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

2018 · 16 Apr · Shift 1 · Q34
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. /2018 · 16 Apr · Shift 1 · Q34

Matrices and Determinants question

2018 · 16 Apr · Shift 1 · Q34

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
The number of values of k for which the system of linear equations, (k + 2)x + 10y = k kx + (k +3)y = k -1 has no solution, is :
  1. A
    1
  2. B
    2
  3. C
    3
  4. D
    infinitely many
View written solutionFree

Correct answer: A

  1. Write the system in standard form

The given system is

(k+2)x+10y=k(k+2)x+10y=k(k+2)x+10y=k kx+(k+3)y=k−1kx+(k+3)y=k-1kx+(k+3)y=k−1

For a pair of linear equations

a1x+b1y=c1,a2x+b2y=c2,a_1x+b_1y=c_1, \qquad a_2x+b_2y=c_2,a1​x+b1​y=c1​,a2​x+b2​y=c2​,

there is no solution when

a1a2=b1b2≠c1c2,\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2},a2​a1​​=b2​b1​​=c2​c1​​,

provided the denominators are nonzero. Equivalently, the coefficient matrix must be singular, and the augmented system must be inconsistent.

  1. Form the coefficient matrix determinant

The coefficient matrix is

(k+210kk+3)\begin{pmatrix} k+2 & 10\\ k & k+3 \end{pmatrix}(k+2k​10k+3​)

Its determinant is

Δ=(k+2)(k+3)−10k.\Delta=(k+2)(k+3)-10k.Δ=(k+2)(k+3)−10k.

Now simplify:

Δ=k2+5k+6−10k=k2−5k+6.\Delta=k^2+5k+6-10k=k^2-5k+6.Δ=k2+5k+6−10k=k2−5k+6.

Factorize:

Δ=(k−2)(k−3).\Delta=(k-2)(k-3).Δ=(k−2)(k−3).

For no solution, first we need

Δ=0⇒k=2 or k=3.\Delta=0 \Rightarrow k=2 \text{ or } k=3.Δ=0⇒k=2 or k=3.
  1. Check each value for inconsistency

Case 1: k=2k=2k=2

The equations become

4x+10y=24x+10y=24x+10y=2 2x+5y=12x+5y=12x+5y=1

The first equation is exactly twice the second, since

2(2x+5y=1)⇒4x+10y=2.2(2x+5y=1) \Rightarrow 4x+10y=2.2(2x+5y=1)⇒4x+10y=2.

So both equations represent the same line.

Hence, for k=2k=2k=2, the system has infinitely many solutions, not no solution.

Case 2: k=3k=3k=3

The equations become

5x+10y=35x+10y=35x+10y=3 3x+6y=23x+6y=23x+6y=2

Now compare ratios:

a1a2=53,b1b2=106=53,c1c2=32.\frac{a_1}{a_2}=\frac{5}{3}, \qquad \frac{b_1}{b_2}=\frac{10}{6}=\frac{5}{3}, \qquad \frac{c_1}{c_2}=\frac{3}{2}.a2​a1​​=35​,b2​b1​​=610​=35​,c2​c1​​=23​.

Since

a1a2=b1b2≠c1c2,\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2},a2​a1​​=b2​b1​​=c2​c1​​,

the lines are parallel and distinct.

Hence, for k=3k=3k=3, the system has no solution.

  1. Count the number of such values

Only one value of kkk gives no solution, namely

k=3.k=3.k=3.

So the number of values is

1.\boxed{1}.1​.

Thus the correct option is A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.

PreviousNext

More from Matrices and Determinants

  • If the system of linear equations x + ky + 3z = 0 3x + ky - 2z = 0 2x + 4y - 3z = 0 has a non-zero solution (x, y, z), then y2xz​ is equal to2018 · MCQ
  • If ​x−42x2x​2xx−42x​2x2xx−4​​=(A+Bx)(x−A)2 then the ordered pair (A, B) is…2018 · MCQ
  • The number of real values of λ for which the system of linear equations 2x + 4y −λ z = 0 4x + λ y + 2z = 0 λ x + 2y + 2z = 0 has infinitely many solutions, is :2017 · MCQ
  • Let A be any 3 × 3 invertible matrix. Then which one of the following is not always true ?2017 · MCQ
  • If S=⎩⎨⎧​x∈[0,2π]:​0sinxcosx​cosx0sinx​−sinxcosx0​​=0⎭⎬⎫​, then x∈S∑​tan(3π​+x)…2017 · MCQ
  • For two 3 × 3 matrices A and B, let A + B = 2BT and 3A + 2B = I3, where BT is the transpose of B and I3 is 3 × 3 identity matrix. Then :2017 · MCQ
  • If S is the set of distinct values of 'b' for which the following system of linear equations x + y + z = 1 x + ay + z = 1 ax + by + z = 0 has no solution, then S is :2017 · MCQ
  • If A=[2−4​−31​], then adj(3A2 + 12A) is equal to2017 · MCQ