Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Matrices and Determinants question

2024 · 29 Jan · Shift 2 · Q56
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Matrices and Determinants
  5. /2024 · 29 Jan · Shift 2 · Q56

Matrices and Determinants question

2024 · 29 Jan · Shift 2 · Q56

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
Let for any three distinct consecutive terms a,b,ca, b, ca,b,c of an A.P, the lines ax+by+c=0a x+b y+c=0ax+by+c=0 be concurrent at the point PPP and Q(α,β)Q(\alpha, \beta)Q(α,β) be a point such that the system of equations x+y+z=6,2x+5y+αz=β and \begin{aligned} & x+y+z=6, \\ & 2 x+5 y+\alpha z=\beta \text { and } \end{aligned}​x+y+z=6,2x+5y+αz=β and ​ x+2y+3z=4x+2 y+3 z=4x+2y+3z=4, has infinitely many solutions. Then (PQ)2(P Q)^2(PQ)2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 113

  1. Family of concurrent lines from three consecutive A.P. terms

Let three distinct consecutive terms of an A.P. be a=d−r,b=d,c=d+r,a=d-r,\quad b=d,\quad c=d+r,a=d−r,b=d,c=d+r, with r≠0r\neq 0r=0.

The given line is ax+by+c=0.ax+by+c=0.ax+by+c=0. So for any three consecutive terms, the line becomes (d−r)x+dy+(d+r)=0.(d-r)x+dy+(d+r)=0.(d−r)x+dy+(d+r)=0. Rearrange: d(x+y+1)+r(−x+1)=0.d(x+y+1)+r(-x+1)=0.d(x+y+1)+r(−x+1)=0.

Since this line is said to be concurrent at a fixed point P(x0,y0)P(x_0,y_0)P(x0​,y0​) for all such choices of consecutive terms, the above must hold for arbitrary ddd and rrr. Hence both coefficients must vanish: x0+y0+1=0,x_0+y_0+1=0,x0​+y0​+1=0, −x0+1=0.-x_0+1=0.−x0​+1=0. Thus,

\qquad y_0=-2.$$ So, $$P=(1,-2).$$ --- 2. **Condition for infinitely many solutions of the system** The system is $$x+y+z=6,$$ $$2x+5y+\alpha z=\beta,$$ $$x+2y+3z=4.$$ For infinitely many solutions, the three equations must be dependent and consistent. So one row must be a linear combination of the other two. Let $$\lambda(x+y+z=6)+\mu(x+2y+3z=4)=(2x+5y+\alpha z=\beta).$$ Comparing coefficients: $$\lambda+\mu=2,$$ $$\lambda+2\mu=5.$$ Subtracting, $$\mu=3,$$ so $$\lambda= -1.$$ Now compare the coefficient of $z$: $$\alpha=\lambda+3\mu=-1+9=8.$$ And the constant term: $$\beta=6\lambda+4\mu=6(-1)+4(3)=6.$$ Hence, $$Q=(\alpha,\beta)=(8,6).$$ --- 3. **Compute $PQ^2$** $$P=(1,-2),\qquad Q=(8,6).$$ Therefore, $$PQ^2=(8-1)^2+(6-(-2))^2=7^2+8^2=49+64=113.$$ --- 4. **Final answer** $$\boxed{113}$$ The derived answer matches the stored correct answer.
PreviousNext

More from Matrices and Determinants

  • Consider the system of linear equations x+y+z=4μ,x+2y+2λz=10μ,x+3y+4λ2z=μ2+15 where λ,μ∈R. Which one of the following statements is NOT correct ?2024 · MCQ
  • Let R=​x00​0y0​00z​​ be a non-zero 3×3 matrix, where xsinθ=ysin(θ+32π​)=zsin(θ+34π​)eq0,θ∈(0,2π)…2024 · MCQ
  • Consider the system of linear equations x+y+z=5,x+2y+λ2z=9,x+3y+λz=μ, where λ,μ∈R. Then, which of the following statement is NOT correct?2024 · MCQ
  • If the system of linear equations ​x−2y+z=−42x+αy+3z=53x−y+βz=3​ has infinitely many solutions, then 12α+13β is equal to2024 · MCQ
  • Let A be a 3×3 real matrix such that A​101​​=2​101​​,A​−101​​=4​−101​​,A​010​​=2​010​​. …2024 · MCQ
  • Let A be a 3×3 matrix and det(A)=2. If n=det(2024− times adj(adj(…..(adjA))​​)), then the remainder when n is divided…2024 · Numerical
  • Let S denote the set of all real values of λ such that the system of equations λx+y+z=1x+λy+z=1x+y+λz=1 is inconsistent, then ∑λ∈S​(∣λ∣2+∣λ∣) is equal to2023 · MCQ
  • For the system of linear equations αx+y+z=1,x+αy+z=1,x+y+αz=β, which one of the following statements is NOT correct?2023 · MCQ