Question

Solve the sets of homogeneous equations by row reducing the matrix. $$\left\\{\begin{array}{r}2 x-3 y+5 z=0 \\\x+2 y-z=0 \\\x-5 y+6 z=0 \\\4 x+y+3 z=0\end{array}\right.$$

Short answer

Solution is \( x = -z \), \( y = z \), and \( z = z\). General solution: \( (-z, z, z) \) where \( z \in \mathbb{R} \).

Step-by-step solution

- Write Augmented Matrix

First, write the system of equations as an augmented matrix:draw the augmented matrix for the system:\[\begin{pmatrix}2 & -3 & 5 & | & 0 \ 1 & 2 & -1 & | & 0 \ 1 & -5 & 6 & | & 0 \ 4 & 1 & 3 & | & 0o\end{pmatrix}\]

- Transform to Row Echelon Form (REF)

Apply row operations to transform the matrix into Row Echelon Form (REF). Begin by making the first element of the first row a 1 if possible. If the leading entry isn't a 1, swap rows if necessary.New matrix state:\[\begin{pmatrix}1 & 2 & -1 & | & 0 \ 2 & -3 & 5 & | & 0 \ 1 & -5 & 6 & | & 0 \ 4 & 1 & 3 & | & 0o\end{pmatrix}\]

- Eliminate below Pivot

Eliminate the elements below the pivot in the first column by row operations. specifically, subtract 2 times row 1 from row 2 and subtract row 1 from row 3 and subtract 4 times row 1 from row 4:New matrix state:\[\begin{pmatrix}1 & 2 & -1 & | & 0 \ 0 & -7 & 7 & | & 0 \ 0 & -7 & 7 & | & 0 \ 0 & -7 & 7 & | & 0o\end{pmatrix}\]

- Further Reduction

Recognize rows 2, 3, and 4 are identical. Therefore, replace rows 3 and 4 with zero rows, then simplify the next pivot:New matrix state:\[\begin{pmatrix}1 & 2 & -1 & | & 0 \ 0 & -7 & 7 & | & 0 \ 0 & 0 & 0 & | & 0 \ 0 & 0 & 0 & | & 0o\end{pmatrix}\]

- Reduced Row Echelon Form

Lastly, make the pivot in the second row a 1 by dividing by -7:New matrix state:\[\begin{pmatrix}1 & 2 & -1 & | & 0 \ 0 & 1 & -1 & | & 0 \ 0 & 0 & 0 & | & 0 \ 0 & 0 & 0 & | & 0o\end{pmatrix}\]

- Write Systems of Equations

Convert the reduced matrix back into a system of equations:\[\begin{cases}x + 2y - z = 0 \ y - z = 0o\end{cases}\]

- Solve for Variables

From second equation, solve for \(y\) in terms of \(z\):\[ y = z\]Substitute \(y = z\) into first equation and simplify:\[x + 2z - z = 0 \ x + z = 0 \ x = -z\]

Key concepts

Row Reduction

The process of row reduction, also known as row operations, is essential for solving systems of linear equations. When you perform row reduction, you use three primary operations:

• Swapping two rows
• Multiplying a row by a nonzero constant
• Adding or subtracting a multiple of one row to another row

These operations help to simplify the matrix, making it easier to find solutions. The ultimate goal is to reach a form where the values reveal the solutions directly or simplify back substitution. Let's start by transforming the given system of equations into an augmented matrix, where the last column represents the constants on the right side of the equations.

Row Echelon Form (REF)

Row Echelon Form (REF) is a specific arrangement of a matrix that helps us to solve systems of linear equations more efficiently. To achieve REF:

• All nonzero rows are above any rows of all zeros.
• The leading entry of each nonzero row (also called a pivot) is 1.
• The pivot in any row is to the right of the pivot in the row above it.

To transform our matrix into REF, we begin with the first pivot and use row operations to eliminate the entries below it. Look at the first column, make the first non-zero element a 1, and then perform row operations to make all entries below it zero. This significantly simplifies the matrix and reduces the complexity of solving the system.

Reduced Row Echelon Form (RREF)

Reduced Row Echelon Form (RREF) goes a step further than REF by ensuring even more specific conditions are met:

• Each pivot is 1, and it is the only nonzero entry in its column.
• All rows with pivots (leading 1s) appear above rows without pivots.
• Every leading 1 is to the right of any leading 1s in the rows above it.

For our system, once we've achieved REF, we'll adjust the matrix further. For example, making sure each row has only one non-zero value in the columns describing pivot positions. This means we'll divide rows by the pivot value and perform additional row operations to clear the other elements in the pivot columns. The result is a matrix that makes back substitution straightforward, revealing the solution directly with minimal additional computation.

Gaussian Elimination

Gaussian Elimination is a method that incorporates row reduction to simplify a matrix into REF or RREF. Here's the step-by-step approach:

• Construct the augmented matrix from the system of equations.
• Use row operations to achieve REF by eliminating entries below the pivots.
• If needed, proceed to RREF by cleaning above the pivots as well.

For our exercise, Gaussian Elimination helps solve the system by providing a clear and systematic procedure to simplify the matrix step by step. Once the matrix is in its final simplified form, whether REF or RREF, the solution can be directly interpreted or easily solved by back substitution. This convergence of approaches gives us a powerful tool for tackling linear systems.