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Matrices and Determinants question

2020 · 2 Sep · Shift 1 · Q20
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  5. /2020 · 2 Sep · Shift 1 · Q20

Matrices and Determinants question

2020 · 2 Sep · Shift 1 · Q20

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let S be the set of all λ∈\lambda \inλ∈ R for which the system of linear equations 2x – y + 2z = 2 x – 2y +λ\lambdaλ z = –4 x + λ\lambdaλ y + z = 4 has no solution. Then the set S :
  1. A
    contains more than two elements.
  2. B
    contains exactly two elements.
  3. C
    is a singleton.
  4. D
    is an empty set.
View written solutionFree

Correct answer: B

We need the values of λ∈R\lambda \in \mathbb{R}λ∈R for which the system

{2x−y+2z=2x−2y+λz=−4x+λy+z=4\begin{cases} 2x-y+2z=2 \\ x-2y+\lambda z=-4 \\ x+\lambda y+z=4 \end{cases}⎩⎨⎧​2x−y+2z=2x−2y+λz=−4x+λy+z=4​

has no solution.

1. Write the augmented matrix

The coefficient matrix is

A=(2−121−2λ1λ1),A=\begin{pmatrix} 2 & -1 & 2\\ 1 & -2 & \lambda\\ 1 & \lambda & 1 \end{pmatrix},A=​211​−1−2λ​2λ1​​,

and the augmented matrix is

[2−1221−2λ−41λ14].\left[ \begin{array}{ccc|c} 2 & -1 & 2 & 2\\ 1 & -2 & \lambda & -4\\ 1 & \lambda & 1 & 4 \end{array} \right].​211​−1−2λ​2λ1​2−44​​.

A system has no solution only when the coefficient matrix is singular and the augmented system is inconsistent.

So first find when det⁡(A)=0\det(A)=0det(A)=0.

2. Compute the determinant

det⁡(A)=∣2−121−2λ1λ1∣.\det(A)= \begin{vmatrix} 2 & -1 & 2\\ 1 & -2 & \lambda\\ 1 & \lambda & 1 \end{vmatrix}.det(A)=​211​−1−2λ​2λ1​​.

Expanding along the first row:

det⁡(A)=2∣−2λλ1∣−(−1)∣1λ11∣+2∣1−21λ∣.\det(A)=2\begin{vmatrix}-2 & \lambda\\ \lambda & 1\end{vmatrix} -(-1)\begin{vmatrix}1 & \lambda\\ 1 & 1\end{vmatrix} +2\begin{vmatrix}1 & -2\\ 1 & \lambda\end{vmatrix}.det(A)=2​−2λ​λ1​​−(−1)​11​λ1​​+2​11​−2λ​​.

Now,

∣−2λλ1∣=−2−λ2,\begin{vmatrix}-2 & \lambda\\ \lambda & 1\end{vmatrix}=-2-\lambda^2,​−2λ​λ1​​=−2−λ2, ∣1λ11∣=1−λ,\begin{vmatrix}1 & \lambda\\ 1 & 1\end{vmatrix}=1-\lambda,​11​λ1​​=1−λ, ∣1−21λ∣=λ+2.\begin{vmatrix}1 & -2\\ 1 & \lambda\end{vmatrix}=\lambda+2.​11​−2λ​​=λ+2.

Hence,

det⁡(A)=2(−2−λ2)+(1−λ)+2(λ+2).\det(A)=2(-2-\lambda^2)+(1-\lambda)+2(\lambda+2).det(A)=2(−2−λ2)+(1−λ)+2(λ+2).

Simplify:

det⁡(A)=−4−2λ2+1−λ+2λ+4\det(A)=-4-2\lambda^2+1-\lambda+2\lambda+4det(A)=−4−2λ2+1−λ+2λ+4 =1+λ−2λ2.=1+\lambda-2\lambda^2.=1+λ−2λ2.

So,

det⁡(A)=1+λ−2λ2.\det(A)=1+\lambda-2\lambda^2.det(A)=1+λ−2λ2.

Set this equal to zero:

1+λ−2λ2=01+\lambda-2\lambda^2=01+λ−2λ2=0 2λ2−λ−1=02\lambda^2-\lambda-1=02λ2−λ−1=0 (2λ+1)(λ−1)=0.(2\lambda+1)(\lambda-1)=0.(2λ+1)(λ−1)=0.

Thus,

λ=1orλ=−12.\lambda=1 \quad \text{or} \quad \lambda=-\frac12.λ=1orλ=−21​.

Only these two values can possibly give no solution.

3. Check consistency for λ=1\lambda=1λ=1

Substitute λ=1\lambda=1λ=1:

{2x−y+2z=2x−2y+z=−4x+y+z=4\begin{cases} 2x-y+2z=2 \\ x-2y+z=-4 \\ x+y+z=4 \end{cases}⎩⎨⎧​2x−y+2z=2x−2y+z=−4x+y+z=4​

From the last two equations,

(x+y+z)−(x−2y+z)=4−(−4)(x+y+z)-(x-2y+z)=4-(-4)(x+y+z)−(x−2y+z)=4−(−4) 3y=8⇒y=83.3y=8 \Rightarrow y=\frac83.3y=8⇒y=38​.

Then from x+y+z=4x+y+z=4x+y+z=4,

x+z=4−83=43.x+z=4-\frac83=\frac43.x+z=4−38​=34​.

From the first equation,

2x+2z=2+y=2+83=1432x+2z=2+y=2+\frac83=\frac{14}{3}2x+2z=2+y=2+38​=314​ x+z=73.x+z=\frac73.x+z=37​.

But this contradicts x+z=43x+z=\frac43x+z=34​.

So for λ=1\lambda=1λ=1, the system is inconsistent, hence no solution.

4. Check consistency for λ=−12\lambda=-\frac12λ=−21​

Substitute λ=−12\lambda=-\frac12λ=−21​:

{2x−y+2z=2x−2y−12z=−4x−12y+z=4\begin{cases} 2x-y+2z=2 \\ x-2y-\frac12 z=-4 \\ x-\frac12 y+z=4 \end{cases}⎩⎨⎧​2x−y+2z=2x−2y−21​z=−4x−21​y+z=4​

Multiply the second equation by 222:

2x−4y−z=−8.2x-4y-z=-8.2x−4y−z=−8.

Multiply the third equation by 222:

2x−y+2z=8.2x-y+2z=8.2x−y+2z=8.

But the first equation is

2x−y+2z=2.2x-y+2z=2.2x−y+2z=2.

So we get

2x−y+2z=2and2x−y+2z=8,2x-y+2z=2 \quad \text{and} \quad 2x-y+2z=8,2x−y+2z=2and2x−y+2z=8,

which is impossible.

Hence for λ=−12\lambda=-\frac12λ=−21​, the system is also inconsistent, so no solution.

5. Form the set SSS

Therefore,

S={1,−12}.S=\left\{1,-\frac12\right\}.S={1,−21​}.

So SSS contains exactly two elements.

Hence the correct option is:

B\boxed{\text{B}}B​

6. Compare with stored correct answer

Stored correct answer: B

Our derived answer: B

They agree.

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