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Matrices and Determinants question

2021 · 24 Feb · Shift 2 · Q26
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  5. /2021 · 24 Feb · Shift 2 · Q26

Matrices and Determinants question

2021 · 24 Feb · Shift 2 · Q26

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
For the system of linear equations: x−2y=1,x−y+kz=−2,ky+4z=6,k∈Rx - 2y = 1,x - y + kz = - 2,ky + 4z = 6,k \in Rx−2y=1,x−y+kz=−2,ky+4z=6,k∈R, consider the following statements : (A) The system has unique solution if ke2,ke−2k e 2,k e - 2ke2,ke−2. (B) The system has unique solution if k = −-− 2 (C) The system has unique solution if k = 2 (D) The system has no solution if k = 2 (E) The system has infinite number of solutions if k e−e -e− 2. Which of the following statements are correct?
  1. A
    (B) and (E) only
  2. B
    (C) and (D) only
  3. C
    (A) and (E) only
  4. D
    (A) and (D) only
View written solutionFree

Correct answer: D

  1. Write the system in matrix form

The equations are:

{x−2y=1x−y+kz=−2ky+4z=6\begin{cases} x-2y=1 \\ x-y+kz=-2 \\ ky+4z=6 \end{cases}⎩⎨⎧​x−2y=1x−y+kz=−2ky+4z=6​

Coefficient matrix:

A=(1−201−1k0k4)A=\begin{pmatrix} 1 & -2 & 0\\ 1 & -1 & k\\ 0 & k & 4 \end{pmatrix}A=​110​−2−1k​0k4​​

A linear system has a unique solution iff det⁡(A)≠0\det(A)\neq 0det(A)=0.

  1. Compute the determinant
det⁡(A)=∣1−201−1k0k4∣\det(A)= \begin{vmatrix} 1 & -2 & 0\\ 1 & -1 & k\\ 0 & k & 4 \end{vmatrix}det(A)=​110​−2−1k​0k4​​

Expanding along the first row:

det⁡(A)=1∣−1kk4∣−(−2)∣1k04∣+0\det(A)=1\begin{vmatrix}-1 & k\\ k & 4\end{vmatrix}-(-2)\begin{vmatrix}1 & k\\ 0 & 4\end{vmatrix}+0det(A)=1​−1k​k4​​−(−2)​10​k4​​+0 =1((−1)(4)−k2)+2(4)=1((-1)(4)-k^2)+2(4)=1((−1)(4)−k2)+2(4) =−4−k2+8=4−k2=-4-k^2+8=4-k^2=−4−k2+8=4−k2

So,

det⁡(A)=4−k2=(2−k)(2+k)\det(A)=4-k^2=(2-k)(2+k)det(A)=4−k2=(2−k)(2+k)

Hence the system has a unique solution when

4−k2≠0  ⟺  k≠2,−24-k^2\neq 0 \iff k\neq 2,-24−k2=0⟺k=2,−2

Therefore, statement (A) is true.

  1. Check the special cases k=2k=2k=2 and k=−2k=-2k=−2

Case 1: k=2k=2k=2

The system becomes

{x−2y=1x−y+2z=−22y+4z=6\begin{cases} x-2y=1 \\ x-y+2z=-2 \\ 2y+4z=6 \end{cases}⎩⎨⎧​x−2y=1x−y+2z=−22y+4z=6​

From the first equation,

x=1+2yx=1+2yx=1+2y

Substitute into the second:

(1+2y)−y+2z=−2(1+2y)-y+2z=-2(1+2y)−y+2z=−2 y+2z=−3y+2z=-3y+2z=−3

But the third equation gives

2y+4z=6  ⟺  y+2z=32y+4z=6 \iff y+2z=32y+4z=6⟺y+2z=3

So we get

y+2z=−3andy+2z=3y+2z=-3 \quad \text{and} \quad y+2z=3y+2z=−3andy+2z=3

which is inconsistent.

Hence, for k=2k=2k=2, the system has no solution.

Therefore:

  • (C) false
  • (D) true

Case 2: k=−2k=-2k=−2

The system becomes

{x−2y=1x−y−2z=−2−2y+4z=6\begin{cases} x-2y=1 \\ x-y-2z=-2 \\ -2y+4z=6 \end{cases}⎩⎨⎧​x−2y=1x−y−2z=−2−2y+4z=6​

From the first equation,

x=1+2yx=1+2yx=1+2y

Substitute into the second:

(1+2y)−y−2z=−2(1+2y)-y-2z=-2(1+2y)−y−2z=−2 y−2z=−3y-2z=-3y−2z=−3

Third equation gives

−2y+4z=6  ⟺  y−2z=−3-2y+4z=6 \iff y-2z=-3−2y+4z=6⟺y−2z=−3

Both are the same equation, so the system is consistent and dependent. Thus there are infinitely many solutions.

Therefore:

  • (B) false
  • (E) false, because infinite solutions occur only when k=−2k=-2k=−2, not when k≠−2k\neq -2k=−2.
  1. Evaluate all statements
  • (A) True
  • (B) False
  • (C) False
  • (D) True
  • (E) False

So the correct combination is:

(A) and (D) only\boxed{\text{(A) and (D) only}}(A) and (D) only​

This corresponds to Option D.

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