Linear Algebra Examples

$2 d + 3 g - h = a$ , $d - g + 3 h = b$ , $3 d + 7 g - 5 h = c$

Step 1

Subtract $a$ from both sides of the equation.

$2 d + 3 g - h - a = 0, d - g + 3 h = b, 3 d + 7 g - 5 h = c$

Step 2

Subtract $b$ from both sides of the equation.

$2 d + 3 g - h - a = 0, d - g + 3 h - b = 0, 3 d + 7 g - 5 h = c$

Step 3

Subtract $c$ from both sides of the equation.

$2 d + 3 g - h - a = 0, d - g + 3 h - b = 0, 3 d + 7 g - 5 h - c = 0$

Step 4

Write the system of equations in matrix form.

$[\begin{array}{c} - 1 & 0 & 0 & 2 & 3 & - 1 & 0 \\ 0 & - 1 & 0 & 1 & - 1 & 3 & 0 \\ 0 & 0 & - 1 & 3 & 7 & - 5 & 0 \end{array}]$

Step 5

Find the reduced row echelon form.

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Step 5.1

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

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Step 5.1.1

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

$[\begin{array}{c} - - 1 & - 0 & - 0 & - 1 \cdot 2 & - 1 \cdot 3 & - - 1 & - 0 \\ 0 & - 1 & 0 & 1 & - 1 & 3 & 0 \\ 0 & 0 & - 1 & 3 & 7 & - 5 & 0 \end{array}]$

Step 5.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - 2 & - 3 & 1 & 0 \\ 0 & - 1 & 0 & 1 & - 1 & 3 & 0 \\ 0 & 0 & - 1 & 3 & 7 & - 5 & 0 \end{array}]$

Step 5.2

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

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Step 5.2.1

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

$[\begin{array}{c} 1 & 0 & 0 & - 2 & - 3 & 1 & 0 \\ - 0 & - - 1 & - 0 & - 1 \cdot 1 & - - 1 & - 1 \cdot 3 & - 0 \\ 0 & 0 & - 1 & 3 & 7 & - 5 & 0 \end{array}]$

Step 5.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 0 & - 2 & - 3 & 1 & 0 \\ 0 & 1 & 0 & - 1 & 1 & - 3 & 0 \\ 0 & 0 & - 1 & 3 & 7 & - 5 & 0 \end{array}]$

Step 5.3

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

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Step 5.3.1

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

$[\begin{array}{c} 1 & 0 & 0 & - 2 & - 3 & 1 & 0 \\ 0 & 1 & 0 & - 1 & 1 & - 3 & 0 \\ - 0 & - 0 & - - 1 & - 1 \cdot 3 & - 1 \cdot 7 & - - 5 & - 0 \end{array}]$

Step 5.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 0 & - 2 & - 3 & 1 & 0 \\ 0 & 1 & 0 & - 1 & 1 & - 3 & 0 \\ 0 & 0 & 1 & - 3 & - 7 & 5 & 0 \end{array}]$

Step 6

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

$a - 2 d - 3 g + h = 0$

$b - d + g - 3 h = 0$

$c - 3 d - 7 g + 5 h = 0$

Step 7

Move all terms not containing $a$ to the right side of the equation.

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Step 7.1

Add $2 d$ to both sides of the equation.

$a - 3 g + h = 2 d$

$b - d + g - 3 h = 0$

$c - 3 d - 7 g + 5 h = 0$

Step 7.2

Add $3 g$ to both sides of the equation.

$a + h = 2 d + 3 g$

$b - d + g - 3 h = 0$

$c - 3 d - 7 g + 5 h = 0$

Step 7.3

Subtract $h$ from both sides of the equation.

$a = 2 d + 3 g - h$

$b - d + g - 3 h = 0$

$c - 3 d - 7 g + 5 h = 0$

$a = 2 d + 3 g - h$

$b - d + g - 3 h = 0$

$c - 3 d - 7 g + 5 h = 0$

Step 8

Move all terms not containing $b$ to the right side of the equation.

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Step 8.1

Add $d$ to both sides of the equation.

$b + g - 3 h = d$

$a = 2 d + 3 g - h$

$c - 3 d - 7 g + 5 h = 0$

Step 8.2

Subtract $g$ from both sides of the equation.

$b - 3 h = d - g$

$a = 2 d + 3 g - h$

$c - 3 d - 7 g + 5 h = 0$

Step 8.3

Add $3 h$ to both sides of the equation.

$b = d - g + 3 h$

$a = 2 d + 3 g - h$

$c - 3 d - 7 g + 5 h = 0$

$b = d - g + 3 h$

$a = 2 d + 3 g - h$

$c - 3 d - 7 g + 5 h = 0$

Step 9

Move all terms not containing $c$ to the right side of the equation.

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Step 9.1

Add $3 d$ to both sides of the equation.

$c - 7 g + 5 h = 3 d$

$a = 2 d + 3 g - h$

$b = d - g + 3 h$

Step 9.2

Add $7 g$ to both sides of the equation.

$c + 5 h = 3 d + 7 g$

$a = 2 d + 3 g - h$

$b = d - g + 3 h$

Step 9.3

Subtract $5 h$ from both sides of the equation.

$c = 3 d + 7 g - 5 h$

$a = 2 d + 3 g - h$

$b = d - g + 3 h$

$c = 3 d + 7 g - 5 h$

$a = 2 d + 3 g - h$

$b = d - g + 3 h$

Step 10

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

$(2 d + 3 g - h, d - g + 3 h, 3 d + 7 g - 5 h, d, g, h)$

Step 11

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.

$X = [\begin{array}{c} a \\ b \\ c \\ d \\ g \\ h \end{array}] = [\begin{array}{c} 2 d + 3 g - h \\ d - g + 3 h \\ 3 d + 7 g - 5 h \\ d \\ g \\ h \end{array}]$