If the system of linear equations 2 x + 3 y - z = - 2 x + y + z = 4 x - y + λ z = 4 λ - 4 where…

If the system of linear equations

2x+3y-z=-2

x+y+z=4

x-y+λz=4λ-4 where λ,

has no solution, then

  1. $\lambda = 7$
  2. $\lambda = -7$
  3. $\lambda = 8$
  4. $\lambda^{2} = 1$

Solution

If we have three equations in three variables, $\begin{aligned} a_1 x + b_1 y + c_1 z &= d_1 \\ a_2 x + b_2 y + c_2 z &= d_2 \\ a_3 x + b_3 y + c_3 z &= d_3 \end{aligned}$ then $\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$, $\Delta_1 = \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{vmatrix}$, $\Delta_2 = \begin{vmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{vmatrix}$, $\Delta_3 = \begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix}$. If the set of equations do not have a solution then $\Delta = 0$ and at least one among $\Delta_1$, $\Delta_2$, $\Delta_3$ is not equal to zero. $\Delta = \begin{vmatrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & \lambda \end{vmatrix} = 0$ $\Rightarrow |\lambda| = 7 \Rightarrow \lambda = \pm 7 \ldots 1$ Case 1: $\lambda = 7$

The system of equations are 

2x+3y-z=-2x+y+z=4x-y+7z=24

a2x+3y-z+2+bx+y+z-4=x-y+7z-24

Comparing the coefficients of x and y we get, 

2a+b=13a+b=-1a=-2, b=5

For these values of a and b, coefficients of z and the constant values are matching. So the set of equations will have infinitely many solutions.

Case 2: λ=-7

The system of equations are 

2x+3y-z=-2x+y+z=4x-y-7z=-32

a2x+3y-z+2+bx+y+z-4=x-y-7z+32

Comparing the coefficients of x and y we get, 

2a+b=13a+b=-1a=-2, b=5

For these values of a and b, coefficients of z and the constant values are not matching. So the set of equations will have no solutions.

Asked in: JEE Main 2022 (28 Jun Shift 1)

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