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 .
