Algebra

Matrices & Determinants

Medium

core

The value of k for which the system of equations

$x + y + z = 1$

$x - ky + z = 1$

$x - y + z = 1$

has more than one solutions is

only 1

only - 1

1 and - 1 only

any real number

✅ Correct Option: 4

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

The three equations are:

$x + y + z = 1$

$x - ky + z = 1$

$x - y + z = 1$

Subtracting the first equation from the third equation:

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

$-2y = 0$

$y = 0$

Subtracting the first equation from the second equation:

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

$-ky - y = 0$

$-y(k + 1) = 0$

Substituting $y = 0$ :

$-0(k + 1) = 0$

$0 = 0$

This is true for any value of $k$ .

Substituting $y = 0$ into all equations:

$x + z = 1$

All three equations become identical.

The system reduces to:

$y = 0$

$x + z = 1$

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

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

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

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Choose the correct answer from the options given below:

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The value of k for which the system of equations x+y+z=1x + y + z = 1x+y+z=1 x−ky+z=1x - ky + z = 1x−ky+z=1 x−y+z=1x - y + z = 1x−y+z=1 has more than one solutions is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The value of k for which the system of equations

$x + y + z = 1$

$x - ky + z = 1$

$x - y + z = 1$

has more than one solutions is