Linear Algebra Examples

$b - d + a g = 0$ , $b + d + 2 g = 0$ , $- b + 3 d - 6 g = 0$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} g & 1 & - 1 & 0 & 0 \\ 0 & 1 & 1 & 2 & 0 \\ 0 & - 1 & 3 & - 6 & 0 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} \frac{g}{g} & \frac{1}{g} & \frac{- 1}{g} & \frac{0}{g} & \frac{0}{g} \\ 0 & 1 & 1 & 2 & 0 \\ 0 & - 1 & 3 & - 6 & 0 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{g} & - \frac{1}{g} & 0 & 0 \\ 0 & 1 & 1 & 2 & 0 \\ 0 & - 1 & 3 & - 6 & 0 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & \frac{1}{g} & - \frac{1}{g} & 0 & 0 \\ 0 & 1 & 1 & 2 & 0 \\ 0 + 0 & - 1 + 1 \cdot 1 & 3 + 1 \cdot 1 & - 6 + 1 \cdot 2 & 0 + 0 \end{array}]$

Step 2.2.2

Simplify $R_{3}$ .

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

Step 2.3

Multiply each element of $R_{3}$ by $\frac{1}{4}$ to make the entry at $3, 3$ a $1$ .

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Step 2.3.1

Multiply each element of $R_{3}$ by $\frac{1}{4}$ to make the entry at $3, 3$ a $1$ .

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

Step 2.3.2

Simplify $R_{3}$ .

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

Simplify $R_{2}$ .

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

Step 2.5

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

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Step 2.5.1

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

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

Step 2.5.2

Simplify $R_{1}$ .

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

Step 2.6

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

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Step 2.6.1

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

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

Step 2.6.2

Simplify $R_{1}$ .

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

Step 3

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

$a - \frac{4 g}{g} = 0$

$b + 3 g = 0$

$d - g = 0$

Step 4

Solve the equation for $a$ .

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Step 4.1

Simplify each term.

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Step 4.1.1

Cancel the common factor of $g$ .

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Step 4.1.1.1

Cancel the common factor.

$a - \frac{4 g}{g} = 0$

$b + 3 g = 0$

$d - g = 0$

Step 4.1.1.2

Divide $4$ by $1$ .

$a - 1 \cdot 4 = 0$

$b + 3 g = 0$

$d - g = 0$

$a - 1 \cdot 4 = 0$

$b + 3 g = 0$

$d - g = 0$

Step 4.1.2

Multiply $- 1$ by $4$ .

$a - 4 = 0$

$b + 3 g = 0$

$d - g = 0$

$a - 4 = 0$

$b + 3 g = 0$

$d - g = 0$

Step 4.2

Add $4$ to both sides of the equation.

$a = 4$

$b + 3 g = 0$

$d - g = 0$

$a = 4$

$b + 3 g = 0$

$d - g = 0$

Step 5

Subtract $3 g$ from both sides of the equation.

$b = - 3 g$

$a = 4$

$d - g = 0$

Step 6

Add $g$ to both sides of the equation.

$d = g$

$a = 4$

$b = - 3 g$

Step 7

The solution is the set of ordered pairs that makes the system true.

$(4, - 3 g, g, g)$

Step 8

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

$X = [\begin{array}{c} a \\ b \\ d \\ g \end{array}] = [\begin{array}{c} 4 \\ - 3 g \\ g \\ g \end{array}]$