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MCQMediumJEE 2023Solving Linear Equations (Matrix Method)

JEE Mathematics 2023 Question with Solution

If the system of equations x+y+az=bx+y+az = b, 2x+5y+2z=62x + 5y + 2z = 6, x+2y+3z=3x+2y+3z = 3 has infinitely many solutions, then 2a+3b2a+3b is equal to:

  • A

    2828

  • B

    2020

  • C

    2525

  • D

    2323

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The system of equations is

x+y+az=b2x+5y+2z=6x+2y+3z=3\begin{aligned} x+y+az &= b \\ 2x + 5y + 2z &= 6 \\ x+2y+3z &= 3 \end{aligned}

Find: The value of 2a+3b2a+3b when the system has infinitely many solutions.

For the system of equations to have infinitely many solutions, the determinant of the coefficient matrix Δ\Delta and the determinants of the matrices obtained by replacing each column with the constant terms Δx\Delta_x, Δy\Delta_y, Δz\Delta_z must all be equal to zero.

The coefficient matrix is

(11a252123)\begin{pmatrix} 1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3 \end{pmatrix}

So,

Δ=∣11a252123∣=1(15−4)−1(6−2)+a(4−5)=11−4−a=7−a\Delta = \begin{vmatrix} 1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3 \end{vmatrix} = 1(15 - 4) - 1(6 - 2) + a(4 - 5) = 11 - 4 - a = 7 - a

For infinitely many solutions,

Δ=0\Delta = 0

Hence,

7−a=0⇒a=77-a=0 \Rightarrow a=7

Now calculate Δx\Delta_x:

Δx=∣b17652323∣=b(15−4)−1(18−6)+7(12−15)=11b−12−21=11b−33\Delta_x = \begin{vmatrix} b & 1 & 7 \\ 6 & 5 & 2 \\ 3 & 2 & 3 \end{vmatrix} = b(15 - 4) - 1(18 - 6) + 7(12 - 15) = 11b - 12 - 21 = 11b - 33

For infinitely many solutions,

Δx=0\Delta_x = 0

Thus,

11b−33=0⇒b=311b - 33 = 0 \Rightarrow b = 3

Now,

2a+3b=2(7)+3(3)=14+9=232a + 3b = 2(7) + 3(3) = 14 + 9 = 23

Therefore, the correct option is D.

Common mistakes

  • Using only Δ=0\Delta = 0 and stopping there is incorrect because infinitely many solutions require consistency as well. After finding aa, also check a replaced-column determinant such as Δx=0\Delta_x = 0 to determine bb.

  • Expanding the determinant with the wrong signs is a common error. In cofactor expansion along the first row, the signs are +,−,++,-,+, so the middle term must be subtracted.

  • Confusing 'infinitely many solutions' with 'no unique solution' is incorrect. A zero determinant alone may also indicate no solution; for infinitely many solutions, the system must be dependent and consistent.

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