Step-by-Step Guidance: Analyzing Solutions of Systems from Augmented Matrices (RREF/REF) LO10

Q1(a). Given the following augmented matrix, determine the type of solution and write all solutions if they exist:

$A = \begin{pmatrix} 1 & -1 & 0 & 2 & 4 \\ 0 & 0 & 1 & 3 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}$

Background

Topic: Systems of Linear Equations, Row-Reduced Echelon Form (RREF)

This question tests your ability to interpret an augmented matrix in (R)REF form, determine the nature of the solution set (unique, infinite, or none), and express all solutions if they exist.

Key Terms and Formulas

• Augmented Matrix: A matrix representing a system of linear equations, including the constants on the right side.

• Row-Reduced Echelon Form (RREF): A matrix form where each leading entry is 1, each leading 1 is the only nonzero entry in its column, and each leading 1 is to the right of the leading 1 in the row above.

• Pivot Variable: A variable corresponding to a leading 1 in a row.

• Free Variable: A variable that does not correspond to a leading 1 in any row.

Step-by-Step Guidance

1. Write out the system of equations represented by the matrix. Each row corresponds to an equation in variables $x_1, x_2, x_3, x_4$ (since there are 4 columns before the augmented column):

   • Row 1: $1x_1 - 1x_2 + 0x_3 + 2x_4 = 4$

   • Row 2: $0x_1 + 0x_2 + 1x_3 + 3x_4 = 2$

   • Row 3: $0x_1 + 0x_2 + 0x_3 + 0x_4 = 0$ (which is always true)

2. Identify the pivot columns (columns with leading 1s): In this matrix, the first and third columns are pivot columns, corresponding to $x_1$ and $x_3$.

3. Determine which variables are free: Variables not corresponding to pivot columns are free. Here, $x_2$ and $x_4$ are free variables.

4. Express the pivot variables in terms of the free variables using the equations from Step 1. Rearrange each equation to solve for the pivot variable:

   • From Row 1: $x_1 = 4 + x_2 - 2x_4$

   • From Row 2: $x_3 = 2 - 3x_4$

Try solving on your own before revealing the answer!

Final Answer:

The system has infinitely many solutions. The general solution is:

• $x_1 = 4 + x_2 - 2x_4$

• $x_2 = s$ (free parameter)

• $x_3 = 2 - 3x_4$

• $x_4 = t$ (free parameter)

Where $s$ and $t$ are real numbers. This expresses all solutions in terms of the free variables $x_2$ and $x_4$.

Q1(b). Given the following augmented matrix, determine the type of solution and write all solutions if they exist:

$A = \begin{pmatrix} 1 & 2 & -1 & 9 \\ 0 & 1 & 4 & -1 \\ 0 & 0 & 1 & -1 \end{pmatrix}$

Background

Topic: Systems of Linear Equations, Row-Reduced Echelon Form (RREF)

This question tests your ability to interpret an augmented matrix in (R)REF form, determine the nature of the solution set (unique, infinite, or none), and express all solutions if they exist.

Key Terms and Formulas

• Augmented Matrix

• Row-Reduced Echelon Form (RREF)

• Pivot Variable

• Free Variable

Step-by-Step Guidance

1. Write out the system of equations represented by the matrix. Each row corresponds to an equation in variables $x_1, x_2, x_3$:

   • Row 1: $x_1 + 2x_2 - x_3 = 9$

   • Row 2: $x_2 + 4x_3 = -1$

   • Row 3: $x_3 = -1$

2. Identify the pivot columns: All variables $x_1, x_2, x_3$ are pivot variables (each row has a leading 1 in a different column).

3. Since all variables are pivot variables and there are no contradictory rows (like $0 = 1$), the system has a unique solution.

4. Back-substitute starting from the last equation to find the values of each variable. Start with $x_3$ from Row 3, then use that value in Row 2 to find $x_2$, and finally use both in Row 1 to find $x_1$.

Try solving on your own before revealing the answer!

Final Answer:

• $x_3 = -1$

• $x_2 = -1 - 4(-1) = 3$

• $x_1 = 9 - 2(3) + 1 = 4$

The system has a unique solution: $x_1 = 4$, $x_2 = 3$, $x_3 = -1$.