91Ó°ÊÓ

In mathematics, a system of equations is a set of two or more equations with the same variables. Solving a system means finding a set of values for the variables that simultaneously satisfy all the equations in the system. For example, in the given exercise:
\[\begin{aligned} x+y-4z &=-4 \ 5x-3y-2z &=0 \ 2x+6y+7z &=30 \end{aligned} \]
We have a system of three equations with three variables \(x, y,\) and \(z\). The solution to this system is a specific set of values \((x, y, z)\) that satisfies all three equations. Finding solutions to such systems is crucial in fields like physics, engineering, and economics, where multiple conditions must often be satisfied simultaneously.

To solve systems of equations like the one in the exercise, we often convert them into matrix form. The augmented matrix combines the coefficients of the variables and the constants from the equations into a single matrix. For the system:
\[\begin{aligned} x + y - 4z &= -4 \ 5x - 3y - 2z &= 0 \ 2x + 6y + 7z &= 30 \end{aligned} \]
The augmented matrix representation is:
\[\begin{bmatrix} 1 & 1 & -4 & | & -4 \ 5 & -3 & -2 & | & 0 \ 2 & 6 & 7 & | & 30 \end{bmatrix} \]
The vertical divider '|' separates the coefficients on the left from the constants on the right. This matrix form simplifies the process of applying row operations, which are steps involved in Gaussian elimination.

Row operations are systematic manipulations of the rows in an augmented matrix to transform it into a simpler form, often an upper triangular form. There are three types of row operations:

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

These operations are used to create zeros below the pivot elements (leading coefficients). For instance, in the exercise, this method led us from:
\[\begin{bmatrix} 1 & 1 & -4 & | & -4 \5 & -3 & -2 & | & 0 \2 & 6 & 7 & | & 30 \end{bmatrix} \]to the simplified form:\[\begin{bmatrix} 1 & 1 & -4 & | & -4 \0 & -8 & 18 & | & 20 \0 & 0 & 24 & | & 48 \end{bmatrix}\]Reaching this form allows us to easily perform back substitution to find the solutions.

Back substitution is the final step in solving a system using Gaussian elimination after achieving an upper triangular matrix form. It involves solving the system from the last row upwards. Starting with the simplest equation, which usually has only one variable, we find its value and then substitute it into the preceding equations.
In our example, the matrix was reduced to:
\[\begin{bmatrix} 1 & 1 & -4 & | & -4 \0 & -8 & 18 & | & 20 \0 & 0 & 24 & | & 48 \end{bmatrix}\]
Here, we start from the bottom:

• From the last equation: \(24z = 48\), we find \(z = 2\).
• Moving upwards: substitute \(z = 2\) into the second equation, \(-8y + 18 \cdot 2 = 20\), giving \(y = 2\).
• Finally, substitute \(y = 2\) and \(z = 2\) into the first equation, \(x + 2 - 4 \cdot 2 = -4\), resulting in \(x = 2\).

Thus, we derive the solution \((x, y, z) = (2, 2, 2)\), fully solving the system of equations.