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Matrices and Determinants question

2017 · Shift 0 · Q28
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Matrices and Determinants question

2017 · Shift 0 · Q28

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If S is the set of distinct values of 'b' for which the following system of linear equations x + y + z = 1 x + ay + z = 1 ax + by + z = 0 has no solution, then S is :
  1. A
    an empty set
  2. B
    an infinite set
  3. C
    a finite set containing two or more elements
  4. D
    a singleton
View written solutionFree

Correct answer: D

  1. Write the system in matrix form

The system is

{x+y+z=1x+ay+z=1ax+by+z=0\begin{cases} x+y+z=1 \\ x+ay+z=1 \\ ax+by+z=0 \end{cases}⎩⎨⎧​x+y+z=1x+ay+z=1ax+by+z=0​

Coefficient matrix:

A=(1111a1ab1)A=\begin{pmatrix} 1&1&1\\ 1&a&1\\ a&b&1 \end{pmatrix}A=​11a​1ab​111​​

Augmented matrix:

(11111a11ab10)\left(\begin{array}{ccc|c} 1&1&1&1\\ 1&a&1&1\\ a&b&1&0 \end{array}\right)​11a​1ab​111​110​​

For the system to have no solution, we need

  • det⁡(A)=0\det(A)=0det(A)=0 and
  • the augmented matrix to be inconsistent.
  1. Compute the determinant
det⁡(A)=∣1111a1ab1∣\det(A)= \begin{vmatrix} 1&1&1\\ 1&a&1\\ a&b&1 \end{vmatrix}det(A)=​11a​1ab​111​​

Apply row operations:

R2→R2−R1,R3→R3−aR1R_2\to R_2-R_1, \qquad R_3\to R_3-aR_1R2​→R2​−R1​,R3​→R3​−aR1​

Then

det⁡(A)=∣1110a−100b−a1−a∣\det(A)= \begin{vmatrix} 1&1&1\\ 0&a-1&0\\ 0&b-a&1-a \end{vmatrix}det(A)=​100​1a−1b−a​101−a​​

Expand along the second row:

det⁡(A)=(a−1)∣1101−a∣=(a−1)(1−a)=−(a−1)2\det(A)=(a-1) \begin{vmatrix} 1&1\\ 0&1-a \end{vmatrix} =(a-1)(1-a)=-(a-1)^2det(A)=(a−1)​10​11−a​​=(a−1)(1−a)=−(a−1)2

So,

det⁡(A)=0  ⟺  a=1\det(A)=0 \iff a=1det(A)=0⟺a=1

Thus, no solution is possible only when a=1a=1a=1.

  1. Substitute a=1a=1a=1 into the system

The equations become

{x+y+z=1x+y+z=1x+by+z=0\begin{cases} x+y+z=1 \\ x+y+z=1 \\ x+by+z=0 \end{cases}⎩⎨⎧​x+y+z=1x+y+z=1x+by+z=0​

Subtract the first equation from the third:

(x+by+z)−(x+y+z)=0−1(x+by+z)-(x+y+z)=0-1(x+by+z)−(x+y+z)=0−1 (b−1)y=−1(b-1)y=-1(b−1)y=−1

Now analyze:

  • If b=1b=1b=1, then left side is 000, so we get 0=−10=-10=−1 which is impossible. Hence the system has no solution.

  • If b≠1b\ne 1b=1, then y=−1b−1y=\frac{-1}{b-1}y=b−1−1​ and solutions exist.

Therefore, the only value of bbb for which the system has no solution is

b=1 b=1b=1

So,

S={1}S=\{1\}S={1}

which is a singleton.

  1. Check options
  • A: empty set →\to→ false
  • B: infinite set →\to→ false
  • C: finite set with two or more elements →\to→ false
  • D: singleton →\to→ true

Hence, the correct option is D.

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