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 :
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
This question is part of PrepWiser's free JEE Main question bank. 274 more solved questions on Matrices and Determinants are available — start with the harder ones if your accuracy is >70%.