If the system of equations
x+y+z=2
2x+4y–z=6
3x+2y+$\lambda$z=$\mu$
has infinitely many solutions, then
Solution
$$D = 0\,\left| {\matrix{
1 & 1 & 1 \cr
2 & 4 & { - 1} \cr
3 & 2 & \lambda \cr
} } \right| = 0$$<br><br>$\Rightarrow$ $(4\lambda + 2) - 1(2\lambda + 3) + 1(4 - 12) = 0$<br><br>$\Rightarrow$ $4\lambda + 2$ $- 2\lambda - 3$$-$8$= 0$<br><br>$\Rightarrow$ $2\lambda = 9 \Rightarrow \lambda = {9 \over 2}$<br><br>$${D_x} = \left| {\matrix{
2 & 1 & 1 \cr
6 & 4 & { - 1} \cr
\mu & 2 & { - 9/2} \cr
} } \right| = 0$$<br><br>$\Rightarrow \mu = 5$<br><br>By checking all the options we find (C) is correct <br><br>$2\lambda + \mu = 14$
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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