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MCQMediumJEE 2023Solving Linear Equations (Matrix Method)

JEE Mathematics 2023 Question with Solution

For the system of linear equations 2x+4y+2az=b2x + 4y + 2az = b, x+2y+3z=4x + 2y + 3z = 4, 2x−5y+2z=82x - 5y + 2z = 8, which of the following is NOT correct?

  • A

    It has infinitely many solutions if a=3,b=8a = 3, b = 8

  • B

    It has unique solution if a=b=8a = b = 8

  • C

    It has unique solution if a=b=6a = b = 6

  • D

    It has infinitely many solutions if a=3,b=6a = 3, b = 6

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

2x+4y+2az=b2x + 4y + 2az = b x+2y+3z=4x + 2y + 3z = 4 2x−5y+2z=82x - 5y + 2z = 8

Find: Which statement is not correct.

Let the coefficient matrix be

A=∣242a1232−52∣,Δ=det⁡(A)A = \begin{vmatrix} 2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & -5 & 2 \end{vmatrix}, \quad \Delta = \det(A)

Expanding along the first row,

Δ=2∣23−52∣−4∣1322∣+2a∣122−5∣\Delta = 2\begin{vmatrix} 2 & 3 \\ -5 & 2 \end{vmatrix} - 4\begin{vmatrix} 1 & 3 \\ 2 & 2 \end{vmatrix} + 2a\begin{vmatrix} 1 & 2 \\ 2 & -5 \end{vmatrix}

Evaluating the minors,

∣23−52∣=4+15=19\begin{vmatrix} 2 & 3 \\ -5 & 2 \end{vmatrix} = 4 + 15 = 19 ∣1322∣=1⋅2−3⋅2=2−6=−4\begin{vmatrix} 1 & 3 \\ 2 & 2 \end{vmatrix} = 1 \cdot 2 - 3 \cdot 2 = 2 - 6 = -4 ∣122−5∣=(1)(−5)−(2)(2)=−5−4=−9\begin{vmatrix} 1 & 2 \\ 2 & -5 \end{vmatrix} = (1)(-5) - (2)(2) = -5 - 4 = -9

So,

Δ=2⋅19−4⋅(−4)+2a⋅(−9)=38+16−18a=54−18a=18(3−a)\Delta = 2 \cdot 19 - 4 \cdot (-4) + 2a \cdot (-9) = 38 + 16 - 18a = 54 - 18a = 18(3 - a)

For a unique solution, we need

Δ≠0\Delta \neq 0

Hence,

a≠3a \neq 3

Therefore statements B and C are correct, since for a=8a = 8 and a=6a = 6, respectively, the determinant is non-zero.

For infinitely many solutions, we require

Δ=0\Delta = 0

so

a=3a = 3

Then consistency conditions on the augmented matrix must also hold. From the given working, when a=3a = 3 and b=8b = 8, the system has infinitely many solutions, so statement A is correct.

For statement D, when a=3a = 3 and b=6b = 6, the system does not have infinitely many solutions. Hence statement D is the one which is not correct.

Therefore, the correct option is D.

Using determinant criterion

Given: A system of three linear equations in three variables with parameters aa and bb.

Find: The incorrect statement among the four options.

The determinant of the coefficient matrix decides whether the system has a unique solution:

Δ=18(3−a)\Delta = 18(3-a)

If

Δ≠0\Delta \neq 0

then the system has a unique solution, and if

Δ=0\Delta = 0

then we must further check consistency.

So for

a=8a = 8

and for

a=6a = 6

we get

Δ≠0\Delta \neq 0

Therefore options B and C are true.

When

a=3a = 3

we get

Δ=0\Delta = 0

Now the nature of the solution depends on bb. The extracted solution states that b=8b = 8 gives infinitely many solutions, but b=6b = 6 does not. Therefore option A is true and option D is false.

Hence, the statement that is not correct is D.

Common mistakes

  • Students often conclude that Δ=0\Delta = 0 automatically means infinitely many solutions. This is wrong because Δ=0\Delta = 0 can also correspond to an inconsistent system. After getting a=3a = 3, check the augmented matrix or consistency condition as well.

  • A common mistake is evaluating the determinant incorrectly by missing the sign of the middle cofactor term. In expansion along the first row, the second term carries a negative sign. Preserve the cofactor signs carefully.

  • Some students verify only the value of aa for options involving unique solutions and ignore that the options state both aa and bb. Here uniqueness depends on the coefficient matrix, so once a≠3a \neq 3, the value of bb does not affect uniqueness.

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