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The system of equations x+y−z=1,3x+y−2z=3,x−y+λz=1x + y - z = 1, 3x + y - 2z = 3, x - y + \lambda z = 1 has infinite number of solutions if λ\lambda is equal to

Solution

✅ Correct Option: 1

For the system:

x+y−z=1x + y - z = 1

3x+y−2z=33x + y - 2z = 3

x−y+λz=1x - y + \lambda z = 1

The coefficient matrix is:

∣11−131−21−1λ∣\begin{vmatrix} 1 & 1 & -1 \\ 3 & 1 & -2 \\ 1 & -1 & \lambda \end{vmatrix}


For infinite solutions, the determinant of the coefficient matrix must equal zero.

Det =1(1⋅λ−(−2)⋅(−1))−1(3⋅λ−(−2)⋅1)+(−1)(3⋅(−1)−1⋅1)= 1(1 \cdot \lambda - (-2) \cdot (-1)) - 1(3 \cdot \lambda - (-2) \cdot 1) + (-1)(3 \cdot (-1) - 1 \cdot 1)

Det =1(λ−2)−1(3λ+2)−1(−3−1)= 1(\lambda - 2) - 1(3\lambda + 2) - 1(-3 - 1)

Det =λ−2−3λ−2+4= \lambda - 2 - 3\lambda - 2 + 4

Det =−2λ= -2\lambda


Setting the determinant equal to zero:

−2λ=0-2\lambda = 0

λ=0\lambda = 0


With λ=0\lambda = 0, the system becomes:

x+y−z=1x + y - z = 1 ... (1)

3x+y−2z=33x + y - 2z = 3 ... (2)

x−y=1x - y = 1 ... (3)

From equation (3): x=1+yx = 1 + y

Substituting in equation (1):

(1+y)+y−z=1(1 + y) + y - z = 1

2y−z=02y - z = 0

z=2yz = 2y

Checking equation (2):

3(1+y)+y−2(2y)=3+3y+y−4y=33(1 + y) + y - 2(2y) = 3 + 3y + y - 4y = 3 ✓

The system is consistent. Any value can be chosen for yy, with corresponding xx and zz values.


Therefore, λ=0\lambda = 0

When λ≠0\lambda \neq 0, the determinant ≠0\neq 0, making the system have a unique solution rather than infinite solutions.

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