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Matrices and Determinants question

2023 · 29 Jan · Shift 1 · Q38
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Matrices and Determinants question

2023 · 29 Jan · Shift 1 · Q38

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Consider the following system of equations αx+2y+z=12αx+3y+z=13x+αy+2z=β\alpha x+2y+z=12\alpha x+3y+z=13x+\alpha y+2z=\betaαx+2y+z=12αx+3y+z=13x+αy+2z=β for some α,β∈R\alpha,\beta\in \mathbb{R}α,β∈R. Then which of the following is NOT correct.
  1. A
    It has a solution for all αe−1\alpha e-1αe−1 and β=2\beta=2β=2
  2. B
    It has no solution if α=−1\alpha=-1α=−1 and βe2\beta e2βe2
  3. C
    It has no solution for α=−1\alpha=-1α=−1 and for all β∈R\beta \in \mathbb{R}β∈R
  4. D
    It has no solution for α=3\alpha=3α=3 and for all βe2\beta e2βe2
View written solutionFree

Correct answer: C

Let the system be

{αx+2y+z=12αx+3y+z=13x+αy+2z=β\begin{cases} \alpha x+2y+z=1 \\ 2\alpha x+3y+z=1 \\ 3x+\alpha y+2z=\beta \end{cases}⎩⎨⎧​αx+2y+z=12αx+3y+z=13x+αy+2z=β​

We must find which statement is NOT correct.

1. Write the coefficient matrix

The coefficient matrix is

A=(α212α313α2).A=\begin{pmatrix} \alpha & 2 & 1\\ 2\alpha & 3 & 1\\ 3 & \alpha & 2 \end{pmatrix}.A=​α2α3​23α​112​​.

Its determinant will tell us when the system has a unique solution.

Compute:

det⁡(A)=α∣31α2∣−2∣2α132∣+1∣2α33α∣.\det(A)= \alpha\begin{vmatrix}3&1\\ \alpha&2\end{vmatrix} -2\begin{vmatrix}2\alpha&1\\ 3&2\end{vmatrix} +1\begin{vmatrix}2\alpha&3\\ 3&\alpha\end{vmatrix}.det(A)=α​3α​12​​−2​2α3​12​​+1​2α3​3α​​.

So,

det⁡(A)=α(6−α)−2(4α−3)+(2α2−9).\det(A)=\alpha(6-\alpha)-2(4\alpha-3)+(2\alpha^2-9).det(A)=α(6−α)−2(4α−3)+(2α2−9).

Simplifying,

det⁡(A)=6α−α2−8α+6+2α2−9=α2−2α−3.\det(A)=6\alpha-\alpha^2-8\alpha+6+2\alpha^2-9 =\alpha^2-2\alpha-3.det(A)=6α−α2−8α+6+2α2−9=α2−2α−3.

Hence

det⁡(A)=(α−3)(α+1).\det(A)=(\alpha-3)(\alpha+1).det(A)=(α−3)(α+1).

Therefore:

  • if α≠3,−1\alpha\neq 3,-1α=3,−1, the determinant is nonzero, so the system has a unique solution for every β\betaβ.
  • special cases are α=−1\alpha=-1α=−1 and α=3\alpha=3α=3.

2. Case: α=−1\alpha=-1α=−1

Then the system becomes

{−x+2y+z=1...(1)−2x+3y+z=1...(2)3x−y+2z=β...(3)\begin{cases} -x+2y+z=1 \quad ...(1)\\ -2x+3y+z=1 \quad ...(2)\\ 3x-y+2z=\beta \quad ...(3) \end{cases}⎩⎨⎧​−x+2y+z=1...(1)−2x+3y+z=1...(2)3x−y+2z=β...(3)​

Subtract (1) from (2):

(−2x+3y+z)−(−x+2y+z)=0(-2x+3y+z)-(-x+2y+z)=0(−2x+3y+z)−(−x+2y+z)=0 −x+y=0⇒y=x.-x+y=0 \Rightarrow y=x.−x+y=0⇒y=x.

Put y=xy=xy=x in (1):

−x+2x+z=1⇒x+z=1⇒z=1−x.-x+2x+z=1 \Rightarrow x+z=1 \Rightarrow z=1-x.−x+2x+z=1⇒x+z=1⇒z=1−x.

Now use (3):

3x−x+2(1−x)=β3x-x+2(1-x)=\beta3x−x+2(1−x)=β 2x+2−2x=β2x+2-2x=\beta2x+2−2x=β β=2.\beta=2.β=2.

So for α=−1\alpha=-1α=−1:

  • if β=2\beta=2β=2, the system is consistent (in fact infinitely many solutions),
  • if β≠2\beta\neq 2β=2, the system has no solution.

Thus:

  • A is correct.
  • B is correct.
  • C says no solution for α=−1\alpha=-1α=−1 and for all β∈R\beta\in\mathbb Rβ∈R, which is false because β=2\beta=2β=2 gives solutions.

3. Case: α=3\alpha=3α=3

Then the system becomes

{3x+2y+z=1...(1)6x+3y+z=1...(2)3x+3y+2z=β...(3)\begin{cases} 3x+2y+z=1 \quad ...(1)\\ 6x+3y+z=1 \quad ...(2)\\ 3x+3y+2z=\beta \quad ...(3) \end{cases}⎩⎨⎧​3x+2y+z=1...(1)6x+3y+z=1...(2)3x+3y+2z=β...(3)​

From (2)−2×-2\times−2×(1):

(6x+3y+z)−2(3x+2y+z)=1−2(6x+3y+z)-2(3x+2y+z)=1-2(6x+3y+z)−2(3x+2y+z)=1−2 −y−z=−1⇒y+z=1.-y-z=-1 \Rightarrow y+z=1.−y−z=−1⇒y+z=1.

From (3)−-−(1):

(3x+3y+2z)−(3x+2y+z)=β−1(3x+3y+2z)-(3x+2y+z)=\beta-1(3x+3y+2z)−(3x+2y+z)=β−1 y+z=β−1.y+z=\beta-1.y+z=β−1.

For consistency,

β−1=1⇒β=2.\beta-1=1 \Rightarrow \beta=2.β−1=1⇒β=2.

So for α=3\alpha=3α=3:

  • if β=2\beta=2β=2, system is consistent,
  • if β≠2\beta\neq 2β=2, system has no solution.

Thus D is correct.

4. Check all options

  • A: For all α≠−1\alpha\neq -1α=−1 and β=2\beta=2β=2 → true. If α≠3,−1\alpha\neq 3,-1α=3,−1, unique solution exists; if α=3\alpha=3α=3, still consistent when β=2\beta=2β=2.
  • B: No solution if α=−1\alpha=-1α=−1 and β≠2\beta\neq 2β=2 → true.
  • C: No solution for α=−1\alpha=-1α=−1 and for all β∈R\beta\in\mathbb Rβ∈R → false, because for β=2\beta=2β=2 there are solutions.
  • D: No solution for α=3\alpha=3α=3 and for all β≠2\beta\neq 2β=2 → true.

Hence the statement which is NOT correct is

C.\boxed{\text{C}}.C​.
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