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Matrices and Determinants question

2018 · 15 Apr · Shift 2 · Q24
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  5. /2018 · 15 Apr · Shift 2 · Q24

Matrices and Determinants question

2018 · 15 Apr · Shift 2 · Q24

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If the system of linear equations x + ay + z = 3 x + 2y + 2z = 6 x + 5y + 3z = b has no solution, then :
  1. A
    a = −-− 1, b = 9
  2. B
    a = −-− 1, b eee 9
  3. C
    a e−e -e− 1, b = 9
  4. D
    a = 1, b eee 9
View written solutionFree

Correct answer: B

  1. Write the system in augmented matrix form:
{x+ay+z=3x+2y+2z=6x+5y+3z=b\begin{cases} x+ay+z=3\\ x+2y+2z=6\\ x+5y+3z=b \end{cases}⎩⎨⎧​x+ay+z=3x+2y+2z=6x+5y+3z=b​

Coefficient matrix is

A=(1a1122153)A=\begin{pmatrix} 1 & a & 1\\ 1 & 2 & 2\\ 1 & 5 & 3 \end{pmatrix}A=​111​a25​123​​

For the system to have no solution, the coefficient matrix must be singular and the augmented system must be inconsistent.

  1. First find when the coefficient matrix is singular.
det⁡(A)=∣1a1122153∣\det(A)=\begin{vmatrix} 1 & a & 1\\ 1 & 2 & 2\\ 1 & 5 & 3 \end{vmatrix}det(A)=​111​a25​123​​

Apply row operations: replace R2→R2−R1R_2\to R_2-R_1R2​→R2​−R1​ and R3→R3−R1R_3\to R_3-R_1R3​→R3​−R1​:

det⁡(A)=∣1a102−a105−a2∣\det(A)=\begin{vmatrix} 1 & a & 1\\ 0 & 2-a & 1\\ 0 & 5-a & 2 \end{vmatrix}det(A)=​100​a2−a5−a​112​​

So,

det⁡(A)=∣2−a15−a2∣=(2−a)⋅2−(5−a)\det(A)=\begin{vmatrix} 2-a & 1\\ 5-a & 2 \end{vmatrix} =(2-a)\cdot 2-(5-a)det(A)=​2−a5−a​12​​=(2−a)⋅2−(5−a) =4−2a−5+a=−1−a=4-2a-5+a=-1-a=4−2a−5+a=−1−a

Hence,

det⁡(A)=−(a+1)\det(A)=-(a+1)det(A)=−(a+1)

For singularity,

det⁡(A)=0⇒a=−1\det(A)=0 \Rightarrow a=-1det(A)=0⇒a=−1
  1. Now substitute a=−1a=-1a=−1 into the system:
{x−y+z=3x+2y+2z=6x+5y+3z=b\begin{cases} x-y+z=3\\ x+2y+2z=6\\ x+5y+3z=b \end{cases}⎩⎨⎧​x−y+z=3x+2y+2z=6x+5y+3z=b​

Subtract the first equation from the second:

(x+2y+2z)−(x−y+z)=6−3(x+2y+2z)-(x-y+z)=6-3(x+2y+2z)−(x−y+z)=6−3 3y+z=3...(i)3y+z=3 \qquad ...(i)3y+z=3...(i)

Subtract the second equation from the third:

(x+5y+3z)−(x+2y+2z)=b−6(x+5y+3z)-(x+2y+2z)=b-6(x+5y+3z)−(x+2y+2z)=b−6 3y+z=b−6...(ii)3y+z=b-6 \qquad ...(ii)3y+z=b−6...(ii)

For consistency, equations (i) and (ii) must match, so we need

3=b−6⇒b=93=b-6 \Rightarrow b=93=b−6⇒b=9

Thus:

  • if b=9b=9b=9, the system is consistent (infinitely many solutions, since determinant is zero),
  • if b≠9b\ne 9b=9, the system is inconsistent, hence no solution.
  1. Therefore, the condition for no solution is
a=−1,b≠9a=-1, \quad b\ne 9a=−1,b=9

So the correct option is B.

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