If $a,b,c$ are the positive integers, then the determinant $\Delta = \left| {\begin{array}{*{20}{c}} {{a^2} + x}&{ab}&{ac}\\ {ab}&{{b^2} + x}&{bc}\\ {ac}&{bc}&{{c^2} + x} \end{array}} \right|$ is divisible by

$\left( {{a^2} + {b^2} + {c^2}} \right)$

JEE MATHEMATICS MCQ: MATRICES

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Analyzing +4 | -1

a,b,c

are the positive integers, then the determinant

\Delta = \left| {\begin{array}{cccccccccccccccccccc} {{a^2} + x}&{ab}&{ac}\\ {ab}&{{b^2} + x}&{bc}\\ {ac}&{bc}&{{c^2} + x} \end{array}} \right|

is divisible by

View answer and explanation

Correct option: D

None of these

Option B (Divisible by both

x

and

y

) is correct because after applying column operations, the determinant simplifies to

xy

, which is divisible by both

x

and

y

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\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0

a, b, c

a x+b y+c=0

{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3

2x + 3y + 1 = 0,3x + y - 2 = 0

ax + 2y - b = 0

\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|

\omega

\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right|

A

2 \times 2

A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]

|A|