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Matrices and Determinants question

2018 · 15 Apr · Shift 1 · Q28
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  5. /2018 · 15 Apr · Shift 1 · Q28

Matrices and Determinants question

2018 · 15 Apr · Shift 1 · Q28

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let S be the set of all real values of k for which the systemof linear equations x + y + z = 2 2x + y −-− z = 3 3x + 2y + kz = 4 has a unique solution. Then S is :
  1. A
    an empty set
  2. B
    equal to {0}
  3. C
    equal to R
  4. D
    equal to R −-− {0}
View written solutionFree

Correct answer: D

  1. For a system of three linear equations in three variables to have a unique solution, the determinant of its coefficient matrix must be non-zero.

  2. Write the coefficient matrix:

A=(11121−132k)A = \begin{pmatrix} 1 & 1 & 1\\ 2 & 1 & -1\\ 3 & 2 & k \end{pmatrix}A=​123​112​1−1k​​
  1. Compute det⁡(A)\det(A)det(A):
det⁡(A)=∣11121−132k∣\det(A)= \begin{vmatrix} 1 & 1 & 1\\ 2 & 1 & -1\\ 3 & 2 & k \end{vmatrix}det(A)=​123​112​1−1k​​

Expand along the first row:

det⁡(A)=1∣1−12k∣−1∣2−13k∣+1∣2132∣\det(A)=1\begin{vmatrix}1 & -1\\ 2 & k\end{vmatrix} -1\begin{vmatrix}2 & -1\\ 3 & k\end{vmatrix} +1\begin{vmatrix}2 & 1\\ 3 & 2\end{vmatrix}det(A)=1​12​−1k​​−1​23​−1k​​+1​23​12​​
  1. Evaluate each minor:
∣1−12k∣=1⋅k−(−1)⋅2=k+2\begin{vmatrix}1 & -1\\ 2 & k\end{vmatrix}=1\cdot k-(-1)\cdot 2=k+2​12​−1k​​=1⋅k−(−1)⋅2=k+2 ∣2−13k∣=2k−(−1)⋅3=2k+3\begin{vmatrix}2 & -1\\ 3 & k\end{vmatrix}=2k-(-1)\cdot 3=2k+3​23​−1k​​=2k−(−1)⋅3=2k+3 ∣2132∣=4−3=1\begin{vmatrix}2 & 1\\ 3 & 2\end{vmatrix}=4-3=1​23​12​​=4−3=1

So,

det⁡(A)=(k+2)−(2k+3)+1=−k\det(A)=(k+2)-(2k+3)+1 = -kdet(A)=(k+2)−(2k+3)+1=−k
  1. For a unique solution,
det⁡(A)≠0  ⟹  −k≠0  ⟹  k≠0\det(A)\ne 0 \implies -k\ne 0 \implies k\ne 0det(A)=0⟹−k=0⟹k=0

Hence,

S=R∖{0}S=\mathbb{R}\setminus\{0\}S=R∖{0}
  1. Compare with the options:
  • A: empty set — incorrect
  • B: {0}\{0\}{0} — incorrect
  • C: R\mathbb{R}R — incorrect
  • D: R−{0}\mathbb{R}-\{0\}R−{0} — correct

Therefore, the correct option is D.

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