Problem 51 - Elementary and Intermediate Algebra

Question

Solve. $$\begin{aligned} w+x-y+z &=0 ,\\\ w-2 x-2 y-z &=-5 ,\\\ w-3 x-y+z &=4 ,\\\ 2 w-x-y+3 z &=7 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is w=3, x=2, y=1, z=3.

Step 1 - Write the system of equations

The system of equations can be written as:\[ \begin{aligned} w + x - y + z &= 0 ,\ w - 2x - 2y - z &= -5 ,\ w - 3x - y + z &= 4 ,\ 2w - x - y + 3z &= 7 \end{aligned} \]

Step 2 - Form the augmented matrix

Write the augmented matrix for the given system:\[ \begin{pmatrix} 1 & 1 & -1 & 1 & | & 0 \ 1 & -2 & -2 & -1 & | & -5 \ 1 & -3 & -1 & 1 & | & 4 \ 2 & -1 & -1 & 3 & | & 7 \end{pmatrix} \]

Step 3 - Apply row operations to reach echelon form

Perform row operations to transform the matrix into echelon form. First, subtract the first row from the second, third, and fourth rows:\[ R_2 \to R_2 - R_1:\begin{pmatrix} 1 & 1 & -1 & 1 & | & 0 \ 0 & -3 & -1 & -2 & | & -5 \ 1 & -3 & -1 & 1 & | & 4 \ 2 & -1 & -1 & 3 & | & 7 \end{pmatrix} \]Subtract the first row from the third row: \[ R_3 \to R_3 - R_1:\begin{pmatrix} 1 & 1 & -1 & 1 & | & 0 \ 0 & -3 & -1 & -2 & | & -5 \ 0 & -4 & 0 & 0 & | & 4 \ 2 & -1 & -1 & 3 & | & 7 \end{pmatrix} \]Subtract twice the first row from the fourth row: \[ R_4 \to R_4 - 2R_1:\begin{pmatrix} 1 & 1 & -1 & 1 & | & 0 \ 0 & -3 & -1 & -2 & | & -5 \ 0 & -4 & 0 & 0 & | & 4 \ 0 & -3 & 1 & 1 & | & 7 \end{pmatrix} \]

Step 4 - Further row operations

Perform more row operations to simplify:\[ R_4 \to R_4 - R_2:\begin{pmatrix} 1 & 1 & -1 & 1 & | & 0 \ 0 & -3 & -1 & -2 & | & -5 \ 0 & -4 & 0 & 0 & | & 4 \ 0 & 0 & 2 & 3 & | & 12 \end{pmatrix} \]Divide row four by 2:\[ R_4 \to \frac{R_4}{2}:\begin{pmatrix} 1 & 1 & -1 & 1 & | & 0 \ 0 & -3 & -1 & -2 & | & -5 \ 0 & -4 & 0 & 0 & | & 4 \ 0 & 0 & 1 & 1.5 & | & 6 \end{pmatrix} \]Add 1/3 of row 2 to row 1:\[ R_1 \to R_1 + \frac{1}{3} R_2:\begin{pmatrix} 1 & 0 & -\frac{4}{3} & \frac{1}{3} & | & -\frac{5}{3} \ 0 & -3 & -1 & -2 & | & -5 \ 0 & -4 & 0 & 0 & | & 4 \ 0 & 0 & 1 & 1.5 & | & 6 \end{pmatrix} \]

Step 5 - Solve for variables

Back substitute to solve for each variable. From row 4: \( z = 6 - 1.5y \), substitute into row 3:\[ 0 - 4y = 4 - (6 - 1.5y) \to y = 1 \]use value for y in row 4: \[ 1.5 - y = 6 \to z = 3 \]Substitute y, z into row 2: \[ x = \frac{-5 + 4 + 3}{-3} \to x = 2 \]use values in row 1 to find w: \[ w = - \frac{5 - 4x - z}{3} \to w = 3 \]

Key Concepts

Augmented Matrix

An augmented matrix is an essential tool in solving systems of linear equations. It's a way to represent the system in a concise format, which combines the coefficients of the variables and the constants from the equations. In our exercise, we have four equations and four variables. The augmented matrix looks like this: $$\begin{pmatrix} 1 & 1 & -1 & 1 & | & 0 \, 1 & -2 & -2 & -1 & | & -5 \, 1 & -3 & -1 & 1 & | & 4 \, 2 & -1 & -1 & 3 & | & 7 \end{pmatrix} $$ The vertical line separates the coefficients of the variables from the constants on the right-hand side of the equations. This allows us to focus on the coefficients for row operations more easily.

Row Operations

Row operations are used to simplify the augmented matrix into a form that makes it easier to solve the system of equations. These operations include:

• Swapping two rows (Row Swapping)
• Multiplying a row by a non-zero scalar (Row Scaling)
• Adding or subtracting a multiple of one row to another row (Row Addition/Subtraction)

For instance, in our solution, we subtract the first row from other rows to eliminate the coefficient of the first variable in the subsequent rows. This step-by-step reduction is crucial in achieving the echelon form, making it simpler to solve for variables.

Echelon Form

Echelon form, or row-echelon form, is a simplified version of the augmented matrix that helps in solving linear equations. It's achieved through strategic row operations. In echelon form:

• All non-zero rows are above the rows of zeros.
• The leading coefficient (first non-zero number) of a non-zero row is always to the right of the leading coefficient of the row above it.
• The leading coefficients are typically 1 for easier calculations.

Our goal in the exercise is to simplify the matrix until we get a triangular shape, like this: $$\begin{pmatrix} 1 & 1 & -1 & 1 & | & 0 \, 0 & -3 & -1 & -2 & | & -5 \, 0 & -4 & 0 & 0 & | & 4 \, 0 & 0 & 1 & 1.5 & | & 6 \end{pmatrix} $$ This echelon form is closer to solving the system than the original augmented matrix.

Back Substitution

Back substitution is the final step in solving the system of equations once the augmented matrix is in echelon form. Starting with the last row, solve for the corresponding variable and substitute it back into the previous rows. For example:

• From the last row, we find that $$z = 6 - 1.5y;$$
• We use this value of $$z$$ in the third row equation to solve for $$y.$$
• Next, we use the values of $$y$$ and $$z$$ to solve for $$x$$ and so on.

This step-by-step substitution helps in finding the values of all variables systematically. In our solution, the values are: $$w = 3, x = 2, y = 1, z = 3.$$ Each step builds on the accuracy of the previous one, which is the power of back substitution.