CUET UG 2025 — Mathematics Algebra

The system has more than one solution when the equations are not all independent.

The three equations are:

$$x + y + z = 1$$x+y+z=1

$$x - ky + z = 1$$x−ky+z=1

$$x - y + z = 1$$x−y+z=1

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Subtracting the first equation from the third equation:

$$(x - y + z) - (x + y + z) = 1 - 1$$(x−y+z)−(x+y+z)=1−1

$$-2y = 0$$−2y=0

$$y = 0$$y=0

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Subtracting the first equation from the second equation:

$$(x - ky + z) - (x + y + z) = 1 - 1$$(x−ky+z)−(x+y+z)=1−1

$$-ky - y = 0$$−ky−y=0

$$-y(k + 1) = 0$$−y(k+1)=0

Substituting $y = 0$y=0:

$$-0(k + 1) = 0$$−0(k+1)=0

$$0 = 0$$0=0

This is true for any value of $k$k.

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Substituting $y = 0$y=0 into all equations:

$$x + z = 1$$x+z=1

$$x + z = 1$$x+z=1

$$x + z = 1$$x+z=1

All three equations become identical.

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The system reduces to:

$$y = 0$$y=0

$$x + z = 1$$x+z=1

With 3 variables and effectively 2 constraints, there are infinitely many solutions. For example: $(1, 0, 0)$(1,0,0), $(0, 0, 1)$(0,0,1), $(2, 0, -1)$(2,0,−1).

Since $y = 0$y=0 is forced by the first and third equations, the second equation is automatically satisfied for any value of $k$k.

Therefore, the value of $k$k is any real number.