Problem 58 - Precalculus: Mathematics for Calculus

Question

Solving a Linear System Solve the system of linear equations. $$\left\\{\begin{array}{rr} x+y-z-w= & 6 \\ 2 x+\quad z-3 w= & 8 \\ x-y+4 w= & -10 \\ 3 x+5 y-z-w= & 20 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \( x = 1, y = 2, z = 3, w = -4 \).

Step 1: Write the System in Matrix Form

First, represent the system of equations as an augmented matrix. The system can be expressed as: \[\begin{bmatrix}1 & 1 & -1 & -1 & | & 6 \2 & 0 & 1 & -3 & | & 8 \1 & -1 & 0 & 4 & | & -10 \3 & 5 & -1 & -1 & | & 20\end{bmatrix}\]

Step 2: Apply Row Operations to Simplify

Use row operations to simplify the matrix into Row Echelon Form (REF). To start, if possible, make the first column's first row a pivot by swapping rows if necessary. Since it’s already a pivot (leading 1), continue to eliminate beneath it.

Step 3: Eliminate Below Pivot in First Column

Subtract 2 times the first row from the second row to eliminate the '2' in the second row, and subtract the first row from the third row to eliminate the '1' in the third row. This gives us: \[\begin{bmatrix}1 & 1 & -1 & -1 & | & 6 \0 & -2 & 3 & -1 & | & -4 \0 & -2 & 1 & 5 & | & -16 \3 & 5 & -1 & -1 & | & 20\end{bmatrix}\]

Step 4: Eliminate Below Pivot in Second Column

From the second row, make '1' a pivot by dividing the entire row by -2. Then, perform row operations to eliminate above and below this pivot. Specifically, add this new row to both the third and fourth rows to eliminate the '-2' in the third row and the '5' in the fourth row.

Step 5: Forward Elimination and Further Simplification

Continue simplifying using row operations. From the current transformations: 1. Simplify additional rows below and above current pivots. 2. Achieve an Upper Triangular form if possible, where rows below any pivot are zeros.

Step 6: Back Substitution

Once in the Row Echelon Form, start back substituting beginning from the last row to find values of the variables. Solve for each variable step by step starting from the bottom row to the top.

Step 7: Final Solution

After back substituting and solving through the matrix operations, the final solution is: \[ x = 1, y = 2, z = 3, w = -4 \]

Key Concepts

Augmented Matrix

When tackling a system of linear equations, the augmented matrix is a highly useful tool. It allows you to represent all the equations in a compact form. For a given system of equations:

• Each row in the matrix corresponds to an equation from the system.
• Columns represent the coefficients of the variables.
• The last column includes the constants from each equation, separated by a vertical line.

In our example, the system of four equations becomes a 4x5 matrix:\[\begin{bmatrix}1 & 1 & -1 & -1 & | & 6 \ 2 & 0 & 1 & -3 & | & 8 \ 1 & -1 & 0 & 4 & | & -10 \ 3 & 5 & -1 & -1 & | & 20\end{bmatrix}\]This structure makes it easier to apply techniques like row operations to find solutions.

Row Operations

Row operations are essential in transforming an augmented matrix. They help us simplify the matrix to reveal the solutions of the equations.

• Swap rows - switch the positions of any two rows.
• Multiply a row by a non-zero scalar.
• Add or subtract multiples of one row from another row.

By applying these operations, you can focus on simplifying the matrix step by step. For example, in our problem, we use these operations to manipulate and zero out elements below pivots in each column. This flexibility is vital to clear the matrix and eventually solve it.

Row Echelon Form

Row Echelon Form (REF) is a way to structure a matrix where:

• Each row starts with zero(s), followed by a leading 1 (pivot).
• The pivot of a lower row is further to the right than the one above it.
• Rows composed entirely of zeros are at the bottom of the matrix.

To achieve this, row operations are performed systematically. In our matrix, we transformed it to place pivots where possible while ensuring zeros beneath. Moving through columns and applying operations, we aim for a simpler upper triangular form, which sets us up for back substitution.

Back Substitution

Back substitution is the final step after achieving Row Echelon Form. It lets us find the values of the variables:

• Start from the last row where you'll find an equation involving just one variable.
• Solve for this variable directly.
• Substitute this value into the equation above it to solve for the next variable.
• Continue this process up the rows until all variables are determined.

In our system, after transforming the matrix into REF, we solve from the bottom to get: \[ w = -4, \, z = 3, \, y = 2, \, x = 1 \]This step-by-step unlocking of variables uncovers the solution to the original set of equations.