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Matrices and Determinants question

2005 · Shift 0 · Q75
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  5. /2005 · Shift 0 · Q75

Matrices and Determinants question

2005 · Shift 0 · Q75

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
The system of equations α x+y+z=α−1x+αy+z=α−1x+y+α z=α−1\begin{matrix} {\alpha \,x + y + z = \alpha - 1} \\ {x + \alpha y + z = \alpha - 1} \\ {x + y + \alpha \,z = \alpha - 1} \\ \end{matrix}αx+y+z=α−1x+αy+z=α−1x+y+αz=α−1​ has no solutions, if α\alphaα is :
  1. A
    −2-2−2
  2. B
    either −2-2−2 or 111
  3. C
    not −2-2−2
  4. D
    111
View written solutionFree

Correct answer: A

  1. Write the system in matrix form

The given system is

{αx+y+z=α−1x+αy+z=α−1x+y+αz=α−1\begin{cases} \alpha x + y + z = \alpha - 1 \\ x + \alpha y + z = \alpha - 1 \\ x + y + \alpha z = \alpha - 1 \end{cases}⎩⎨⎧​αx+y+z=α−1x+αy+z=α−1x+y+αz=α−1​

Its coefficient matrix is

A=(α111α111α)A=\begin{pmatrix} \alpha & 1 & 1\\ 1 & \alpha & 1\\ 1 & 1 & \alpha \end{pmatrix}A=​α11​1α1​11α​​

and the RHS vector is

b=(α−1α−1α−1).\mathbf{b}=\begin{pmatrix}\alpha-1\\ \alpha-1\\ \alpha-1\end{pmatrix}.b=​α−1α−1α−1​​.

We need the value of α\alphaα for which the system has no solution.

  1. Find when the coefficient matrix is singular

A system can have no solution only when the coefficient matrix is singular, i.e.

det⁡(A)=0.\det(A)=0.det(A)=0.

Now,

A=(α−1)I+JA=(\alpha-1)I + JA=(α−1)I+J

where JJJ is the 3×33\times 33×3 matrix of all 1's. For this standard matrix,

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

So,

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

Thus only α=1\alpha=1α=1 or α=−2\alpha=-2α=−2 need to be checked.

  1. Check α=1\alpha=1α=1

Substitute α=1\alpha=1α=1:

{x+y+z=0x+y+z=0x+y+z=0\begin{cases} x+y+z=0\\ x+y+z=0\\ x+y+z=0 \end{cases}⎩⎨⎧​x+y+z=0x+y+z=0x+y+z=0​

This is consistent and has infinitely many solutions.

So for α=1\alpha=1α=1, the system does not have no solution.

  1. Check α=−2\alpha=-2α=−2

Substitute α=−2\alpha=-2α=−2:

{−2x+y+z=−3x−2y+z=−3x+y−2z=−3\begin{cases} -2x+y+z=-3\\ x-2y+z=-3\\ x+y-2z=-3 \end{cases}⎩⎨⎧​−2x+y+z=−3x−2y+z=−3x+y−2z=−3​

Add all three equations:

(−2x+x+x)+(y−2y+y)+(z+z−2z)=−3−3−3(-2x+x+x)+(y-2y+y)+(z+z-2z)=-3-3-3(−2x+x+x)+(y−2y+y)+(z+z−2z)=−3−3−3 0=−9,0=-9,0=−9,

which is impossible.

Hence the system is inconsistent, so it has no solution for

α=−2.\alpha=-2.α=−2.
  1. Evaluate options
  • A: −2-2−2 — Correct
  • B: either −2-2−2 or 111 — Wrong, because α=1\alpha=1α=1 gives infinitely many solutions
  • C: not −2-2−2 — Wrong
  • D: 111 — Wrong
  1. Final answer

The system has no solution when

α=−2.\boxed{\alpha=-2}.α=−2​.
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