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Matrices and Determinants question

2021 · 17 Mar · Shift 1 · Q32
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Matrices and Determinants question

2021 · 17 Mar · Shift 1 · Q32

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
The system of equations kx + y + z = 1, x + ky + z = k and x + y + zk = k2 has no solution if k is equal to :
  1. A
    0
  2. B
    −-− 1
  3. C
    −-− 2
  4. D
    1
View written solutionFree

Correct answer: C

  1. Write the system in matrix form

The equations are:

{kx+y+z=1x+ky+z=kx+y+kz=k2\begin{cases} kx+y+z=1 \\ x+ky+z=k \\ x+y+kz=k^2 \end{cases}⎩⎨⎧​kx+y+z=1x+ky+z=kx+y+kz=k2​

So the coefficient matrix is

A=(k111k111k)A=\begin{pmatrix} k&1&1\\ 1&k&1\\ 1&1&k \end{pmatrix}A=​k11​1k1​11k​​

and the constant vector is

B=(1kk2).B=\begin{pmatrix}1\\k\\k^2\end{pmatrix}.B=​1kk2​​.

For the system to have no solution, we need:

  • det⁡(A)=0\det(A)=0det(A)=0 (so the system is not uniquely solvable), and
  • the augmented system must be inconsistent.
  1. Find det⁡(A)\det(A)det(A)

Observe that

A=(k−1)I+JA=(k-1)I+JA=(k−1)I+J

where JJJ is the 3×33\times 33×3 all-ones matrix.

For this symmetric matrix,

det⁡(A)=(k−1)2(k+2).\det(A)=(k-1)^2(k+2).det(A)=(k−1)2(k+2).

Hence,

det⁡(A)=0  ⟺  k=1 or k=−2.\det(A)=0 \iff k=1 \text{ or } k=-2.det(A)=0⟺k=1 or k=−2.

So only k=1k=1k=1 and k=−2k=-2k=−2 need to be checked.

  1. Check k=1k=1k=1

Substitute k=1k=1k=1 into the system:

{x+y+z=1x+y+z=1x+y+z=1\begin{cases} x+y+z=1 \\ x+y+z=1 \\ x+y+z=1 \end{cases}⎩⎨⎧​x+y+z=1x+y+z=1x+y+z=1​

All three equations are identical, so there are infinitely many solutions.

Thus, for k=1k=1k=1, the system does not have no solution.

  1. Check k=−2k=-2k=−2

Substitute k=−2k=-2k=−2:

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

Add the left-hand sides of all three equations:

(−2x+y+z)+(x−2y+z)+(x+y−2z)=0.(-2x+y+z)+(x-2y+z)+(x+y-2z)=0.(−2x+y+z)+(x−2y+z)+(x+y−2z)=0.

So the sum of the left sides is 000.

But the sum of the right-hand sides is

1+(−2)+4=3.1+(-2)+4=3.1+(−2)+4=3.

So we get

0=3,0=3,0=3,

which is impossible.

Hence the system is inconsistent for k=−2k=-2k=−2.

Therefore, the system has no solution when

k=−2.k=-2.k=−2.
  1. Evaluate options
  • A: 000 → det⁡(A)≠0\det(A)\neq 0det(A)=0, unique solution exists.
  • B: −1-1−1 → det⁡(A)≠0\det(A)\neq 0det(A)=0, unique solution exists.
  • C: −2-2−2 → inconsistent, no solution.
  • D: 111 → infinitely many solutions, not no solution.

So the correct option is:

C:−2\boxed{\text{C}:-2}C:−2​
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