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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Determinants

BITSAT 2024 · Paper 2 · Mathematics · Determinants

The equations x-y+2 z=43 x+y+4 z=6x+y+z=1 have

The equations x−y+2z=43x+y+4z=6x+y+z=1x-y+2 z=43 x+y+4 z=6x+y+z=1 have

Options

  1. Aunique solution
  2. Binfinitely many solutions
  3. Cno solution
  4. Dtwo solutions

Correct answer

B. infinitely many solutions

Step-by-step solution

Given equations, x−y+2z=43x+y+4z=6x+y+z=1x-y+2 z=43 x+y+4 z=6x+y+z=1 Let \Delta=\left|1−12314111\begin{array}{ccc}1 & -1 & 2 \\ 3 & 1 & 4 \\ 1 & 1 & 1\end{array}\right|=1(1-4)+1(3-4)+2(3-1)=-3-1+4=0 and \Delta_1=\left|4−12614111\begin{array}{ccc}4 & -1 & 2 \\ 6 & 1 & 4 \\ 1 & 1 & 1\end{array}\right|=4(1-4)+1(6-4)+2(6-1)=-12+2+10=0 Now, Δ=0\Delta=0 and Δ1=0∴\Delta_1=0\therefore These equations have infinitely many solutions.
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