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Matrices and Determinants question

2022 · 26 Jun · Shift 1 · Q23
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Matrices and Determinants question

2022 · 26 Jun · Shift 1 · Q23

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
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 :
  1. A
    (3,13)\left( {3,{1 \over 3}} \right)(3,31​)
  2. B
    (−3,13)\left( { - 3,{1 \over 3}} \right)(−3,31​)
  3. C
    (−3,−13)\left( { - 3, - {1 \over 3}} \right)(−3,−31​)
  4. D
    (3,−13)\left( {3, - {1 \over 3}} \right)(3,−31​)
View written solutionFree

Correct answer: C

  1. Write the system in matrix form

The system is

{3x−2y+z=b5x−8y+9z=32x+y+az=−1\begin{cases} 3x-2y+z=b \\ 5x-8y+9z=3 \\ 2x+y+az=-1 \end{cases}⎩⎨⎧​3x−2y+z=b5x−8y+9z=32x+y+az=−1​

Its coefficient matrix is

A=(3−215−8921a).A=\begin{pmatrix} 3 & -2 & 1\\ 5 & -8 & 9\\ 2 & 1 & a \end{pmatrix}.A=​352​−2−81​19a​​.

For the system to have no solution, we need:

  • det⁡(A)=0\det(A)=0det(A)=0 so that the system is not uniquely solvable, and
  • the augmented system must be inconsistent.
  1. Find when det⁡(A)=0\det(A)=0det(A)=0

Compute

det⁡(A)=∣3−215−8921a∣.\det(A)=\begin{vmatrix} 3 & -2 & 1\\ 5 & -8 & 9\\ 2 & 1 & a \end{vmatrix}.det(A)=​352​−2−81​19a​​.

Expanding along the first row,

det⁡(A)=3∣−891a∣−(−2)∣592a∣+1∣5−821∣.\det(A)=3\begin{vmatrix}-8 & 9\\ 1 & a\end{vmatrix}-(-2)\begin{vmatrix}5 & 9\\ 2 & a\end{vmatrix}+1\begin{vmatrix}5 & -8\\ 2 & 1\end{vmatrix}.det(A)=3​−81​9a​​−(−2)​52​9a​​+1​52​−81​​.

So,

det⁡(A)=3(−8a−9)+2(5a−18)+(5+16).\det(A)=3(-8a-9)+2(5a-18)+(5+16).det(A)=3(−8a−9)+2(5a−18)+(5+16). =−24a−27+10a−36+21= -24a-27+10a-36+21=−24a−27+10a−36+21 =−14a−42= -14a-42=−14a−42 =−14(a+3).= -14(a+3).=−14(a+3).

Thus,

det⁡(A)=0  ⟺  a=−3.\det(A)=0 \iff a=-3.det(A)=0⟺a=−3.

So only options with a=−3a=-3a=−3 are possible: B and C.

  1. Substitute a=−3a=-3a=−3 and test consistency

Now the system becomes

{3x−2y+z=b5x−8y+9z=32x+y−3z=−1\begin{cases} 3x-2y+z=b \\ 5x-8y+9z=3 \\ 2x+y-3z=-1 \end{cases}⎩⎨⎧​3x−2y+z=b5x−8y+9z=32x+y−3z=−1​

We check whether the third equation is a linear combination of the first two on the left-hand side.

Let

λ(3x−2y+z)+μ(5x−8y+9z)=2x+y−3z.\lambda(3x-2y+z)+\mu(5x-8y+9z)=2x+y-3z.λ(3x−2y+z)+μ(5x−8y+9z)=2x+y−3z.

Then coefficients must satisfy

3λ+5μ=2,3\lambda+5\mu=2,3λ+5μ=2, −2λ−8μ=1,-2\lambda-8\mu=1,−2λ−8μ=1, λ+9μ=−3.\lambda+9\mu=-3.λ+9μ=−3.

Solve the first two:

From

3λ+5μ=2,3\lambda+5\mu=2,3λ+5μ=2, −2λ−8μ=1,-2\lambda-8\mu=1,−2λ−8μ=1,

we get μ=−1\mu=-1μ=−1 and λ=73\lambda=\frac{7}{3}λ=37​.

Check in the third:

λ+9μ=73−9=73−273=−203≠−3.\lambda+9\mu=\frac{7}{3}-9=\frac{7}{3}-\frac{27}{3}=-\frac{20}{3}\neq -3.λ+9μ=37​−9=37​−327​=−320​=−3.

So this direct combination does not match; instead, reduce rows to find the dependence relation among equations.

  1. Row reduction of the coefficient matrix for a=−3a=-3a=−3
(3−215−8921−3)\begin{pmatrix} 3 & -2 & 1\\ 5 & -8 & 9\\ 2 & 1 & -3 \end{pmatrix}​352​−2−81​19−3​​

Apply row operations:

R2→3R2−5R1,R3→3R3−2R1R_2 \to 3R_2-5R_1, \qquad R_3 \to 3R_3-2R_1R2​→3R2​−5R1​,R3​→3R3​−2R1​

Then

R2=(0,−14,22),R3=(0,7,−11).R_2=(0,-14,22), \qquad R_3=(0,7,-11).R2​=(0,−14,22),R3​=(0,7,−11).

So the matrix becomes

(3−210−142207−11).\begin{pmatrix} 3 & -2 & 1\\ 0 & -14 & 22\\ 0 & 7 & -11 \end{pmatrix}.​300​−2−147​122−11​​.

Now observe

R2=−2R3.R_2=-2R_3.R2​=−2R3​.

Hence the second and third equations become dependent when a=−3a=-3a=−3.

So for consistency, the constants must satisfy the same relation.

  1. Apply the same operations to the constants

Original constants are

(b3−1).\begin{pmatrix} b\\ 3\\ -1 \end{pmatrix}.​b3−1​​.

Under the same operations,

R2→3R2−5R1=9−5b,R_2 \to 3R_2-5R_1 = 9-5b,R2​→3R2​−5R1​=9−5b, R3→3R3−2R1=−3−2b.R_3 \to 3R_3-2R_1 = -3-2b.R3​→3R3​−2R1​=−3−2b.

For consistency, since R2=−2R3R_2=-2R_3R2​=−2R3​ on the left side, we must also have

9−5b=−2(−3−2b)=6+4b.9-5b=-2(-3-2b)=6+4b.9−5b=−2(−3−2b)=6+4b.

Thus,

9−5b=6+4b9-5b=6+4b9−5b=6+4b 3=9b3=9b3=9b b=13.b=\frac{1}{3}.b=31​.

So:

  • if b=13b=\frac13b=31​, the system is consistent with infinitely many solutions,
  • if b≠13b\neq \frac13b=31​, the system is inconsistent and has no solution.

Therefore for no solution,

a=−3,b≠13.a=-3, \quad b\neq \frac13.a=−3,b=31​.

Among the options, this gives

(a,b)=(−3,−13).(a,b)=\left(-3,-\frac13\right).(a,b)=(−3,−31​).
  1. Check options
  • A: (3,13)\left(3,\frac13\right)(3,31​) → a≠−3a\neq -3a=−3, so unique solution, not no solution.
  • B: (−3,13)\left(-3,\frac13\right)(−3,31​) → consistent, infinitely many solutions.
  • C: (−3,−13)\left(-3,-\frac13\right)(−3,−31​) → inconsistent, no solution.
  • D: (3,−13)\left(3,-\frac13\right)(3,−31​) → a≠−3a\neq -3a=−3, so unique solution.

Hence the correct option is C.

  1. Comparison with stored answer

Stored correct answer: C

My derived answer: C

They agree.

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