91Ó°ÊÓ

When dealing with systems of equations, the augmented matrix is a powerful tool that allows us to handle multiple linear equations simultaneously. An augmented matrix combines the coefficients of the variables and the constants from the equations into a single matrix. Here, we start by writing the system of equations, which can often look like:

• Equations: \( y + 2z = -2 \), \( x + y = 1 \), \( 2x - z = 0 \)

Before forming the augmented matrix, align each equation’s terms so that corresponding variables are in the same columns:

• System: \[ \begin{align*} 0x + y + 2z &= -2 \ x + y + 0z &= 1 \ 2x + 0y - z &= 0 \end{align*} \]

This organization allows us to construct the augmented matrix:\[\begin{bmatrix} 0 & 1 & 2 & | & -2 \ 1 & 1 & 0 & | & 1 \ 2 & 0 & -1 & | & 0 \end{bmatrix}\]This matrix represents the linear system efficiently for further manipulation using matrix techniques.

Row operations are steps we perform on matrices to simplify them. These operations are crucial in matrix equations to solve for variables systematically. They include:

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

Using row operations, we transform a matrix into simpler forms, like row-echelon form, which is closer to a solution. In our system, we perform operations like swapping rows to position leading coefficients (numbers in front of variables) for easy elimination of other entries.In our example, we start with swapping Row 1 and Row 2:\[\begin{bmatrix} 1 & 1 & 0 & | & 1 \ 0 & 1 & 2 & | & -2 \ 2 & 0 & -1 & | & 0 \end{bmatrix}\]This step places a leading 1 at the start, allowing us to move forward with eliminations and continue towards row-echelon form.

Achieving the row-echelon form is a significant step in solving systems using matrices. This form is characterized by having a staircase-like structure, where each leading entry of a row is to the right of the leading entry of the row above. It simplifies the process of back-substitution.To transform our matrix into this form, we perform operations to create zeros below leading coefficients. Following our row swaps and subsequent operations, we achieve a matrix like:\[\begin{bmatrix} 1 & 1 & 0 & | & 1 \ 0 & 1 & 2 & | & -2 \ 0 & 0 & 3 & | & -6 \end{bmatrix}\]Notice how the leading coefficients form a triangular pattern, helping isolate each variable step by step. Once this is accomplished, the system is ready for the final step: back-substitution.

Back-substitution is the process of solving for variables in a system once it has been simplified to upper triangular form through row operations. Using row-echelon form, we can easily identify and solve for one variable at a time, starting with the one in the last row.In the example, the last row helps solve for \(z\), as we have the equation: \( 0x + 0y + 3z = -6 \). From this, it's straightforward to find:

• \(z = -2\)

Substitute \(z = -2\) into the preceding row to solve for \(y\):

• Row 2: \( y + 2(-2) = -2 \) leading to \( y = 2 \)

Finally, substitute \(y = 2\) and \(z = -2\) into the top row:

• Row 1: \( x + 2 = 1 \), resulting in \( x = -1 \)

This method simplifies each step, confirming the solution by solving equations one variable at a time, ensuring accuracy without complex calculations all at once.