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Matrices and Determinants question

2021 · 27 Aug · Shift 2 · Q28
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  5. /2021 · 27 Aug · Shift 2 · Q28

Matrices and Determinants question

2021 · 27 Aug · Shift 2 · Q28

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let [λ\lambdaλ] be the greatest integer less than or equal to λ\lambdaλ. The set of all values of λ\lambdaλ for which the system of linear equations x + y + z = 4, 3x + 2y + 5z = 3, 9x + 4y + (28 + [λ\lambdaλ])z = [λ\lambdaλ] has a solution is :
  1. A
    R
  2. B
    (−∞-\infty−∞, −-− 9) ∪\cup∪ (−-− 9, ∞\infty∞)
  3. C
    [−-− 9, −-− 8)
  4. D
    (−∞-\infty−∞, −-− 9) ∪\cup∪ [−-− 8, ∞\infty∞)
View written solutionFree

Correct answer: A

  1. Write the system in matrix form

The system is

{x+y+z=43x+2y+5z=39x+4y+(28+[λ])z=[λ]\begin{cases} x+y+z=4 \\ 3x+2y+5z=3 \\ 9x+4y+(28+[\lambda])z=[\lambda] \end{cases}⎩⎨⎧​x+y+z=43x+2y+5z=39x+4y+(28+[λ])z=[λ]​

Let k=[λ]k=[\lambda]k=[λ] (greatest integer function). Then the system becomes

{x+y+z=43x+2y+5z=39x+4y+(28+k)z=k\begin{cases} x+y+z=4 \\ 3x+2y+5z=3 \\ 9x+4y+(28+k)z=k \end{cases}⎩⎨⎧​x+y+z=43x+2y+5z=39x+4y+(28+k)z=k​

We need the values of λ\lambdaλ for which this system is consistent.

  1. Coefficient matrix

The coefficient matrix is

A=(1113259428+k)A=\begin{pmatrix} 1&1&1\\ 3&2&5\\ 9&4&28+k \end{pmatrix}A=​139​124​1528+k​​

Its determinant is

det⁡(A)=∣1113259428+k∣\det(A)= \begin{vmatrix} 1&1&1\\ 3&2&5\\ 9&4&28+k \end{vmatrix}det(A)=​139​124​1528+k​​

Expand along the first row:

det⁡(A)=1∣25428+k∣−1∣35928+k∣+1∣3294∣\det(A)=1\begin{vmatrix}2&5\\4&28+k\end{vmatrix}-1\begin{vmatrix}3&5\\9&28+k\end{vmatrix}+1\begin{vmatrix}3&2\\9&4\end{vmatrix}det(A)=1​24​528+k​​−1​39​528+k​​+1​39​24​​

Now compute each minor:

∣25428+k∣=2(28+k)−20=56+2k−20=36+2k\begin{vmatrix}2&5\\4&28+k\end{vmatrix}=2(28+k)-20=56+2k-20=36+2k​24​528+k​​=2(28+k)−20=56+2k−20=36+2k ∣35928+k∣=3(28+k)−45=84+3k−45=39+3k\begin{vmatrix}3&5\\9&28+k\end{vmatrix}=3(28+k)-45=84+3k-45=39+3k​39​528+k​​=3(28+k)−45=84+3k−45=39+3k ∣3294∣=12−18=−6\begin{vmatrix}3&2\\9&4\end{vmatrix}=12-18=-6​39​24​​=12−18=−6

So,

det⁡(A)=(36+2k)−(39+3k)−6=−9−k\det(A)=(36+2k)-(39+3k)-6=-9-kdet(A)=(36+2k)−(39+3k)−6=−9−k

Thus,

det⁡(A)=−(k+9)\det(A)=-(k+9)det(A)=−(k+9)
  1. Case 1: det⁡(A)≠0\det(A)\neq 0det(A)=0

If k≠−9k\neq -9k=−9, i.e. [λ]≠−9[\lambda]\neq -9[λ]=−9, then det⁡(A)≠0\det(A)\neq 0det(A)=0. Hence the system has a unique solution, so it is consistent.

This gives all λ\lambdaλ except those for which

[λ]=−9[\lambda]=-9[λ]=−9

which means

λ∈[−9,−8)\lambda\in[-9,-8)λ∈[−9,−8)
  1. Case 2: det⁡(A)=0\det(A)=0det(A)=0

Now check k=−9k=-9k=−9. Then the third equation becomes

9x+4y+(28−9)z=−99x+4y+(28-9)z=-99x+4y+(28−9)z=−9

that is,

9x+4y+19z=−99x+4y+19z=-99x+4y+19z=−9

Let us see whether this is compatible with the first two equations.

From

x+y+z=4x+y+z=4x+y+z=4

and

3x+2y+5z=33x+2y+5z=33x+2y+5z=3

we compute

  • Multiply the first equation by 222:
2x+2y+2z=82x+2y+2z=82x+2y+2z=8
  • Subtract from the second:
(3x+2y+5z)−(2x+2y+2z)=3−8(3x+2y+5z)-(2x+2y+2z)=3-8(3x+2y+5z)−(2x+2y+2z)=3−8 x+3z=−5x+3z=-5x+3z=−5

So,

x=−5−3zx=-5-3zx=−5−3z

Using x+y+z=4x+y+z=4x+y+z=4,

y=4−x−z=4−(−5−3z)−z=9+2zy=4-x-z=4-(-5-3z)-z=9+2zy=4−x−z=4−(−5−3z)−z=9+2z

Now substitute into the third equation:

9x+4y+19z=9(−5−3z)+4(9+2z)+19z9x+4y+19z=9(-5-3z)+4(9+2z)+19z9x+4y+19z=9(−5−3z)+4(9+2z)+19z =−45−27z+36+8z+19z=−9=-45-27z+36+8z+19z=-9=−45−27z+36+8z+19z=−9 =−9=-9=−9

which is exactly the RHS.

So the third equation is automatically satisfied. Therefore, even when [λ]=−9[\lambda]=-9[λ]=−9, the system is consistent.

  1. Conclusion
  • If [λ]≠−9[\lambda]\neq -9[λ]=−9, determinant is nonzero ⇒\Rightarrow⇒ unique solution.
  • If [λ]=−9[\lambda]=-9[λ]=−9, determinant is zero but system is still consistent.

Hence the system has a solution for every real λ\lambdaλ.

Therefore, the correct option is

A: R\boxed{\text{A: }\mathbb{R}}A: R​
  1. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

They agree.

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