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

2022 · 26 Jul · Shift 1 · Q27
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. /2022 · 26 Jul · Shift 1 · Q27

Matrices and Determinants question

2022 · 26 Jul · Shift 1 · Q27

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If the system of linear equations. 8x+y+4z=−2x+y+z=0λx−3y=μ8x + y + 4z = - 2x + y + z = 0\lambda x - 3y = \mu8x+y+4z=−2x+y+z=0λx−3y=μ has infinitely many solutions, then the distance of the point (λ,μ,−12)\left( {\lambda ,\mu , - {1 \over 2}} \right)(λ,μ,−21​) from the plane 8x+y+4z+2=08x + y + 4z + 2 = 08x+y+4z+2=0 is :
  1. A
    353\sqrt 535​
  2. B
    4
  3. C
    269{{26} \over 9}926​
  4. D
    103{{10} \over 3}310​
View written solutionFree

Correct answer: NONE OF THE GIVEN OPTIONS; THE CORRECT DISTANCE IS \(\FRAC{128}{9}\).

  1. Interpret the system correctly

The given system is

{8x+y+4z=0−2x+y+z=0λx−3y=μ\begin{cases} 8x+y+4z=0\\ -2x+y+z=0\\ \lambda x-3y=\mu \end{cases}⎩⎨⎧​8x+y+4z=0−2x+y+z=0λx−3y=μ​

We are told that this system has infinitely many solutions.

For a system of 3 linear equations in 3 variables to have infinitely many solutions, the third equation must be dependent on the first two, and the coefficient matrix must be singular.

  1. Use dependence of equations

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

α(8x+y+4z)+β(−2x+y+z)=λx−3y+0⋅z\alpha(8x+y+4z)+\beta(-2x+y+z)=\lambda x-3y+0\cdot zα(8x+y+4z)+β(−2x+y+z)=λx−3y+0⋅z

Also, since the first two equations have RHS zero, the RHS of the third must also be zero for dependence. Hence,

μ=0.\mu=0.μ=0.

Now compare coefficients.

From

α(8x+y+4z)+β(−2x+y+z)\alpha(8x+y+4z)+\beta(-2x+y+z)α(8x+y+4z)+β(−2x+y+z)

we get coefficients:

  • of xxx: 8α−2β8\alpha-2\beta8α−2β
  • of yyy: α+β\alpha+\betaα+β
  • of zzz: 4α+β4\alpha+\beta4α+β

These must match the third equation

λx−3y+0z.\lambda x-3y+0z.λx−3y+0z.

So,

α+β=−3...(1)\alpha+\beta=-3 \qquad ...(1)α+β=−3...(1) 4α+β=0...(2)4\alpha+\beta=0 \qquad ...(2)4α+β=0...(2)

Subtract (1) from (2):

3α=3⇒α=13\alpha=3 \Rightarrow \alpha=13α=3⇒α=1

Then from (2):

4(1)+β=0⇒β=−44(1)+\beta=0 \Rightarrow \beta=-44(1)+β=0⇒β=−4

Hence,

λ=8α−2β=8−2(−4)=8+8=16.\lambda=8\alpha-2\beta=8-2(-4)=8+8=16.λ=8α−2β=8−2(−4)=8+8=16.

So,

(λ,μ,−1/2)=(16,0,−1/2).(\lambda,\mu,-1/2)=(16,0,-1/2).(λ,μ,−1/2)=(16,0,−1/2).
  1. Distance from point to plane

Plane is

8x+y+4z+2=0.8x+y+4z+2=0.8x+y+4z+2=0.

Distance of point P(x1,y1,z1)P(x_1,y_1,z_1)P(x1​,y1​,z1​) from plane Ax+By+Cz+D=0Ax+By+Cz+D=0Ax+By+Cz+D=0 is

∣Ax1+By1+Cz1+D∣A2+B2+C2.\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}.A2+B2+C2​∣Ax1​+By1​+Cz1​+D∣​.

Here,

A=8,  B=1,  C=4,  D=2,A=8,\; B=1,\; C=4,\; D=2,A=8,B=1,C=4,D=2,

and point is

(16,0,−1/2).(16,0,-1/2).(16,0,−1/2).

So distance

=∣8(16)+1(0)+4(−1/2)+2∣82+12+42=\frac{|8(16)+1(0)+4(-1/2)+2|}{\sqrt{8^2+1^2+4^2}}=82+12+42​∣8(16)+1(0)+4(−1/2)+2∣​ =∣128−2+2∣64+1+16=\frac{|128-2+2|}{\sqrt{64+1+16}}=64+1+16​∣128−2+2∣​ =12881=1289.=\frac{128}{\sqrt{81}}=\frac{128}{9}.=81​128​=9128​.
  1. Check against options

The computed distance is

1289\boxed{\frac{128}{9}}9128​​

which is not present among the options.

So the stored correct answer 103\boxed{\frac{10}{3}}310​​ does not match the result.

  1. Verification using determinant method

Coefficient matrix:

A=(814−211λ−30)A=\begin{pmatrix} 8&1&4\\ -2&1&1\\ \lambda&-3&0 \end{pmatrix}A=​8−2λ​11−3​410​​

For infinitely many solutions, determinant must be zero:

∣814−211λ−30∣=0\begin{vmatrix} 8&1&4\\ -2&1&1\\ \lambda&-3&0 \end{vmatrix}=0​8−2λ​11−3​410​​=0

Expanding,

8∣11−30∣−1∣−21λ0∣+4∣−21λ−3∣=08\begin{vmatrix}1&1\\-3&0\end{vmatrix}-1\begin{vmatrix}-2&1\\\lambda&0\end{vmatrix}+4\begin{vmatrix}-2&1\\\lambda&-3\end{vmatrix}=08​1−3​10​​−1​−2λ​10​​+4​−2λ​1−3​​=0 =8(3)−1(−λ)+4(6−λ)=0=8(3)-1(-\lambda)+4(6-\lambda)=0=8(3)−1(−λ)+4(6−λ)=0 24+λ+24−4λ=024+\lambda+24-4\lambda=024+λ+24−4λ=0 48−3λ=048-3\lambda=048−3λ=0 λ=16.\lambda=16.λ=16.

Again, since first two equations are homogeneous, for consistency in dependent case we need

μ=0.\mu=0.μ=0.

Thus same result follows.

Hence the correct distance is

1289.\boxed{\frac{128}{9}}.9128​​.
PreviousNext

More from Matrices and Determinants

  •  Let A=​111​​ and B=​92122−152​−102132162​112−142172​​, then the value of A′BA is: …2022 · MCQ
  • The number of matrices A=(ac​bd​), where a,b,c,d∈{−1,0,1,2,3,……,10}, such that A=A−1, is ​.2022 · Numerical
  • The ordered pair (a, b), for which the system of linear equations 3x − 2y + z = b 5x − 8y + 9z = 3 2x + y + az =− 1 has no solution, is :2022 · MCQ
  • If the system of equations α x + y + z = 5, x + 2y + 3z = 4, x + 3y + 5z = β has infinitely many solutions, then the ordered pair (α, β) is equal to :2022 · MCQ
  • Let X=​000​100​010​​,Y=αI+βX+γX2 and $$Z = {\alpha ^2}I - \alpha \beta X + ({\beta ^2} - \alpha \gamma…2022 · Numerical
  • Let A=(1−2​2−5​). Let α,β∈R be such that αA2+βA=2I. Then α+β is equal to2022 · MCQ
  • Let S be the set containing all 3×3 matrices with entries from {−1,0,1}. The total number of matrices A∈S such that the sum of all the diagonal elements of ATA is 6 is ​.2022 · Numerical
  • Let the system of linear equations x+2y+z=2, αx+3y−z=α, −αx+y+2z=−α be inconsistent. Then α is equal to :2022 · MCQ