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Matrices and Determinants question

2025 · 7 Apr · Shift 2 · Q44
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Matrices and Determinants question

2025 · 7 Apr · Shift 2 · Q44

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let the system of equations x + 5y - z = 1 4x + 3y - 3z = 7 24x + y + λz = μ λ, μ ∈ ℝ, have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy 7 ≤ x + y + z ≤ 77, is :
  1. A
    4
  2. B
    5
  3. C
    3
  4. D
    6
View written solutionFree

Correct answer: C

  1. For infinitely many solutions, the third equation must be dependent on the first two.

The system is:

{x+5y−z=14x+3y−3z=724x+y+λz=μ\begin{cases} x+5y-z=1 \\ 4x+3y-3z=7 \\ 24x+y+\lambda z=\mu \end{cases}⎩⎨⎧​x+5y−z=14x+3y−3z=724x+y+λz=μ​

Let the third equation be a linear combination of the first two:

α(x+5y−z)+β(4x+3y−3z)=24x+y+λz\alpha(x+5y-z)+\beta(4x+3y-3z)=24x+y+\lambda zα(x+5y−z)+β(4x+3y−3z)=24x+y+λz

Comparing coefficients:

α+4β=24\alpha+4\beta=24α+4β=24 5α+3β=15\alpha+3\beta=15α+3β=1

Solve these: From α=24−4β\alpha=24-4\betaα=24−4β,

5(24−4β)+3β=15(24-4\beta)+3\beta=15(24−4β)+3β=1 120−20β+3β=1120-20\beta+3\beta=1120−20β+3β=1 −17β=−119⇒β=7-17\beta=-119 \Rightarrow \beta=7−17β=−119⇒β=7

Then

α=24−28=−4\alpha=24-28=-4α=24−28=−4

Now coefficient of zzz:

λ=−α−3β=−(−4)−21=4−21=−17\lambda = -\alpha-3\beta = -(-4)-21=4-21=-17λ=−α−3β=−(−4)−21=4−21=−17

And RHS:

μ=α(1)+β(7)=−4+49=45\mu=\alpha(1)+\beta(7)=-4+49=45μ=α(1)+β(7)=−4+49=45

So for infinitely many solutions,

λ=−17,μ=45\lambda=-17,\quad \mu=45λ=−17,μ=45
  1. Reduce the system to two independent equations.

We only need to solve:

x+5y−z=1...(1)x+5y-z=1 \quad ...(1)x+5y−z=1...(1) 4x+3y−3z=7...(2)4x+3y-3z=7 \quad ...(2)4x+3y−3z=7...(2)

From (1):

x=1−5y+zx=1-5y+zx=1−5y+z

Substitute into (2):

4(1−5y+z)+3y−3z=74(1-5y+z)+3y-3z=74(1−5y+z)+3y−3z=7 4−20y+4z+3y−3z=74-20y+4z+3y-3z=74−20y+4z+3y−3z=7 4−17y+z=74-17y+z=74−17y+z=7 z=3+17yz=3+17yz=3+17y

Then

x=1−5y+(3+17y)=4+12yx=1-5y+(3+17y)=4+12yx=1−5y+(3+17y)=4+12y

Hence all solutions are:

x=4+12y,z=3+17y,x=4+12y,\quad z=3+17y,x=4+12y,z=3+17y,

with y∈Zy\in \mathbb Zy∈Z for integer solutions.

  1. Apply the condition 7≤x+y+z≤777\le x+y+z\le 777≤x+y+z≤77.

Compute:

x+y+z=(4+12y)+y+(3+17y)=7+30yx+y+z=(4+12y)+y+(3+17y)=7+30yx+y+z=(4+12y)+y+(3+17y)=7+30y

So,

7≤7+30y≤777\le 7+30y\le 777≤7+30y≤77

Subtract 7:

0≤30y≤700\le 30y\le 700≤30y≤70 0≤y≤730\le y\le \frac{7}{3}0≤y≤37​

Since yyy is an integer,

y=0,1,2y=0,1,2y=0,1,2

Thus the corresponding integer solutions are exactly 3.

  1. Check options.
  • A: 444 ❌
  • B: 555 ❌
  • C: 333 ✅
  • D: 666 ❌

Therefore, the correct option is:

C\boxed{\text{C}}C​
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