Let P be a matrix of order 3 × 3 such that all the entries in P are from the set {?1, 0, 1}. Then, the maximum possible value of the determinant of P is _____. [2018]

Question

Let P be a matrix of order 3 × 3 such that all the entries in P are from the set {−1, 0, 1}. Then, the maximum possible value of the determinant of P is _____. [2018]

Smart Achievers · Accepted Answer

(4)

$Let Det (P) =$ $| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} |$

$= a_{1} (b_{2} c_{3} - b_{3} c_{2}) - a_{2} (b_{1} c_{3} - b_{3} c_{1}) + a_{3} (b_{1} c_{2} - b_{2} c_{1})$

If $a_{1} = 1,$ $a_{2} = - 1,$ $a_{3} = 1,$ $b_{2} c_{3} = b_{1} c_{3} = b_{1} c_{2} = 1$ and $b_{3} c_{2} = b_{3} c_{1} = b_{2} c_{1} = - 1$

then the maximum value of $Det (P) = 6$

But it is not possible, as

$(b_{2} c_{3}) (b_{3} c_{1}) (b_{1} c_{2}) = - 1$ and $(b_{1} c_{3}) (b_{3} c_{2}) (b_{2} c_{1}) = 1,$

i.e., $b_{1} b_{2} b_{3} c_{1} c_{2} c_{3} = 1$ and $b_{1} b_{2} b_{3} c_{1} c_{2} c_{3} = - 1$

This is a contradiction.

Similarly, contradiction occurs when

$a_{1} = 1,$ $a_{2} = 1,$ $a_{3} = 1,$ $b_{2} c_{3} = b_{3} c_{1} = b_{1} c_{2} = 1$ and $b_{3} c_{2} = b_{1} c_{3} = b_{2} c_{1} = - 1$

Now, for the value to be 5, one of the terms must be zero, but that will make two terms zero, which means the answer cannot be 5.

Now $| \begin{matrix} 1 & 1 & 1 \\ - 1 & 1 & 1 \\ 1 & - 1 & 1 \end{matrix} |$ $= 4;$

Therefore, the maximum value is 4

Solution

Q.
Let P be a matrix of order 3 × 3 such that all the entries in P are from the set {−1, 0, 1}. Then, the maximum possible value of the determinant of P is _____. [2018]

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Solution

Q. Let P be a matrix of order 3 × 3 such that all the entries in P are from the set {−1, 0, 1}. Then, the maximum possible value of the determinant of P is _____. [2018]

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Q.
Let P be a matrix of order 3 × 3 such that all the entries in P are from the set {−1, 0, 1}. Then, the maximum possible value of the determinant of P is _____. [2018]