"Let the system of linear equations x + y + $\alpha$z = 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"

Question

Accepted Answer

"

Given system of equations

$x + y + az = 2$ ..... (i)

$3x + y + z = 4$ ..... (ii)

$x + 2z = 1$ ..... (iii)

Solving (i), (ii) and (iii), we get

x = 1, y = 1, z = 0 (and for unique solution a $\ne$ $-$3)

Now, ($\alpha$, 1), (1, $\alpha$) and (1, $-$1) are collinear

$\therefore$ $$\left| {\matrix{ \alpha & 1 & 1 \cr 1 & \alpha & 1 \cr 1 & { - 1} & 1 \cr } } \right| = 0$$

$\Rightarrow \alpha (\alpha + 1) - 1(0) + 1( - 1 - \alpha ) = 0$

$\Rightarrow {\alpha ^2} - 1 = 0$

$\therefore$ $\alpha = \, \pm \,1$

$\therefore$ Sum of absolute values of $\alpha = 1 + 1 = 2$

"

Let the system of linear equations

x + y + $\alpha$z = 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

Solution

About this question

Drill 25 more like these. Every day. Free.

Let the system of linear equations x + y + $\alpha$z = 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

Solution

About this question

More Matrices and Determinants questions

Drill 25 more like these. Every day. Free.

Let the system of linear equations

x + y + $\alpha$z = 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