Let the system of linear equations
$x+y+\alpha z=2$
$3 x+y+z=4$
$x+2 z=1$
have a unique solution $\left(x^{*}, y^{*}, z^{*}\right)$. If $\left(\alpha, x^{*}\right),\left(y^{*}, \alpha\right)$ and $\left(x^{*},-y^{*}\right)$ are collinear points, then the sum of absolute values of all possible values of $\alpha$ is
Let the system of linear equations
$x+y+\alpha z=2$
$3 x+y+z=4$
$x+2 z=1$
have a unique solution $\left(x^{*}, y^{*}, z^{*}\right)$. If $\left(\alpha, x^{*}\right),\left(y^{*}, \alpha\right)$ and $\left(x^{*},-y^{*}\right)$ are collinear points, then the sum of absolute values of all possible values of $\alpha$ is
- [JEE MAIN 2022]
- A
$4$
- B
$3$
- C
$2$
- D
$1$
Similar Questions
Statement $1$ : If the system of equations $x + ky + 3z = 0, 3x+ ky - 2z = 0, 2x + 3y - 4z = 0$ has a nontrivial solution, then the value of $k$ is $\frac{31}{2}$
Statement $2$ : A system of three homogeneous equations in three variables has a non trivial solution if the determinant of the coefficient matrix is zero.
Statement $1$ : If the system of equations $x + ky + 3z = 0, 3x+ ky - 2z = 0, 2x + 3y - 4z = 0$ has a nontrivial solution, then the value of $k$ is $\frac{31}{2}$
Statement $2$ : A system of three homogeneous equations in three variables has a non trivial solution if the determinant of the coefficient matrix is zero.
- [AIEEE 2012]
If system of equations $kx + 2y - z = 2,$$\left( {k - 1} \right)x + ky + z = 1,x + \left( {k - 1} \right)y + kz = 3$ has only one solution, then number of possible real value$(s)$ of $k$ is -
If system of equations $kx + 2y - z = 2,$$\left( {k - 1} \right)x + ky + z = 1,x + \left( {k - 1} \right)y + kz = 3$ has only one solution, then number of possible real value$(s)$ of $k$ is -
If for some $\alpha$ and $\beta$ in $R,$ the intersection of the following three planes $x+4 y-2 z=1$ ; $x+7 y-5 z=\beta$ ; $x+5 y+\alpha z=5$ is a line in $\mathrm{R}^{3},$ then $\alpha+\beta$ is equal to
If for some $\alpha$ and $\beta$ in $R,$ the intersection of the following three planes $x+4 y-2 z=1$ ; $x+7 y-5 z=\beta$ ; $x+5 y+\alpha z=5$ is a line in $\mathrm{R}^{3},$ then $\alpha+\beta$ is equal to
- [JEE MAIN 2020]
The equation $\left| {\begin{array}{*{20}{c}}{{{(1 + x)}^2}}&{{{(1 - x)}^2}}&{ - \,(2 + {x^2})}\\{2x + 1}&{3x}&{1 - 5x}\\{x + 1}&{2x}&{2 - 3x}\end{array}} \right|$ $+$ $\left| {\begin{array}{*{20}{c}}{{{(1 + x)}^2}}&{2x + 1}&{x + 1}\\{{{(1 - x)}^2}}&{3x}&{2x}\\{1 - 2x}&{3x - 2}&{2x - 3}\end{array}} \right|$ $= 0$
$\left| {\,\begin{array}{*{20}{c}}1&a&b\\{ - a}&1&c\\{ - b}&{ - c}&1\end{array}\,} \right| = $
$\left| {\,\begin{array}{*{20}{c}}1&a&b\\{ - a}&1&c\\{ - b}&{ - c}&1\end{array}\,} \right| = $