Question

If 3x+2y+z=0, x+4y+z=0, 2x+y+4z=0 be a system of equation, then * 1 point a. It has only the trivial solution x=0, y=0, z=0 b. It is consistent c. It can be reduced to a single equation and so a solution does not exist d. Determinant of the matrix of coefficient is zero

Answer 1

Given:

A system of equations 3x+2y+z=0, x+4y+z=0, 2x+y+4z=0  

To find:

a. It has only the trivial solution x=0, y=0, z=0  

b. It is consistent  

c. It can be reduced to a single equation and so a solution does not exist  

d. Determinant of the matrix of coefficient is zero

Solution:

From given, we have,

A system of equations 3x + 2y + z = 0, x + 4y + z = 0, 2x + y + 4z = 0  

So, we have,

$A = \left[\begin{array}{ccc}3&2&1\\1&4&1\\2&1&4\end{array}\right]$

$X = \left[\begin{array}{ccc}x\\y\\z\end{array}\right]$

$O = \left[\begin{array}{ccc}0\\0\\0\end{array}\right]$

The determinant of A,

|A| =

3   2   1

1   4   1

2   1   4

3 (16 - 1) - 2 (4 - 2) + 1 (1 - 8) = 3(15) - 2(2) + 1(-7) = 45 - 4 - 7  = 34

|A| = 34  ≠ 0

∴ ρ (A) = 3

∴ ρ (A)  is equal to the number of varaibles.

∴ Option (a) There is only trivial solution x = 0, y = 0, z = 0.