91Ó°ÊÓ

In solving systems of linear equations, an augmented matrix is a powerful tool. An augmented matrix serves as a simplified representation of a system of linear equations. This matrix includes both the coefficients of the variables and the constants from each equation, arranged in a structured format.

For example, the system given in the exercise is:

• \(2x - y + z = -1\)
• \(x + 3y - 2z = 2\)
• \(-5x + 6y - 5z = 5\)

The augmented matrix incorporates these equations into a single matrix form:\[\begin{bmatrix}2 & -1 & 1 & | & -1 \1 & 3 & -2 & | & 2 \-5 & 6 & -5 & | & 5\end{bmatrix}\]The vertical line separates the coefficients from the constants of the equations. This compact representation makes it easier to apply further calculations and manipulations, such as row operations.

Row operations are essential techniques used to simplify augmented matrices and eventually solve systems of equations. These operations include swapping rows, multiplying a row by a non-zero constant, and adding or subtracting rows. They help transform the matrix into a simpler form, like the triangular form, which is much easier to interpret and resolve.

In our case, the goal is to manipulate the initial augmented matrix:\[\begin{bmatrix}2 & -1 & 1 & | & -1 \1 & 3 & -2 & | & 2 \-5 & 6 & -5 & | & 5\end{bmatrix}\]to isolate and solve each variable. Through specific row operations like replacing Row 3 with \(R3 + 5R1\) and Row 1 with \(R1 - 2R2\), the matrix transitions progressively to:\[\begin{bmatrix}0 & -7 & 5 & | & -5 \1 & 3 & -2 & | & 2 \0 & 1 & 0 & | & 0\end{bmatrix}\]Each operation was chosen to simplify the matrix step by step, gradually removing the coefficients below the leading coefficients (the first non-zero number from the left in any row). This method is systematic and helps achieve a form that is straightforward to interpret for solving the variables.

Back-substitution is the final phase in solving a system of equations using augmented matrices. Once an augmented matrix is brought to triangular or row-reduced form through row operations, back-substitution allows us to find the solutions for each variable, working from the bottom row upwards.

With the matrix simplified, it looks like this:\[\begin{bmatrix}0 & -7 & 5 & | & -5 \1 & 3 & -2 & | & 2 \0 & 1 & 0 & | & 0\end{bmatrix}\]In this form, back-substitution is applied by resolving equations step by step, beginning with the last row. We first find \(y\) from \(0y + 1y = 0\), giving us \(y = 0\). Next, substitute \(y = 0\) into the equation from the second row to solve for \(x\), and then substitute back into the first row to find \(z\).

In the exercise's solution, it gives:

• \(y = 0\)
• Substituting \(y = 0\) into \(x + 3y - 2z = 2\), we find \(x = 2 + 2z\)
• Substitute \(y = 0\) into the first row, we find \(-7 \times 0 + 5z = -5\), hence \(z = -1\)
• Then substituting \(z = -1\) into \(x = 2 + 2z\), gives \(x = 0\)

These steps guide us to the solution \((x, y, z) = (0, 0, -1)\), showcasing the importance and efficacy of back-substitution in systematically resolving systems of equations.