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Matrices and Determinants question

2020 · 6 Sep · Shift 2 · Q26
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Matrices and Determinants question

2020 · 6 Sep · Shift 2 · Q26

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
The sum of distinct values of λ\lambdaλ for which the system of equations (λ−1)x+(3λ+1)y+2λz=0(λ−1)x+(4λ−2)y+(λ+3)z=02x+(3λ+1)y+3(λ−1)z=0\left( {\lambda - 1} \right)x + \left( {3\lambda + 1} \right)y + 2\lambda z = 0\left( {\lambda - 1} \right)x + \left( {4\lambda - 2} \right)y + \left( {\lambda + 3} \right)z = 02x + \left( {3\lambda + 1} \right)y + 3\left( {\lambda - 1} \right)z = 0(λ−1)x+(3λ+1)y+2λz=0(λ−1)x+(4λ−2)y+(λ+3)z=02x+(3λ+1)y+3(λ−1)z=0 has non-zero solutions, is ‾\underline{\hspace{2cm}}​ .
Numerical answer
View written solutionFree

Correct answer: 3

We need the values of λ\lambdaλ for which the homogeneous system has a non-zero solution.
For a homogeneous system, non-trivial solutions exist iff the determinant of the coefficient matrix is zero.

1. Form the coefficient matrix

From the question, the system is

(λ−1)x+(3λ+1)y+2λz=0(\lambda-1)x+(3\lambda+1)y+2\lambda z=0(λ−1)x+(3λ+1)y+2λz=0 (λ−1)x+(4λ−2)y+(λ+3)z=0(\lambda-1)x+(4\lambda-2)y+(\lambda+3)z=0(λ−1)x+(4λ−2)y+(λ+3)z=0 2x+(3λ+1)y+3(λ−1)z=02x+(3\lambda+1)y+3(\lambda-1)z=02x+(3λ+1)y+3(λ−1)z=0

So the coefficient matrix is

A=(λ−13λ+12λλ−14λ−2λ+323λ+13(λ−1)).A=\begin{pmatrix} \lambda-1 & 3\lambda+1 & 2\lambda\\ \lambda-1 & 4\lambda-2 & \lambda+3\\ 2 & 3\lambda+1 & 3(\lambda-1) \end{pmatrix}.A=​λ−1λ−12​3λ+14λ−23λ+1​2λλ+33(λ−1)​​.

We require

det⁡(A)=0.\det(A)=0.det(A)=0.

2. Compute the determinant

Apply the row operation

R2→R2−R1,R_2 \to R_2-R_1,R2​→R2​−R1​,

which does not change the determinant. Then

R2=(0,λ−3,3−λ)=(0,λ−3,−(λ−3)).R_2=(0,\lambda-3,3-\lambda)=(0,\lambda-3,-(\lambda-3)).R2​=(0,λ−3,3−λ)=(0,λ−3,−(λ−3)).

Thus

det⁡(A)=∣λ−13λ+12λ0λ−3−(λ−3)23λ+13λ−3∣.\det(A)= \begin{vmatrix} \lambda-1 & 3\lambda+1 & 2\lambda\\ 0 & \lambda-3 & -(\lambda-3)\\ 2 & 3\lambda+1 & 3\lambda-3 \end{vmatrix}.det(A)=​λ−102​3λ+1λ−33λ+1​2λ−(λ−3)3λ−3​​.

Factor (λ−3)(\lambda-3)(λ−3) from the second row:

det⁡(A)=(λ−3)∣λ−13λ+12λ01−123λ+13λ−3∣.\det(A)=(\lambda-3) \begin{vmatrix} \lambda-1 & 3\lambda+1 & 2\lambda\\ 0 & 1 & -1\\ 2 & 3\lambda+1 & 3\lambda-3 \end{vmatrix}.det(A)=(λ−3)​λ−102​3λ+113λ+1​2λ−13λ−3​​.

Now expand along the second row:

∣λ−13λ+12λ01−123λ+13λ−3∣=1⋅∣λ−12λ23λ−3∣+1⋅∣λ−13λ+123λ+1∣.\begin{vmatrix} \lambda-1 & 3\lambda+1 & 2\lambda\\ 0 & 1 & -1\\ 2 & 3\lambda+1 & 3\lambda-3 \end{vmatrix} =1\cdot \begin{vmatrix} \lambda-1 & 2\lambda\\ 2 & 3\lambda-3 \end{vmatrix} +1\cdot \begin{vmatrix} \lambda-1 & 3\lambda+1\\ 2 & 3\lambda+1 \end{vmatrix}.​λ−102​3λ+113λ+1​2λ−13λ−3​​=1⋅​λ−12​2λ3λ−3​​+1⋅​λ−12​3λ+13λ+1​​.

Now calculate each minor:

First minor

∣λ−12λ23λ−3∣=(λ−1)(3λ−3)−4λ.\begin{vmatrix} \lambda-1 & 2\lambda\\ 2 & 3\lambda-3 \end{vmatrix} =(\lambda-1)(3\lambda-3)-4\lambda.​λ−12​2λ3λ−3​​=(λ−1)(3λ−3)−4λ.

Since 3λ−3=3(λ−1)3\lambda-3=3(\lambda-1)3λ−3=3(λ−1),

(λ−1)(3λ−3)−4λ=3(λ−1)2−4λ.(\lambda-1)(3\lambda-3)-4\lambda=3(\lambda-1)^2-4\lambda.(λ−1)(3λ−3)−4λ=3(λ−1)2−4λ.

Second minor

∣λ−13λ+123λ+1∣=(λ−1)(3λ+1)−2(3λ+1)\begin{vmatrix} \lambda-1 & 3\lambda+1\\ 2 & 3\lambda+1 \end{vmatrix} =(\lambda-1)(3\lambda+1)-2(3\lambda+1)​λ−12​3λ+13λ+1​​=(λ−1)(3λ+1)−2(3λ+1) =(3λ+1)(λ−3).=(3\lambda+1)(\lambda-3).=(3λ+1)(λ−3).

So,

det⁡(A)=(λ−3)(3(λ−1)2−4λ+(3λ+1)(λ−3)).\det(A)=(\lambda-3)\Big(3(\lambda-1)^2-4\lambda+(3\lambda+1)(\lambda-3)\Big).det(A)=(λ−3)(3(λ−1)2−4λ+(3λ+1)(λ−3)).

Simplify the bracket:

3(λ−1)2−4λ=3(λ2−2λ+1)−4λ=3λ2−10λ+3,3(\lambda-1)^2-4\lambda=3(\lambda^2-2\lambda+1)-4\lambda=3\lambda^2-10\lambda+3,3(λ−1)2−4λ=3(λ2−2λ+1)−4λ=3λ2−10λ+3, (3λ+1)(λ−3)=3λ2−8λ−3.(3\lambda+1)(\lambda-3)=3\lambda^2-8\lambda-3.(3λ+1)(λ−3)=3λ2−8λ−3.

Adding,

3λ2−10λ+3+3λ2−8λ−3=6λ2−18λ=6λ(λ−3).3\lambda^2-10\lambda+3+3\lambda^2-8\lambda-3=6\lambda^2-18\lambda=6\lambda(\lambda-3).3λ2−10λ+3+3λ2−8λ−3=6λ2−18λ=6λ(λ−3).

Hence

det⁡(A)=(λ−3)⋅6λ(λ−3)=6λ(λ−3)2.\det(A)=(\lambda-3)\cdot 6\lambda(\lambda-3)=6\lambda(\lambda-3)^2.det(A)=(λ−3)⋅6λ(λ−3)=6λ(λ−3)2.

3. Condition for non-zero solutions

For non-trivial solutions,

6λ(λ−3)2=0.6\lambda(\lambda-3)^2=0.6λ(λ−3)2=0.

Therefore,

λ=0orλ=3.\lambda=0 \quad \text{or} \quad \lambda=3.λ=0orλ=3.

The distinct values are 000 and 333.

4. Sum of distinct values

0+3=3.0+3=3.0+3=3.

5. Compare with stored answer

Derived answer = 333
Stored correct answer = 333

They agree.

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