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ISI 2018 Subjective Problem 8 , A Problem from Matrix

\(a_{i j} \in\{1,-1\}\) 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 \(4 \).


Key Concepts


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 .

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\(a_{i j} \in\{1,-1\}\) 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 \(4 \).


Key Concepts


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 .

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