Matrices and Determinants 2 Question 24
26. Let $P$ be a matrix of order $3 \times 3$ 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
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Answer:
Correct Answer: 26. (4)
Solution:
- Let $\operatorname{Det}(P)=\begin{vmatrix}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{vmatrix}$
$ =a_{1}\left(b_{2} c_{3}-b_{3} c_{2}\right)-a_{2}\left(b_{1} c_{3}-b_{3} c_{1}\right)+a_{3}\left(b_{1} c_{2}-b_{2} c_{1}\right) $
Now, maximum value of $\operatorname{Det}(P)=6$
If $a_{1}=1, a_{2}=-1, a_{3}=1, b_{2} c_{3}=b_{1} c_{3}=b_{1} c_{2}=1$
$ \text { and } b_{3} c_{2}=b_{3} c_{1}=b_{2} c_{1}=-1 $
But it is not possible as
$ \left(b_{2} c_{3}\right)\left(b_{3} c_{1}\right)\left(b_{1} c_{2}\right)=-1 \text { and }\left(b_{1} c_{3}\right)\left(b_{3} c_{2}\right)\left(b_{2} c_{1}\right)=1 $
$ \text { i.e., } b_{1} b_{2} b_{3} c_{1} c_{2} c_{3}=1 \text { and }-1 $
Similar contradiction occurs when $a_{1}=1, a_{2}=1, a_{3}=1, b_{2} c_{1}=b_{3} c_{1}=b_{1} c_{2}=1$
$ \text { and } b_{3} c_{2}=b_{1} c_{3}=b_{1} c_{2}=-1 $
Now, for value to be 5 one of the terms must be zero but that will make 2 terms zero which means answer cannot be 5
Now, $\begin{vmatrix}1 & 1 & 1 \\ -1 & 1 & 1 \\ 1 & -1 & 1\end{vmatrix}=4$
Hence, maximum value is 4 .
