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Matrices and Determinants question

2020 · 8 Jan · Shift 2 · Q30
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Matrices and Determinants question

2020 · 8 Jan · Shift 2 · Q30

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
The system of linear equations λ\lambdaλ x + 2y + 2z = 5 2 λ\lambdaλ x + 3y + 5z = 8 4x + λ\lambdaλ y + 6z = 10 has
  1. A
    a unique solution when λ\lambdaλ = –8
  2. B
    no solution when λ\lambdaλ = 2
  3. C
    infinitely many solutions when λ\lambdaλ = 2
  4. D
    no solution when λ\lambdaλ = 8
View written solutionFree

Correct answer: B

We are given the system

{λx+2y+2z=52λx+3y+5z=84x+λy+6z=10\begin{cases} \lambda x+2y+2z=5 \\ 2\lambda x+3y+5z=8 \\ 4x+\lambda y+6z=10 \end{cases}⎩⎨⎧​λx+2y+2z=52λx+3y+5z=84x+λy+6z=10​

We must determine which option is correct.

1. Coefficient matrix and determinant

The coefficient matrix is

A=(λ222λ354λ6)A=\begin{pmatrix} \lambda & 2 & 2\\ 2\lambda & 3 & 5\\ 4 & \lambda & 6 \end{pmatrix}A=​λ2λ4​23λ​256​​

For a unique solution, we need

det⁡(A)≠0.\det(A)\neq 0.det(A)=0.

Let us compute the determinant:

det⁡(A)=∣λ222λ354λ6∣\det(A)= \begin{vmatrix} \lambda & 2 & 2\\ 2\lambda & 3 & 5\\ 4 & \lambda & 6 \end{vmatrix}det(A)=​λ2λ4​23λ​256​​

Expanding along the first row,

det⁡(A)=λ∣35λ6∣−2∣2λ546∣+2∣2λ34λ∣\det(A)=\lambda \begin{vmatrix} 3 & 5\\ \lambda & 6 \end{vmatrix} -2 \begin{vmatrix} 2\lambda & 5\\ 4 & 6 \end{vmatrix} +2 \begin{vmatrix} 2\lambda & 3\\ 4 & \lambda \end{vmatrix}det(A)=λ​3λ​56​​−2​2λ4​56​​+2​2λ4​3λ​​

Now,

∣35λ6∣=18−5λ\begin{vmatrix} 3 & 5\\ \lambda & 6 \end{vmatrix}=18-5\lambda​3λ​56​​=18−5λ ∣2λ546∣=12λ−20\begin{vmatrix} 2\lambda & 5\\ 4 & 6 \end{vmatrix}=12\lambda-20​2λ4​56​​=12λ−20 ∣2λ34λ∣=2λ2−12\begin{vmatrix} 2\lambda & 3\\ 4 & \lambda \end{vmatrix}=2\lambda^2-12​2λ4​3λ​​=2λ2−12

So,

det⁡(A)=λ(18−5λ)−2(12λ−20)+2(2λ2−12)\det(A)=\lambda(18-5\lambda)-2(12\lambda-20)+2(2\lambda^2-12)det(A)=λ(18−5λ)−2(12λ−20)+2(2λ2−12) =18λ−5λ2−24λ+40+4λ2−24=18\lambda-5\lambda^2-24\lambda+40+4\lambda^2-24=18λ−5λ2−24λ+40+4λ2−24 =−λ2−6λ+16=-\lambda^2-6\lambda+16=−λ2−6λ+16 =−(λ2+6λ−16)=-(\lambda^2+6\lambda-16)=−(λ2+6λ−16) =−(λ+8)(λ−2)=-(\lambda+8)(\lambda-2)=−(λ+8)(λ−2)

Thus,

det⁡(A)=0  ⟺  λ=2 or λ=−8.\det(A)=0 \iff \lambda=2 \text{ or } \lambda=-8.det(A)=0⟺λ=2 or λ=−8.

Therefore:

  • if λ≠2,−8\lambda\neq 2,-8λ=2,−8, the system has a unique solution;
  • for λ=2\lambda=2λ=2 or λ=−8\lambda=-8λ=−8, we must check consistency.

2. Check option A: unique solution when λ=−8\lambda=-8λ=−8

Since

det⁡(A)=0at λ=−8,\det(A)=0 \quad \text{at } \lambda=-8,det(A)=0at λ=−8,

the system cannot have a unique solution.

So A is false.

3. Check λ=2\lambda=2λ=2

Substitute λ=2\lambda=2λ=2:

{2x+2y+2z=54x+3y+5z=84x+2y+6z=10\begin{cases} 2x+2y+2z=5 \\ 4x+3y+5z=8 \\ 4x+2y+6z=10 \end{cases}⎩⎨⎧​2x+2y+2z=54x+3y+5z=84x+2y+6z=10​

Augmented matrix:

[2225435842610]\left[ \begin{array}{ccc|c} 2&2&2&5\\ 4&3&5&8\\ 4&2&6&10 \end{array} \right]​244​232​256​5810​​

Perform row operations:

R2→R2−2R1,R3→R3−2R1R_2\to R_2-2R_1, \qquad R_3\to R_3-2R_1R2​→R2​−2R1​,R3​→R3​−2R1​

Then

[22250−11−20−220]\left[ \begin{array}{ccc|c} 2&2&2&5\\ 0&-1&1&-2\\ 0&-2&2&0 \end{array} \right]​200​2−1−2​212​5−20​​

Now,

R3→R3−2R2R_3\to R_3-2R_2R3​→R3​−2R2​

So we get

[22250−11−20004]\left[ \begin{array}{ccc|c} 2&2&2&5\\ 0&-1&1&-2\\ 0&0&0&4 \end{array} \right]​200​2−10​210​5−24​​

The last row gives

0=4,0=4,0=4,

which is impossible. Hence the system is inconsistent.

Therefore, when λ=2\lambda=2λ=2, the system has no solution.

So:

  • B is true
  • C is false

4. Check option D: no solution when λ=8\lambda=8λ=8

At λ=8\lambda=8λ=8,

det⁡(A)=−(8+8)(8−2)=−16⋅6≠0\det(A)=-(8+8)(8-2)=-16\cdot 6\neq 0det(A)=−(8+8)(8−2)=−16⋅6=0

So the system has a unique solution, not no solution.

Hence D is false.

5. Final conclusion

Only option B is correct.

B\boxed{\text{B}}B​

This matches the stored correct answer.

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