Question

Solve the given system using matrices. $$\left\\{\begin{array}{rr}3 x+y+2 z= & 1 \\\2 x-2 y+5 z= & 5 \\\x+3 y+2 z= & -9\end{array}\right.$$

Short answer

The solution is \(x = 35\), \(y = -1\), and \(z = -15\).

Step-by-step solution

- Write the system of equations in matrix form

Rewrite the given system of linear equations in matrix form as follows:\[ A \mathbf{x} = \mathbf{b} \]Where:\[ A = \begin{bmatrix} 3 & 1 & 2 \ 2 & -2 & 5 \ 1 & 3 & 2 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 1 \ 5 \ -9 \end{bmatrix} \]

- Form the augmented matrix

Combine the matrix \(A\) and vector \(\mathbf{b}\) to form the augmented matrix:\[\left[ A \mathbf{b} \right] = \begin{bmatrix} 3 & 1 & 2 & 1 \ 2 & -2 & 5 & 5 \ 1 & 3 & 2 & -9 \end{bmatrix}\]

- Apply Gaussian elimination

Use Gaussian elimination to transform the augmented matrix to row-echelon form:1. Swap row 1 and row 3 to get a leading 1 in the top-left position:\[\begin{bmatrix} 1 & 3 & 2 & -9 \ 2 & -2 & 5 & 5 \ 3 & 1 & 2 & 1 \end{bmatrix}\]2. Eliminate the entries below the leading 1 in column 1:\[\begin{bmatrix} 1 & 3 & 2 & -9 \ 0 & -8 & 1 & 23 \ 0 & -8 & -4 & 28 \end{bmatrix}\]3. Normalize row 2 (divide by -8):\[\begin{bmatrix} 1 & 3 & 2 & -9 \ 0 & 1 & -\frac{1}{8} & -\frac{23}{8} \ 0 & -8 & -4 & 28 \end{bmatrix}\]

- Continue Gaussian elimination

4. Eliminate the entry below the leading 1 in row 2:\[\begin{bmatrix} 1 & 3 & 2 & -9 \ 0 & 1 & -\frac{1}{8} & -\frac{23}{8} \ 0 & 0 & -3.5 & 51 \end{bmatrix}\]5. Normalize row 3 (divide by -3.5):\[\begin{bmatrix} 1 & 3 & 2 & -9 \ 0 & 1 & -\frac{1}{8} & -\frac{23}{8} \ 0 & 0 & 1 & -15 \end{bmatrix}\]

- Back-substitution

Use back-substitution to find the values of \(x\), \(y\), and \(z\):1. From row 3: \(z = -15\)2. Substitute \(z\) into row 2: \(y - \frac{1}{8}(-15) = -\frac{23}{8}\). Simplify to find \(y = -1\)3. Substitute \(y\) and \(z\) into row 1: \(x + 3(-1) + 2(-15) = -9\). Simplify to find \(x = 35\)

Key concepts

Linear Equations

Linear equations are mathematical expressions that represent straight lines when plotted on a graph. Each equation describes a relationship among variables, typically expressed in the form: \(a_1x_1 + a_2x_2 + \text{...} + a_nx_n = b\), where \(a_i\) are coefficients and \(b\) is a constant term.

The goal is to find the values of the variables that satisfy all the equations simultaneously. In a system of linear equations, we can have multiple equations interacting with each other. For our example, we have three linear equations involving three variables: \(x\), \(y\), and \(z\).

Solving such systems can be done through various methods, including graphing, substitution, and using matrices. Using matrices speeds up the solution process, especially for larger systems.

Augmented Matrix

An augmented matrix is a powerful tool for representing a system of linear equations.

It combines the coefficient matrix with the constants from the right side of the equations into one rectangular array. For our system, we initially have:

• The coefficient matrix \(A\): \[ A = \begin{bmatrix} 3 & 1 & 2 \ 2 & -2 & 5 \ 1 & 3 & 2 \end{bmatrix} \]
• The variable vector \(\mathbf{x}\): \[\mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix} \]
• The constants vector \(\mathbf{b}\): \[\mathbf{b} = \begin{bmatrix} 1 \ 5 \ -9 \end{bmatrix} \]

Combining these gives us the augmented matrix:
\[ \left[ A | \mathbf{b} \right] = \begin{bmatrix} 3 & 1 & 2 & 1 \ 2 & -2 & 5 & 5 \ 1 & 3 & 2 & -9 \end{bmatrix} \]

This matrix allows us to apply row operations and facilitates Gaussian elimination.

Row-Echelon Form

Row-echelon form is a special form of a matrix used in solving systems of linear equations. A matrix is in row-echelon form when:

• All zero rows (if any) are at the bottom of the matrix.
• The leading entry of each non-zero row (called a pivot) is to the right of the leading entry of the row above it.
• The entries below each pivot are zeros.

For our augmented matrix, we use Gaussian elimination to convert it into row-echelon form:
Steps:

1. Swap rows to get a 1 in the top-left position.
2. Eliminate entries below the pivot using row operations.
3. Normalize rows as needed.

Finally, we achieve a matrix where every leading term is 1 and zeros below it. This simplifies the process of solving the system.

Back-Substitution

Back-substitution is the final step in the Gaussian elimination method. It involves solving for the variables starting from the bottom row and moving upwards.

Once the augmented matrix is in row-echelon form, you can determine the values of the variables through a step-by-step substitution process:

1. Start with the last row and solve it for the last variable.
2. Substitute the known value back into the previous row to solve for another variable.
3. Continue this process until all variables are determined.

In our example:

• From row 3: \(z = -15\).
• Substitute \(z\) into row 2: \(y - \frac{1}{8}(-15) = -\frac{23}{8}\) simplifies to \(y = -1\).
• Substitute \(y\) and \(z\) into row 1: \(x + 3(-1) + 2(-15) = -9\) simplifies to \(x = 35\).

This systematic approach makes the solution process clear and manageable.