Problem 63 - Precalculus: Mathematics for Calculus

Question

Solving a Linear System Solve the system of linear equations. $$\left\\{\begin{aligned} x+z+w &=4 \\ y-z &=-4 \\ x-2 y+3 z+w &=12 \\ 2 x &-2 z+5 w=-1 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The system has infinitely many solutions, expressed in terms of parameter \(w\).

Step 1: Write the system in matrix form

We will represent the system of linear equations in the form \(AX = B\), where:\[A = \begin{bmatrix} 1 & 0 & 1 & 1 \ 0 & 1 & -1 & 0 \ 1 & -2 & 3 & 1 \ 2 & 0 & -2 & 5 \end{bmatrix}\] \[X = \begin{bmatrix} x \ y \ z \ w \end{bmatrix}\] \[B = \begin{bmatrix} 4 \ -4 \ 12 \ -1 \end{bmatrix}\]

Step 2: Use Gaussian elimination to simplify

To solve the system, we apply row operations to transform matrix \(A\) into an upper triangular matrix. Start by using row 1 to eliminate \(x\) from row 3 and row 4:- Row 3 becomes \(R3 - R1\), giving \([0, -2, 2, 0 | 8]\).- Row 4 becomes \(R4 - 2R1\), giving \([0, 0, -4, 3 | -9]\). So the matrix is:\[\begin{bmatrix} 1 & 0 & 1 & 1 | 4 \ 0 & 1 & -1 & 0 | -4 \ 0 & -2 & 2 & 0 | 8 \ 0 & 0 & -4 & 3 | -9 \end{bmatrix}\]

Step 3: Further simplify using row operations

Next, focus on the 3rd row to eliminate and simplify further:- Divide row 4, the row starting with \([-4..]\), by -4, to get \([0, 0, 1, -\frac{3}{4} | \frac{9}{4}]\).Using the 2nd row, eliminate \(y\) from the 3rd row:- Perform \(R3 + 2R2\), giving \([0, 0, 0, 0 | 0]\).The matrix now is:\[\begin{bmatrix} 1 & 0 & 1 & 1 | 4 \ 0 & 1 & -1 & 0 | -4 \ 0 & 0 & 0 & 0 | 0 \ 0 & 0 & 1 & -\frac{3}{4} | \frac{9}{4} \end{bmatrix}\]

Step 4: Solve for remaining variables using back substitution

Starting from the last non-zero row, express in terms of available variables:- Solve for \(z\) from row 4: \(z - \frac{3}{4}w = \frac{9}{4}\). Thus, \(z = \frac{9}{4} + \frac{3}{4}w\).For row 2:- Solve: \(y - z = -4\); substituting \(z\), get \(y = \frac{9}{4} + \frac{3}{4}w - 4\).For row 1:- Solve: \(x + z + w = 4\); substituting \(z\), get \(x + \frac{9}{4} + \frac{3}{4}w + w = 4\).

Key Concepts

Gaussian Elimination

Gaussian Elimination is a systematic method for solving systems of linear equations. This technique involves performing a series of row operations on an augmented matrix, which represents the system. The main goal is to simplify the matrix into an upper triangular form, a state where all the coefficients below the diagonal are zeros.

• The process starts with the first row and uses it to eliminate the coefficient of the first variable in all subsequent rows.
• Then, it continues with the next row to eliminate the coefficient of the next variable from the rows below it.

This is repeated until you have an upper triangular matrix. The operations used are:

• Swapping two rows
• Multiplying a row by a non-zero constant
• Adding or subtracting the multiples of a row from another row

By the end of this process, the matrix is in a form that can be easily solved through back substitution.

Matrix Representation

Matrix representation is a method used to compactly express a system of linear equations. The system is represented by a matrix equation, typically in the form \(AX = B\), where:

• \(A\) is the coefficient matrix – a matrix formed by aligning all the coefficients of the variables from the equations row-wise.
• \(X\) is the column vector of variables \((x, y, z, w)\).
• \(B\) is the column vector formed by the constants on the opposite side of the equality in each equation.

Representing a system of linear equations in matrix form streamlines the process of applying Gaussian elimination. It organizes all components of the system in a structured manner, making it easier to perform row operations and track changes to each part of the system during the solution process.

Back Substitution

After obtaining an upper triangular matrix through Gaussian elimination, the solution uses back substitution. Back substitution is analogous to peeling an onion layer by layer, where you solve the equations one at a time, starting from the last row and moving upwards.

• Start with the last non-zero row, where typically one variable is isolated. Solve for this variable first.
• Use the solution of the previous step in the equation above it to find another variable.
• Continue solving for each variable by substituting already determined values in previous equations.

This step-by-step approach helps to isolate each variable confidently, finishing with the complete solution to the system. Back substitution complements Gaussian elimination by ensuring that each variable is accounted for accurately.