Question

If the system of simultaneous linear equations $x+\lambda y-2z=1$, $x-y+\lambda z=2$ and $x-2y+3z=3$ is inconsistent for $\lambda = \lambda_1$ and $\lambda_2$, then $\lambda_1 + \lambda_2 =$

• A. 5
• B. $\sqrt{5}$
• C. 1
• D. -1

Hint:

For a system of 3 linear equations, the condition for having no solution or infinitely many solutions is $\det(A) = 0$. The question asks for values where the system is inconsistent, which directly points to this condition. If the resulting equation in the parameter (here, $\lambda$) is quadratic, you can often find the sum or product of the parameter values using Vieta's formulas without explicitly solving for them.

Answer 1

The Correct Option is C

Step 1: Condition for Inconsistency:
A system of linear equations AX = B is inconsistent (has no solution) if the determinant of the coefficient matrix D = |A| is zero, and at least one of the determinants Dx, Dy, Dz (obtained by replacing a column with the constant vector) is non-zero.

Step 2: Calculate Determinant D:

D = | 1   λ  -2 |
    | 1  -1   λ |
    | 1  -2   3 |

Expanding along the first row:

D = 1[(-1)(3) - (λ)(-2)] - λ[(1)(3) - (λ)(1)] - 2[(1)(-2) - (-1)(1)]

D = (-3 + 2λ) - λ(3 - λ) - 2(-2 + 1)

D = -3 + 2λ - 3λ + λ2 - 2(-1)

D = λ2 - λ - 3 + 2

D = λ2 - λ - 1

Step 3: Solve for D = 0:
λ2 - λ - 1 = 0

The roots of this quadratic equation are λ1 and λ2.
According to properties of quadratic equations ax2 + bx + c = 0, the sum of roots is -b / a.

λ1 + λ2 = -(-1) / 1 = 1

Step 4: Verify Inconsistency (Optional but recommended):
For these values, D = 0. We need to ensure Dx, Dy, Dz are not all zero (which would imply infinite solutions).

Since the roots are irrational (1 ± √5) / 2, it is highly unlikely they make all numerators zero simultaneously in this specific setup.

Calculating Dz:

Dz = | 1   λ   1 |
        | 1  -1   2 |
        | 1  -2   3 |

Dz = 1(1) - λ(1) + 1(-1) = -λ

Since λ ≠ 0 (roots are approximately 1.618 and -0.618), Dz ≠ 0. Thus, the system is inconsistent.

Final Answer: λ1 + λ2 = 1