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Matrices and Determinants question

2021 · 24 Feb · Shift 1 · Q35
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  5. /2021 · 24 Feb · Shift 1 · Q35

Matrices and Determinants question

2021 · 24 Feb · Shift 1 · Q35

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
The system of linear equations 3x - 2y - kz = 10 2x - 4y - 2z = 6 x+2y - z = 5m is inconsistent if :
  1. A
    k eee 3, m ∈\in∈ R
  2. B
    k = 3, m e45e {4 \over 5}e54​
  3. C
    k = 3, m =45= {4 \over 5}=54​
  4. D
    k eee 3, m e45e {4 \over 5}e54​
View written solutionFree

Correct answer: B

  1. Write the system in matrix form

The equations are:

3x−2y−kz=103x-2y-kz=103x−2y−kz=10 2x−4y−2z=62x-4y-2z=62x−4y−2z=6 x+2y−z=5mx+2y-z=5mx+2y−z=5m

Coefficient matrix:

A=(3−2−k2−4−212−1)A=\begin{pmatrix} 3 & -2 & -k\\ 2 & -4 & -2\\ 1 & 2 & -1 \end{pmatrix}A=​321​−2−42​−k−2−1​​

For inconsistency, we need:

  • either det⁡(A)=0\det(A)=0det(A)=0, and
  • augmented matrix rank >>> coefficient matrix rank.

So first compute det⁡(A)\det(A)det(A).

  1. Compute the determinant
det⁡(A)=∣3−2−k2−4−212−1∣\det(A)=\begin{vmatrix} 3 & -2 & -k\\ 2 & -4 & -2\\ 1 & 2 & -1 \end{vmatrix}det(A)=​321​−2−42​−k−2−1​​

Expand along the first row:

det⁡(A)=3∣−4−22−1∣−(−2)∣2−21−1∣+(−k)∣2−412∣\det(A)=3\begin{vmatrix}-4 & -2\\ 2 & -1\end{vmatrix} -(-2)\begin{vmatrix}2 & -2\\ 1 & -1\end{vmatrix} +(-k)\begin{vmatrix}2 & -4\\ 1 & 2\end{vmatrix}det(A)=3​−42​−2−1​​−(−2)​21​−2−1​​+(−k)​21​−42​​

Now,

∣−4−22−1∣=(−4)(−1)−(−2)(2)=4+4=8\begin{vmatrix}-4 & -2\\ 2 & -1\end{vmatrix}=(-4)(-1)-(-2)(2)=4+4=8​−42​−2−1​​=(−4)(−1)−(−2)(2)=4+4=8 ∣2−21−1∣=2(−1)−(−2)(1)=−2+2=0\begin{vmatrix}2 & -2\\ 1 & -1\end{vmatrix}=2(-1)-(-2)(1)=-2+2=0​21​−2−1​​=2(−1)−(−2)(1)=−2+2=0 ∣2−412∣=2(2)−(−4)(1)=4+4=8\begin{vmatrix}2 & -4\\ 1 & 2\end{vmatrix}=2(2)-(-4)(1)=4+4=8​21​−42​​=2(2)−(−4)(1)=4+4=8

Therefore,

det⁡(A)=3(8)−(−k)(?)\det(A)=3(8)-(-k)(? )det(A)=3(8)−(−k)(?)

Carefully,

det⁡(A)=3(8)+2(0)+(−k)(8)=24−8k=8(3−k)\det(A)=3(8)+2(0)+(-k)(8)=24-8k=8(3-k)det(A)=3(8)+2(0)+(−k)(8)=24−8k=8(3−k)

So,

det⁡(A)=0  ⟺  k=3\det(A)=0 \iff k=3det(A)=0⟺k=3

Thus, if k≠3k\neq 3k=3, the system has a unique solution and cannot be inconsistent.

Hence inconsistency is possible only when:

k=3k=3k=3
  1. Substitute k=3k=3k=3 into the system

The equations become:

3x−2y−3z=10...(1)3x-2y-3z=10 \quad ...(1)3x−2y−3z=10...(1) 2x−4y−2z=6...(2)2x-4y-2z=6 \quad ...(2)2x−4y−2z=6...(2) x+2y−z=5m...(3)x+2y-z=5m \quad ...(3)x+2y−z=5m...(3)

Now check whether the left-hand sides are linearly dependent.

Add equations (2)(2)(2) and (3)(3)(3) (only LHS first):

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

which is exactly the LHS of equation (1)(1)(1).

So for consistency, the RHS must also satisfy:

6+5m=106+5m=106+5m=10

Thus,

5m=4  ⟹  m=455m=4 \implies m=\frac455m=4⟹m=54​

Therefore:

  • if k=3k=3k=3 and m=45m=\frac45m=54​, the system is consistent (dependent equations).
  • if k=3k=3k=3 and m≠45m\ne\frac45m=54​, the system is inconsistent.
  1. Check options

  • A: k≠3,m∈Rk\ne 3, m\in \mathbb Rk=3,m∈R
    False, because if k≠3k\ne 3k=3, determinant is non-zero, so system is consistent.

  • B: k=3,m≠45k=3, m\ne \frac45k=3,m=54​
    True.

  • C: k=3,m=45k=3, m=\frac45k=3,m=54​
    False, this gives infinitely many solutions, not inconsistency.

  • D: k≠3,m≠45k\ne 3, m\ne \frac45k=3,m=54​
    False, since k≠3k\ne 3k=3 implies unique solution.

  1. Final answer

The system is inconsistent when

k=3,m≠45k=3,\quad m\ne \frac45k=3,m=54​

So the correct option is B.

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