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Matrices and Determinants question

2025 · 23 Jan · Shift 2 · Q35
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Matrices and Determinants question

2025 · 23 Jan · Shift 2 · Q35

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
The system of equations x+y+z=6,x+2y+5z=9,x+5y+λz=μ,\begin{aligned} & x+y+z=6, \\ & x+2 y+5 z=9, \\ & x+5 y+\lambda z=\mu, \end{aligned}​x+y+z=6,x+2y+5z=9,x+5y+λz=μ,​ has no solution if
  1. A
    λ=17,μ=18\lambda=17, \mu=18λ=17,μ=18
  2. B
    λ=17,μeq18\lambda=17, \mu eq 18λ=17,μeq18
  3. C
    λ=15,μeq17\lambda=15, \mu eq 17λ=15,μeq17
  4. D
    λeq17,μeq18\lambda eq 17, \mu eq 18λeq17,μeq18
View written solutionFree

Correct answer: B

  1. Write the augmented system in matrix form:
{x+y+z=6x+2y+5z=9x+5y+λz=μ\begin{cases} x+y+z=6\\ x+2y+5z=9\\ x+5y+\lambda z=\mu \end{cases}⎩⎨⎧​x+y+z=6x+2y+5z=9x+5y+λz=μ​

Coefficient matrix:

A=(11112515λ)A=\begin{pmatrix} 1&1&1\\ 1&2&5\\ 1&5&\lambda \end{pmatrix}A=​111​125​15λ​​

For the system to have no solution, we need:

  • det⁡(A)=0\det(A)=0det(A)=0 so that the equations are dependent/inconsistent possible,
  • and the augmented matrix must have rank greater than rank of coefficient matrix.
  1. Compute det⁡(A)\det(A)det(A):
det⁡(A)=∣11112515λ∣\det(A)=\begin{vmatrix} 1&1&1\\ 1&2&5\\ 1&5&\lambda \end{vmatrix}det(A)=​111​125​15λ​​

Apply row operations:

R2→R2−R1,R3→R3−R1R_2\to R_2-R_1,\qquad R_3\to R_3-R_1R2​→R2​−R1​,R3​→R3​−R1​

So,

det⁡(A)=∣11101404λ−1∣\det(A)=\begin{vmatrix} 1&1&1\\ 0&1&4\\ 0&4&\lambda-1 \end{vmatrix}det(A)=​100​114​14λ−1​​

Expanding along first column,

det⁡(A)=∣144λ−1∣=(λ−1)−16=λ−17\det(A)=\begin{vmatrix} 1&4\\ 4&\lambda-1 \end{vmatrix} =(\lambda-1)-16=\lambda-17det(A)=​14​4λ−1​​=(λ−1)−16=λ−17

Thus,

det⁡(A)=0  ⟺  λ=17\det(A)=0 \iff \lambda=17det(A)=0⟺λ=17

So only when λ=17\lambda=17λ=17 can the system fail to have a unique solution.

  1. Now substitute λ=17\lambda=17λ=17:
{x+y+z=6x+2y+5z=9x+5y+17z=μ\begin{cases} x+y+z=6\\ x+2y+5z=9\\ x+5y+17z=\mu \end{cases}⎩⎨⎧​x+y+z=6x+2y+5z=9x+5y+17z=μ​

Check whether the third equation is a linear combination of the first two.

Let

a(x+y+z)+b(x+2y+5z)=x+5y+17za(x+y+z)+b(x+2y+5z)=x+5y+17za(x+y+z)+b(x+2y+5z)=x+5y+17z

Comparing coefficients:

a+b=1a+b=1a+b=1 a+2b=5a+2b=5a+2b=5 a+5b=17a+5b=17a+5b=17

From first two equations:

b=4,a=−3b=4,\quad a=-3b=4,a=−3

Check zzz coefficient:

a+5b=−3+20=17a+5b=-3+20=17a+5b=−3+20=17

So indeed,

Eq.3=−3⋅Eq.1+4⋅Eq.2\text{Eq.3}=-3\cdot \text{Eq.1}+4\cdot \text{Eq.2}Eq.3=−3⋅Eq.1+4⋅Eq.2

Therefore, for consistency, the constants must also satisfy:

μ=−3⋅6+4⋅9=−18+36=18\mu=-3\cdot 6+4\cdot 9=-18+36=18μ=−3⋅6+4⋅9=−18+36=18
  1. Hence:
  • If λ=17\lambda=17λ=17 and μ=18\mu=18μ=18, infinitely many solutions.
  • If λ=17\lambda=17λ=17 and μ≠18\mu\ne 18μ=18, inconsistent system, so no solution.
  • If λ≠17\lambda\ne 17λ=17, determinant is nonzero, so unique solution exists.
  1. Evaluate options:

  • A: λ=17, μ=18\lambda=17,\ \mu=18λ=17, μ=18
    Consistent, infinitely many solutions. Not correct.

  • B: λ=17, μ≠18\lambda=17,\ \mu\ne 18λ=17, μ=18
    Inconsistent. Correct.

  • C: λ=15, μ≠17\lambda=15,\ \mu\ne 17λ=15, μ=17
    Since λ≠17\lambda\ne 17λ=17, determinant ≠0\ne 0=0, unique solution exists. Not correct.

  • D: λ≠17, μ≠18\lambda\ne 17,\ \mu\ne 18λ=17, μ=18
    Since λ≠17\lambda\ne 17λ=17, unique solution exists regardless of μ\muμ. Not correct.

Therefore, the system has no solution for:

λ=17, μ≠18\boxed{\lambda=17,\ \mu\ne 18}λ=17, μ=18​
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