a_{k 3}=0 \ldots \ldots$
So proceed.

Hint 4:
Obviously,
Hence, $m$ of the $a_{k 2} a_{k 3}$ are $+1^{\prime}$ s and $m$ of them are $-1$ 's.
Now
$\prod_{k=1}^{n} a_{k 2} a_{k 3}=(1)^{m}(-1)^{m}=(-1)^{m} \ldots(4)$

Hint 5:
But,
$\prod_{k=1}^{n} a_{k 2} a_{k 3}=\prod_{k=1}^{n} a_{k 2} \Pi_{k=1}^{n} a_{k 3}$
we get by (1) and (2)
$=(-1)^{m}(-1)^{m}=1 \ldots \ldots(5)$

Hint 6:
Comparing (4) and (5),
we get $(-1)^{m}=1$
Hence, $m$ is even .
Therefore, $n$ is obviously a multiple of 4 .

ISI Entrance Program at Cheenta

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