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

2020 · 7 Jan · Shift 1 · Q29
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. /2020 · 7 Jan · Shift 1 · Q29

Matrices and Determinants question

2020 · 7 Jan · Shift 1 · Q29

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If the system of linear equations 2x + 2ay + az = 0 2x + 3by + bz = 0 2x + 4cy + cz = 0, where a, b, c ∈\in∈ R are non-zero distinct; has a non-zero solution, then:
  1. A
    1a,1b,1c{1 \over a},{1 \over b},{1 \over c}a1​,b1​,c1​ are in A.P.
  2. B
    a + b + c = 0
  3. C
    a, b, c are in G.P.
  4. D
    a,b,c are in A.P.
View written solutionFree

Correct answer: A

  1. Write the homogeneous system in matrix form

The system is

{2x+2ay+az=02x+3by+bz=02x+4cy+cz=0\begin{cases} 2x+2ay+az=0\\ 2x+3by+bz=0\\ 2x+4cy+cz=0 \end{cases}⎩⎨⎧​2x+2ay+az=02x+3by+bz=02x+4cy+cz=0​

This can be written as

[22aa23bb24cc][xyz]=[000].\begin{bmatrix} 2 & 2a & a\\ 2 & 3b & b\\ 2 & 4c & c \end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}.​222​2a3b4c​abc​​​xyz​​=​000​​.

Since this is a homogeneous system, for it to have a non-zero solution, the determinant of the coefficient matrix must be zero.

  1. Compute the determinant

Let

Δ=∣22aa23bb24cc∣.\Delta= \begin{vmatrix} 2 & 2a & a\\ 2 & 3b & b\\ 2 & 4c & c \end{vmatrix}.Δ=​222​2a3b4c​abc​​.

Take out common factor 222 from the first column:

Δ=2∣12aa13bb14cc∣.\Delta=2 \begin{vmatrix} 1 & 2a & a\\ 1 & 3b & b\\ 1 & 4c & c \end{vmatrix}.Δ=2​111​2a3b4c​abc​​.

Now apply row operations:

R2→R2−R1,R3→R3−R1.R_2\to R_2-R_1,\qquad R_3\to R_3-R_1.R2​→R2​−R1​,R3​→R3​−R1​.

Then

Δ=2∣12aa03b−2ab−a04c−2ac−a∣.\Delta=2 \begin{vmatrix} 1 & 2a & a\\ 0 & 3b-2a & b-a\\ 0 & 4c-2a & c-a \end{vmatrix}.Δ=2​100​2a3b−2a4c−2a​ab−ac−a​​.

Expand along the first column:

Δ=2∣3b−2ab−a4c−2ac−a∣.\Delta=2 \begin{vmatrix} 3b-2a & b-a\\ 4c-2a & c-a \end{vmatrix}.Δ=2​3b−2a4c−2a​b−ac−a​​.

So

Δ=2((3b−2a)(c−a)−(b−a)(4c−2a)).\Delta=2\Big((3b-2a)(c-a)-(b-a)(4c-2a)\Big).Δ=2((3b−2a)(c−a)−(b−a)(4c−2a)).

Expand:

(3b−2a)(c−a)=3bc−3ab−2ac+2a2,(3b-2a)(c-a)=3bc-3ab-2ac+2a^2,(3b−2a)(c−a)=3bc−3ab−2ac+2a2, (b−a)(4c−2a)=4bc−2ab−4ac+2a2.(b-a)(4c-2a)=4bc-2ab-4ac+2a^2.(b−a)(4c−2a)=4bc−2ab−4ac+2a2.

Therefore,

(3b−2a)(c−a)−(b−a)(4c−2a)=3bc−3ab−2ac+2a2−(4bc−2ab−4ac+2a2).(3b-2a)(c-a)-(b-a)(4c-2a) =3bc-3ab-2ac+2a^2-(4bc-2ab-4ac+2a^2).(3b−2a)(c−a)−(b−a)(4c−2a)=3bc−3ab−2ac+2a2−(4bc−2ab−4ac+2a2).

Simplify:

=−bc−ab+2ac.= -bc-ab+2ac.=−bc−ab+2ac.

Hence

Δ=2(2ac−ab−bc).\Delta=2(2ac-ab-bc).Δ=2(2ac−ab−bc).

For a non-zero solution,

Δ=0  ⟹  2ac−ab−bc=0.\Delta=0 \implies 2ac-ab-bc=0.Δ=0⟹2ac−ab−bc=0.
  1. Rewrite the condition

Given a,b,c≠0a,b,c\neq 0a,b,c=0, divide by abcabcabc:

2ac−ab−bcabc=0.\frac{2ac-ab-bc}{abc}=0.abc2ac−ab−bc​=0.

Thus,

2b−1c−1a=0.\frac{2}{b}-\frac{1}{c}-\frac{1}{a}=0.b2​−c1​−a1​=0.

Rearrange:

1a+1c=2b.\frac{1}{a}+\frac{1}{c}=\frac{2}{b}.a1​+c1​=b2​.

This is exactly the condition that

1a,1b,1c\frac{1}{a},\frac{1}{b},\frac{1}{c}a1​,b1​,c1​

are in arithmetic progression.

So Option A is true.

  1. Check the other options
  • B: a+b+c=0a+b+c=0a+b+c=0 is not implied by the determinant condition.
  • C: a,b,ca,b,ca,b,c in G.P. would mean b2=acb^2=acb2=ac, which is not obtained here.
  • D: a,b,ca,b,ca,b,c in A.P. would mean 2b=a+c2b=a+c2b=a+c, also not obtained here.

Hence these are not necessarily true.

  1. Final answer

The correct option is

A\boxed{\text{A}}A​

since

1a,1b,1c\frac{1}{a},\frac{1}{b},\frac{1}{c}a1​,b1​,c1​

are in A.P.

PreviousNext

More from Matrices and Determinants

  • If the system of linear equations, x + y + z = 6 x + 2y + 3z = 10 3x + 2y + λ z = μ has more than two solutions, then μ-λ 2 is equal to ​.2020 · Numerical
  • The number of all 3 × 3 matrices A, with enteries from the set {–1, 0, 1} such that the sum of the diagonal elements of AAT is 3, is2020 · Numerical
  • For which of the following ordered pairs (μ, δ), the system of linear equations x + 2y + 3z = 1 3x + 4y + 5z = μ 4x + 4y + 4z =δ is inconsistent ?2020 · MCQ
  • If A=(29​24​) and I=(10​01​) then 10A–1 is equal to :2020 · MCQ
  • The system of linear equations λ x + 2y + 2z = 5 2 λ x + 3y + 5z = 8 4x + λ y + 6z = 10 has2020 · MCQ
  • If for some α and β in R, the intersection of the following three places x + 4y – 2z = 1 x + 7y – 5z = b x + 5y +α z = 5 is a line in R3, then α+β is equal to :2020 · MCQ
  • The following system of linear equations 7x + 6y – 2z = 0 3x + 4y + 2z = 0 x – 2y – 6z = 0, has2020 · MCQ
  • Let A=(cosαsinα​−sinαcosα​), (α∈ R) such that A32=(01​−10​)…2019 · MCQ