91Ó°ÊÓ

In mathematics, a system of equations is a collection of two or more equations with the same set of unknowns. For this exercise, we have a system of three equations with three unknowns:
\(\begin{aligned} -2 x-2 y+3 z &=4 \ 5 x+7 y-z &=2 \ 2 x+2 y-3 z &=-4 \end{aligned}\).
Each equation represents a plane in 3D space. The solution to the system is a set of values for the variables x, y, and z that satisfy all the equations simultaneously.

Determinants are special numbers that can be calculated from a square matrix. They are very useful in solving systems of equations, matrix inversion, and other applications.
For a 3x3 matrix, the determinant can be calculated using the following formula: \[\begin{vmatrix} a & b & c \ d & e & f \ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg)\].
In our exercise, the coefficient matrix is \(A = \begin{pmatrix} -2 & -2 & 3 \ 5 & 7 & -1 \ 2 & 2 & -3 \end{pmatrix}\).
By calculating the determinant of this matrix, we find that the determinant \(D = 0\). This means that the matrix is singular and Cramer's rule cannot be applied directly to solve the system of equations.

Gaussian elimination is a method for solving systems of linear equations. It works by transforming the system's augmented matrix into row echelon form using a sequence of row operations.
These row operations include:

• Swapping two rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting a multiple of one row to another row

For our exercise, we apply Gaussian elimination to the augmented matrix \[ \begin{pmatrix} -2 & -2 & 3 & | & 4 \ 5 & 7 & -1 & | & 2 \ 2 & 2 & -3 & | & -4 \end{pmatrix} \].
Through successive row operations, we obtain the row echelon form: \[ \begin{pmatrix} 1 & 1 & -\frac{3}{2} & | & -2 \ 0 & 1 & -2 & | & -4 \ 0 & 0 & 0 & | & 0 \end{pmatrix} \].
This indicates that the system has infinitely many solutions.

The solution set of a system of equations refers to all possible solutions that satisfy all equations in the system. Since our determinant \(D = 0\), the system is dependent, and we have infinitely many solutions.
To find the solution set, we use back substitution from the row echelon form matrix:

• Let \(z = t\) (where \(t\) is a parameter)
• From Row 2: \(y - 2z = -4 \Rightarrow y = -4 + 2t\)
• From Row 1: \(x + y - \frac{3}{2}z = -2 \Rightarrow x = 2 + \frac{1}{2}t\)

Thus, the final solution set is \((x, y, z) = \left( 2 + \frac{1}{2}t, -4 + 2t, t \right)\), where \(t\) is any real number.