Question

Describe and compare the solution sets of x_1 + 9x_2 - 4x_3 =0, and x_1 +9x_2 -4x_3 = -2.

Answer 1

The first equation, x_1 + 9x_2 - 4x_3 = 0, represents a plane through the origin in three-dimensional space, while the second, x_1 + 9x_2 - 4x_3 = -2 indicates a parallel plane that does not intersect with the first and does not pass through the origin.

Step-by-step explanation:

To describe and compare the solution sets of the equations x_1 + 9x_2 - 4x_3 = 0, and x_1 + 9x_2 - 4x_3 = -2, we first recognize that these are linear equations in three variables. The solution set of each equation represents a plane in three-dimensional space. The first equation, where the result is zero, denotes a plane passing through the origin, meaning it includes the point (0, 0, 0).

The second equation, where the result is -2, also represents a plane parallel to the first but translated along the vector normal to the plane by the distance that results in the constant term of -2 when plugged into the equation. This means the second plane does not pass through the origin since the constant term on the right side of the equation is non-zero.

Therefore, while both equations have an infinite number of solutions, since planes in three dimensions have an infinite extent, these two planes are distinct and do not share any points: the first plane is a subset of solutions that includes the origin, and the second plane is a parallel plane that is shifted away from the origin.

Answer 2

The solution sets of
`\(x_1 + 9x_2 - 4x_3 = 0\)`and
`\(x_1 + 9x_2 - 4x_3 = -2\)`are different. The first equation represents a plane in three-dimensional space passing through the origin, while the second equation represents a parallel plane shifted two units below the first plane.

Step-by-step explanation:

The given system of equations is
`\(x_1 + 9x_2 - 4x_3 = 0\)` and
`\(x_1 + 9x_2 - 4x_3 = -2\).` To understand the solution sets, we can express these equations in a matrix form \(Ax = B\), where \(A\) is the coefficient matrix, \(x\) is the column vector of variables \([x_1, x_2, x_3]^T\), and \(B\) is the column vector on the right-hand side.

`\[A = \begin{bmatrix} 1 & 9 & -4 \\ 1 & 9 & -4 \end{bmatrix}, \quad x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ -2 \end{bmatrix}\]`

Row-reducing the augmented matrix
`\([A|B]\)` brings us to the form
`\([R|C]\),`where R is the reduced row-echelon form of A, and C is the modified right-hand side.

Now, the first equation,
`\(x_1 + 9x_2 - 4x_3 = 0\)`, represents a plane passing through the origin in three-dimensional space. The second equation,
`\(x_1 + 9x_2 - 4x_3 = -2\),`is parallel to the first plane but shifted two units downward. Therefore, the solution sets of the two equations are distinct, with the first plane passing through the origin, and the second plane running parallel to it but displaced in the negative Z-direction.