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Matrices and Determinants question

2023 · 13 Apr · Shift 2 · Q30
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Matrices and Determinants question

2023 · 13 Apr · Shift 2 · Q30

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If the system of equations 2x+y−z=52x−5y+λz=μx+2y−5z=72 x+y-z=52 x-5 y+\lambda z=\mu x+2 y-5 z=72x+y−z=52x−5y+λz=μx+2y−5z=7 has infinitely many solutions, then (λ+μ)2+(λ−μ)2(\lambda+\mu)^{2}+(\lambda-\mu)^{2}(λ+μ)2+(λ−μ)2 is equal to
  1. A
    916
  2. B
    912
  3. C
    920
  4. D
    904
View written solutionFree

Correct answer: A

Let the system be

{2x+y−z=52x−5y+λz=μx+2y−5z=7\begin{cases} 2x+y-z=5 \\ 2x-5y+\lambda z=\mu \\ x+2y-5z=7 \end{cases}⎩⎨⎧​2x+y−z=52x−5y+λz=μx+2y−5z=7​

For a system of 3 linear equations in 3 variables to have infinitely many solutions, we need

  1. the coefficient matrix to be singular, and
  2. the augmented matrix to have the same rank as the coefficient matrix, which must be <3<3<3.

1. Write the coefficient matrix

A=(21−12−5λ12−5)A=\begin{pmatrix} 2&1&-1\\ 2&-5&\lambda\\ 1&2&-5 \end{pmatrix}A=​221​1−52​−1λ−5​​

For infinitely many solutions,

det⁡(A)=0\det(A)=0det(A)=0

Compute the determinant:

det⁡(A)=2∣−5λ2−5∣−1∣2λ1−5∣+(−1)∣2−512∣\det(A)=2\begin{vmatrix}-5&\lambda\\2&-5\end{vmatrix}-1\begin{vmatrix}2&\lambda\\1&-5\end{vmatrix}+(-1)\begin{vmatrix}2&-5\\1&2\end{vmatrix}det(A)=2​−52​λ−5​​−1​21​λ−5​​+(−1)​21​−52​​ =2(25−2λ)−(−10−λ)−(4+5)=2(25-2\lambda)-\bigl(-10-\lambda\bigr)-\bigl(4+5\bigr)=2(25−2λ)−(−10−λ)−(4+5) =50−4λ+10+λ−9=51−3λ=50-4\lambda+10+\lambda-9=51-3\lambda=50−4λ+10+λ−9=51−3λ

So,

51−3λ=0  ⟹  λ=1751-3\lambda=0 \implies \lambda=1751−3λ=0⟹λ=17

2. Impose consistency for infinitely many solutions

Now substitute λ=17\lambda=17λ=17:

{2x+y−z=52x−5y+17z=μx+2y−5z=7\begin{cases} 2x+y-z=5 \\ 2x-5y+17z=\mu \\ x+2y-5z=7 \end{cases}⎩⎨⎧​2x+y−z=52x−5y+17z=μx+2y−5z=7​

Since rank must be 222, one equation must be a linear combination of the other two.

Let

a(2x+y−z)+b(x+2y−5z)=2x−5y+17za(2x+y-z)+b(x+2y-5z)=2x-5y+17za(2x+y−z)+b(x+2y−5z)=2x−5y+17z

Comparing coefficients:

2a+b=2...(1)2a+b=2 \quad ...(1)2a+b=2...(1) a+2b=−5...(2)a+2b=-5 \quad ...(2)a+2b=−5...(2) −a−5b=17...(3)-a-5b=17 \quad ...(3)−a−5b=17...(3)

Solve (1) and (2):

From (1),

b=2−2ab=2-2ab=2−2a

Substitute into (2):

a+2(2−2a)=−5a+2(2-2a)=-5a+2(2−2a)=−5 a+4−4a=−5a+4-4a=-5a+4−4a=−5 −3a=−9  ⟹  a=3-3a=-9 \implies a=3−3a=−9⟹a=3

Then

b=2−2(3)=−4b=2-2(3)=-4b=2−2(3)=−4

Check (3):

−a−5b=−3−5(−4)=−3+20=17-a-5b=-3-5(-4)=-3+20=17−a−5b=−3−5(−4)=−3+20=17

which is satisfied.

So the second equation is

3×(first)−4×(third)3\times(\text{first})-4\times(\text{third})3×(first)−4×(third)

Hence for consistency, the RHS must satisfy the same relation:

μ=3(5)−4(7)=15−28=−13\mu=3(5)-4(7)=15-28=-13μ=3(5)−4(7)=15−28=−13

Thus,

λ=17,μ=−13\lambda=17,\qquad \mu=-13λ=17,μ=−13

3. Compute the required value

We need

(λ+μ)2+(λ−μ)2(\lambda+\mu)^2+(\lambda-\mu)^2(λ+μ)2+(λ−μ)2

Substitute values:

λ+μ=17+(−13)=4\lambda+\mu=17+(-13)=4λ+μ=17+(−13)=4 λ−μ=17−(−13)=30\lambda-\mu=17-(-13)=30λ−μ=17−(−13)=30

Therefore,

(λ+μ)2+(λ−μ)2=42+302=16+900=916(\lambda+\mu)^2+(\lambda-\mu)^2=4^2+30^2=16+900=916(λ+μ)2+(λ−μ)2=42+302=16+900=916

4. Final answer

916\boxed{916}916​

So the correct option is A.

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