Question

The equations x + 2y + 3z = 1, x – y + 4z = 0,

2x + y + 7z = 1 have –

• A. Only one solution
• B. Two solutions
• C. No solution
• D. Infinitely many solutions

Answer 1

Answer: (D)

Infinitely many solutions

Answer 2

D = $\left| \begin{matrix} 1 & 2 & 3 \\ 1 & - 1 & 4 \\ 2 & 1 & 7 \end{matrix} \right|$ = 0;D₁ = $\left| \begin{matrix} 1 & 2 & 3 \\ 0 & - 1 & 4 \\ 1 & 1 & 7 \end{matrix} \right|$ = 0

D₂ = $\left| \begin{matrix} 1 & 1 & 3 \\ 1 & 0 & 4 \\ 2 & 1 & 7 \end{matrix} \right|$ = 0 ; D₃ = $\left| \begin{matrix} 1 & 2 & 1 \\ 1 & - 1 & 0 \\ 2 & 1 & 1 \end{matrix} \right|$ = 0

Given system of equation may have ly many solution

Check :- Let $\begin{matrix} z = k \end{matrix}$

\ x + 2y = 1 – 3k

x – y = – 4k

– + +

3y = 1 + k

Ž $\begin{matrix} y = \frac{1 + k}{3} \end{matrix}$

$\begin{matrix} x = –4k + \frac{1 + k}{3} = \frac{1 - 11k}{3} \end{matrix}$

Satisfies 3^rd equation

2 $\left( \frac{1–11k}{3} \right)$ + $\left( \frac{1 + k}{3} \right)$ + 7k = 1 Ž 2 – 22k + 1 + k + 21k = 3 Ž 3 = 3

\infinitely many sol.