Let the system of linear equations $x + y + az = 2$ $3x + y + z = 4$ $x + 2z = 1$ have a unique solution $({x^ * },{y^ * },{z^ * }).$ If $(\alpha ,{x^ * }),({y^ * },\alpha )$ and $({x^ * }, - {y^ * })$ are collinear points, then the sum of absolute values of all possible values of $\alpha $ is

JEE MATHEMATICS MCQ: MATRICES

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Analyzing +4 | -1

Let the system of linear equations

x + y + az = 2

3x + y + z = 4

x + 2z = 1

have a unique solution

({x^ * },{y^ * },{z^ * }).

(\alpha ,{x^ * }),({y^ * },\alpha )

and

({x^ * }, - {y^ * })

are collinear points, then the sum of absolute values of all possible values of

\alpha

View answer and explanation

Correct option: C

Option C (2) is correct because solving the system of equations and using the collinearity condition leads to two possible values for

\alpha

, and the sum of their absolute values is 2.

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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|