Finite Math Examples

$[\begin{array}{c} - 1 & 3 & 2 \\ 1 & 1 & 0 \\ 1 & 1 & 0 \end{array}]$

Step 1

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} - 1 & 3 & 2 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of $R_{1}$ by $- 1$ to make the entry at $1, 1$ a $1$ .

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Step 2.1.1

Multiply each element of $R_{1}$ by $- 1$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - - 1 & - 1 \cdot 3 & - 1 \cdot 2 & - 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 3 & - 2 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \end{array}]$

Step 2.2

Perform the row operation $R_{2} = R_{2} - R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 2.2.1

Perform the row operation $R_{2} = R_{2} - R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - 3 & - 2 & 0 \\ 1 - 1 & 1 + 3 & 0 + 2 & 0 - 0 \\ 1 & 1 & 0 & 0 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 3 & - 2 & 0 \\ 0 & 4 & 2 & 0 \\ 1 & 1 & 0 & 0 \end{array}]$

Step 2.3

Perform the row operation $R_{3} = R_{3} - R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 2.3.1

Perform the row operation $R_{3} = R_{3} - R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - 3 & - 2 & 0 \\ 0 & 4 & 2 & 0 \\ 1 - 1 & 1 + 3 & 0 + 2 & 0 - 0 \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 3 & - 2 & 0 \\ 0 & 4 & 2 & 0 \\ 0 & 4 & 2 & 0 \end{array}]$

Step 2.4

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 3 & - 2 & 0 \\ \frac{0}{4} & \frac{4}{4} & \frac{2}{4} & \frac{0}{4} \\ 0 & 4 & 2 & 0 \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 3 & - 2 & 0 \\ 0 & 1 & \frac{1}{2} & 0 \\ 0 & 4 & 2 & 0 \end{array}]$

Step 2.5

Perform the row operation $R_{3} = R_{3} - 4 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.5.1

Perform the row operation $R_{3} = R_{3} - 4 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 3 & - 2 & 0 \\ 0 & 1 & \frac{1}{2} & 0 \\ 0 - 4 \cdot 0 & 4 - 4 \cdot 1 & 2 - 4 (\frac{1}{2}) & 0 - 4 \cdot 0 \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 3 & - 2 & 0 \\ 0 & 1 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 2.6

Perform the row operation $R_{1} = R_{1} + 3 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.6.1

Perform the row operation $R_{1} = R_{1} + 3 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + 3 \cdot 0 & - 3 + 3 \cdot 1 & - 2 + 3 (\frac{1}{2}) & 0 + 3 \cdot 0 \\ 0 & 1 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 2.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1}{2} & 0 \\ 0 & 1 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 3

Use the result matrix to declare the final solution to the system of equations.

$x - \frac{1}{2} z = 0$

$y + \frac{1}{2} z = 0$

$0 = 0$

Step 4

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{z}{2} \\ - \frac{z}{2} \\ z \end{array}]$

Step 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = z [\begin{array}{c} \frac{1}{2} \\ - \frac{1}{2} \\ 1 \end{array}]$

Step 6

Write as a solution set.

${z [\begin{array}{c} \frac{1}{2} \\ - \frac{1}{2} \\ 1 \end{array}] | z \in R}$

Step 7

The solution is the set of vectors created from the free variables of the system.

Basis of $Nul (A)$ : ${[\begin{array}{c} \frac{1}{2} \\ - \frac{1}{2} \\ 1 \end{array}]}$

Dimension of $Nul (A)$ : $1$

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