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Question: The number of solutions of the equations \(x + 4y - z = 0,\) \(3x - 4y - z = 0,x - 3y + z = 0\) is....

The number of solutions of the equations x+4y−z=0,x + 4y - z = 0, 3x−4y−z=0,x−3y+z=03x - 4y - z = 0,x - 3y + z = 0 is.

A

0

B

1

C

2

D

Infinite

Answer

1

Explanation

Solution

The given system of homogeneous equations has Δ=∣14−13−4−11−31∣=1(−4−3)−4(3+1)−1(−9+4)\Delta = \left| \begin{matrix} 1 & 4 & - 1 \\ 3 & - 4 & - 1 \\ 1 & - 3 & 1 \end{matrix} \right| = 1( - 4 - 3) - 4(3 + 1) - 1( - 9 + 4)

=−7−16+5≠0= - 7 - 16 + 5 \neq 0.

There exists only one trivial solution.

Δ=∣14−13−4−11−31∣=1(−4−3)−4(3+1)−1(−9+4)\Delta = \left| \begin{array} { r r r } 1 & 4 & - 1 \\ 3 & - 4 & - 1 \\ 1 & - 3 & 1 \end{array} \right| = 1 ( - 4 - 3 ) - 4 ( 3 + 1 ) - 1 ( - 9 + 4 )

=−7−16+5≠0= - 7 - 16 + 5 \neq 0 .