Medium MCQ +4 / -1 PYQ · JEE Mains 2023

Let the system of linear equations

$x+y+kz=2$

$2x+3y-z=1$

$3x+4y+2z=k$

have infinitely many solutions. Then the system

$(k+1)x+(2k-1)y=7$

$(2k+1)x+(k+5)y=10$

has :

  1. A unique solution satisfying $x-y=1$
  2. B infinitely many solutions
  3. C no solution
  4. D unique solution satisfying $x+y=1$ Correct answer

Solution

<p>$x + y + kz = 2$ ............(i)</p> <p>$2x + 3y - z = 1$ ..........(ii)</p> <p>$3x + 4y + 2z = k$ ......(iii)</p> <p>(1) + (2)</p> <p>$3x + 4y + z(k - 1) = 3$</p> <p>Comparing with (3)</p> <p>$k = 3$</p> <p>Now, $4x + 5y = 7$</p> <p>$\Rightarrow 3x + 3y = 3$</p> <p>$7x + 8y = 10$</p> <p>as ${4 \over 7} \ne {5 \over 8}$</p> <p>$\therefore$ Unique solution satisfying $x + y = 1$</p>

About this question

Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations

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