Let P be a matrix of order 3 × 3 such that all entries in P are from set {-1,0,1} . Then, maximum p

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3 and 4 .Determinants and Matrices

normal

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. . . . . . .

$7$

$6$

$5$

$4$

(IIT-2018)

Solution

$\Delta=\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|=\left(a_1 b_2 c_3+a_3 c_2 b_1+a_2 c_1 b_3\right)-\left(a_1 b_3 c_2+a_2 c_3 b_1+a_3 c_1 b_2\right)$

$\left\{ a _{ i }, b _{ i }, c _{ i }\right\} \equiv\{-1,0,1\} i =1,2,3$

Maximum value can be $6$, but that's not possible.

(For $\Delta=6,3$ entries should be $1$ and $6$ entries should be -$1$ in first bracket and $3$ entries should be -$1$ and $6$ entries should be $1$ in second bracket. That's a contradiction!)

$\Delta \neq 5$. (Even if one element is $0$ )

$\therefore \Delta_{\max }=4$ which is possible.

$\operatorname{Ex}:\left[\begin{array}{ccc}1 & -1 & 0 \\ 1 & 1 & 1 \\ -1 & -1 & 1\end{array}\right]$

Standard 12

Mathematics

If $D =$ $\left| {\,\begin{array}{*{20}{c}}{\frac{1}{z}}&{\frac{1}{z}}&{ – \frac{{(x + y)}}{{{z^2}}}}\\{ – \frac{{(y + z)}}{{{x^2}}}}&{\frac{1}{x}}&{\frac{1}{x}}\\{ – \frac{{y(y + z)}}{{{x^2}z}}}&{\frac{{x + 2y + z}}{{xz}}}&{ – \frac{{y(x +y)}}{{x{z^2}}}}\end{array}\,} \right|$ then, the incorrect statement is

normal

If $f(x) = \left| {\begin{array}{*{20}{c}}{x – 3}&{2{x^2} – 18}&{3{x^3} – 81}\\{x – 5}&{2{x^2} – 50}&{4{x^3} – 500}\\1&2&3\end{array}} \right|$ then $f(1).f(3) + f(3).f(5) + f(5).f(1)$=

hard

The determinant $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha + b}\\b&c&{b\alpha + c}\\{a\alpha + b}&{b\alpha + c}&0\end{array}\,} \right| = 0$, if $a,b,c$ are in

medium

(IIT-1987)

If $a,b,c$ are unequal what is the condition that the value of the following determinant is zero $\Delta = \left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3} + 1}\\b&{{b^2}}&{{b^3} + 1}\\c&{{c^2}}&{{c^3} + 1}\end{array}\,} \right|$

medium

(IIT-1985)

Evaluate $\Delta=\left|\begin{array}{lll}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|$

medium

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. . . . . . .

Solution

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