Question

The equation x + 2y + 3z = 1, x – y + 4z = 0, 2x + y + 7z = 1 have –

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

Answer 1

Answer: (D)

Infinitely many solution

Answer 2

Q ∆ = $\left| \begin{matrix} 1 & 2 & 3 \\ 1 & –1 & 4 \\ 2 & 1 & 7 \end{matrix} \right|$

= +1 (– 11) – 2 (– 1) + 3 (3) = – 11 + 2 + 9 = 0

Now ∆_x = $\left| \begin{matrix} 1 & 2 & 3 \\ 0 & –1 & 4 \\ 1 & 1 & 7 \end{matrix} \right|$ = + 1 (– 11) + 1(11) = 0

& ∆_y = $\left| \begin{matrix} 1 & 1 & 3 \\ 1 & 0 & 4 \\ 2 & 1 & 7 \end{matrix} \right|$ = – 1 (– 1) – 1 (1) = 0

& ∆_z = $\left| \begin{matrix} 1 & 2 & 1 \\ 1 & –1 & 0 \\ 2 & 1 & 1 \end{matrix} \right|$ = + 1 (3) + 1 (– 3) = 0

∴ ∞ solution.