Let a,x,y,z be real numbers satisfying the equations ax+ay=z x+ay=z x+ay=az, where x,y,z are not all zero, then the number of the possible values of a is

Question

Let a,x,y,z be real numbers satisfying the equations ax+ay=z x+ay=z x+ay=az, where x,y,z are not all zero, then the number of the possible values of a is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

As x,y,z are not all zero, thus the system has a non-trivial solution. Thus, D=0 ⇒ | a a -1 1 a -1 1 a -a |=0 ⇒ a(a2 - a)-a(a - 1)+1(a - a)=0 or a3-a2-a2+a=0 ⇒ a(a2 - 2 a + 1)=0 ⇒ a=0,1

Q. Let $a, x, y, z$ be real numbers satisfying the equations

$a x + a y = z$

$x + a y = z$

$x + a y = a z,$ where

$x, y, z$ are not all zero, then the number of the possible values of $a$ is

As $x, y, z$ are not all zero, thus the system has a non-trivial solution.
Thus, $D = 0$
$\Rightarrow ∣ ∣ a 11 a a a - 1 - 1 - a ∣ ∣ = 0$
$\Rightarrow a (a^{2} - a) - a (a - 1) + 1 (a - a) = 0$ or $a^{3} - a^{2} - a^{2} + a = 0$
$\Rightarrow a (a^{2} - 2 a + 1) = 0$
$\Rightarrow a = 0, 1$

Q. Let a,x,y,z be real numbers satisfying the equations ax+ay=z x+ay=z x+ay=az, where x,y,z are not all zero, then the number of the possible values of a is

As x,y,z are not all zero, thus the system has a non-trivial solution. Thus, D=0 ⇒∣∣​a11​aaa​−1−1−a​∣∣​=0 ⇒a(a2−a)−a(a−1)+1(a−a)=0 or a3−a2−a2+a=0 ⇒a(a2−2a+1)=0 ⇒a=0,1

Q. Let $a, x, y, z$ be real numbers satisfying the equations

$a x + a y = z$

$x + a y = z$

$x + a y = a z,$ where

$x, y, z$ are not all zero, then the number of the possible values of $a$ is

As $x, y, z$ are not all zero, thus the system has a non-trivial solution.
Thus, $D = 0$
$\Rightarrow ∣ ∣ a 11 a a a - 1 - 1 - a ∣ ∣ = 0$
$\Rightarrow a (a^{2} - a) - a (a - 1) + 1 (a - a) = 0$ or $a^{3} - a^{2} - a^{2} + a = 0$
$\Rightarrow a (a^{2} - 2 a + 1) = 0$
$\Rightarrow a = 0, 1$