For which real values of x do the following vectors form a linearly dependent set in R3?
v1= (x, -1/9, -1/9)
v2= (-1/9, x, -1/9)
v3= (-1/9, -1/9, x)
The vectors form a linearly dependent set in ℜ3 if the determinant of the matrix formed by the three vectors is zero.
Evaluate the determinant:
|x,k,k|
|k,x,k|
|k,k,x|
in terms of x and k, where k can eventually substituted for -1/9.
Solve the resulting cubic equation for x in terms of k. It actually factors nicely into (x+?)(x-??)².
In fact, you should be able to determine ?? by inspection.
Post if you need more help.
To determine if the given vectors form a linearly dependent set, we need to find a non-trivial solution to the equation:
c1 * v1 + c2 * v2 + c3 * v3 = 0
where c1, c2, and c3 are constants, not all zero.
Let's set up the equation and solve it:
c1 * (x, -1/9, -1/9) + c2 * (-1/9, x, -1/9) + c3 * (-1/9, -1/9, x) = (0, 0, 0)
Multiplying each term by their respective constant:
(c1 * x, -c1/9, -c1/9) + (-c2/9, c2 * x, -c2/9) + (-c3/9, -c3/9, c3 * x) = (0, 0, 0)
Combining the terms:
(c1 * x - c2/9 - c3/9, -c1/9 + c2 * x - c3/9, -c1/9 - c2/9 + c3 * x) = (0, 0, 0)
To find a non-trivial solution, the determinant of the coefficient matrix must be zero:
| x -1/9 -1/9 |
| -1/9 x -1/9 |
| -1/9 -1/9 x | = 0
Expanding the determinant:
x((x/x) - 1/81) - (-1/9)((-1/9)(x) - (-1/9)(-1/9)) + (-1/9)((-1/9)(-1/9) - (-1/9)(x))
= x(x - 1/81) - (1/81) + (1/81) = 0
Simplifying the equation:
x^2 - 1/81 = 0
To solve this quadratic equation, we can take the square root of both sides:
x = ±(1/9)
Therefore, the vectors form a linearly dependent set in R3 for x = 1/9 and x = -1/9.