Question

Solve the system of equations. $$\left\\{\begin{aligned} 4 x+y-3 z &=11 \\ 2 x-3 y+2 z &=9 \\ x+y+z &=-3 \end{aligned}\right.$$

Short answer

The solution to the system of equations is \(x=15\), \(y = -3\), \(z = -15\).

Step-by-step solution

Re-write the given system in matrix form

We can re-write the system of linear equations in the form of a matrix: \( \begin{bmatrix} 4 & 1 & -3 & 11 \ 2 & -3 & 2 & 9 \ 1 & 1 & 1 & -3 \end{bmatrix} \). The matrix represents the coefficients of the variables in the system, hence, the first column refers to the coefficients of x, the second column refers to the coefficients of y, the third column refers to the coefficients of z, etc. The last column represents the constant terms on the right side of the equations.

Perform Gaussian elimination

To make the system easier to solve, we want to transform the matrix into row echelon form. This can be achieved by Gaussian elimination. First, we can swap row 1 and row 3, and then subtract twice of the last row from the first row and then add three times the last row to the second row. So, the new matrix becomes \( \begin{bmatrix} 2 & -1 & 1 & 17 \ 5 & 0 & 5 & 0 \ 1 & 1 & 1 & -3 \end{bmatrix} \). Second, subtract 5 times of the second row from the last row, we get \( \begin{bmatrix} 2 & -1 & 1 & 17 \ 5 & 0 & 5 & 0 \ 0 & 1 & 0 & -3 \end{bmatrix} \). At last, swap the second row and the third row and divide the second row by 2, our matrix becomes row echelon form \( \begin{bmatrix} 2 & 0 & 1 & 15 \ 0 & 1 & 0 & -3 \ 5 & 0 & 5 & 0 \end{bmatrix} \).

Solve for x, y, and z using back-substitution

The matrix now corresponds to the system: \(2x + z = 15\), \(y = -3\), and \(5x + 5z = 0\). Solving the third equation we get \(x = -z\), substituting \(x\) in the first equation, we have \(2*(-z) + z = 15\), \(-z = 15\), we get \(z = -15\). Now, substitute \(z = -15\) into the equation \(x = -z\), we get \(x = 15\). Therefore, the solution to the system of equations is \(x=15\), \(y = -3\), \(z = -15\).

Key concepts

Matrix Representation

When solving a system of linear equations, the first step is to represent the system in matrix form. This is called the matrix representation.
By doing so, we take each equation's coefficients and constants and place them into a structured array called an augmented matrix. For example, if we have a system of equations:

• 4x + y - 3z = 11
• 2x - 3y + 2z = 9
• x + y + z = -3

This can be written as: \[ \begin{bmatrix} 4 & 1 & -3 & | & 11 \ 2 & -3 & 2 & | & 9 \ 1 & 1 & 1 & | & -3 \end{bmatrix} \] Here, the vertical line separates the coefficients of the variables on the left from the constants on the right. Each row matches a single equation, and each column up to the line corresponds to a variable. This structured form makes it easier to apply mathematical operations to find the variables' values.

Gaussian Elimination

Gaussian elimination is a method we use to solve systems of linear equations. It's a process that manipulates the rows of matrices to simplify the solution finding. The goal is to transform the original matrix into an upper triangular form, better known as the row echelon form.
We achieve this through elementary row operations that include:

• Swapping two rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting the multiples of another row to the current row

For instance, in our given matrix, we first swap rows or rearrange to make further steps simpler. Then, we perform operations like multiplying and adding rows to eliminate variables from specific rows, gradually simplifying the matrix. Each step of the process brings us closer to a form that is easy to interpret.

Row Echelon Form

Row echelon form is a transformed state of a matrix that makes solving systems of equations much easier. To reach this form, one must use Gaussian elimination effectively. A matrix is in row echelon form when the following conditions are met:

• All non-zero rows are above any rows of all zeros.
• The leading coefficient (first non-zero number from the left) of a non-zero row is always to the right of the leading coefficient of the row directly above it.
• The leading coefficients are one (can be achieved through row operations).

For example, from our system, after applying Gaussian elimination, the matrix might become:\[ \begin{bmatrix} 1 & 1 & 0 & | & 2 \ 0 & 1 & 0 & | & -3 \ 0 & 0 & 1 & | & 4 \end{bmatrix} \] This form allows us to quickly spot solutions as it isolates the equations, simplifying the back-substitution step.

Back-Substitution

Back-substitution is the decisive step after converting a system's matrix into row echelon form.
It involves solving the equations from the bottom upwards, starting with the last row, which should easily yield the value for one variable. From our example system, after reaching row echelon form, consider the last equation from the matrix:

• Find the value from the last row (e.g., if it reads 0x + 0y + z = c, then z = c)
• Substitute this value back into previous equations to solve for other variables.
• Proceed upwards, using known variable values to simplify and solve each equation.

For instance, once you find values for z in the lower row, use it in the preceding rows to find y and subsequently x. Thus, each variable is uncovered step by step, based on the solutions of prior equations, ensuring an accurate solution for all variables in the system.