Linear Algebra Examples

$2 x + b - 3 c = 12$ , $5 a - 4 b + 7 c = 27$ , $10 a + 3 b - c = 40$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 0 & 1 & - 3 & 2 & 12 \\ 5 & - 4 & 7 & 0 & 27 \\ 10 & 3 & - 1 & 0 & 40 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 5 & - 4 & 7 & 0 & 27 \\ 0 & 1 & - 3 & 2 & 12 \\ 10 & 3 & - 1 & 0 & 40 \end{array}]$

Step 2.2

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

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Step 2.2.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{- 4}{5} & \frac{7}{5} & \frac{0}{5} & \frac{27}{5} \\ 0 & 1 & - 3 & 2 & 12 \\ 10 & 3 & - 1 & 0 & 40 \end{array}]$

Step 2.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 & 1 & - 3 & 2 & 12 \\ 10 & 3 & - 1 & 0 & 40 \end{array}]$

Step 2.3

Perform the row operation $R_{3} = R_{3} - 10 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 2.3.1

Perform the row operation $R_{3} = R_{3} - 10 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 & 1 & - 3 & 2 & 12 \\ 10 - 10 \cdot 1 & 3 - 10 (- \frac{4}{5}) & - 1 - 10 (\frac{7}{5}) & 0 - 10 \cdot 0 & 40 - 10 (\frac{27}{5}) \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 & 1 & - 3 & 2 & 12 \\ 0 & 11 & - 15 & 0 & - 14 \end{array}]$

Step 2.4

Perform the row operation $R_{3} = R_{3} - 11 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.4.1

Perform the row operation $R_{3} = R_{3} - 11 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 & 1 & - 3 & 2 & 12 \\ 0 - 11 \cdot 0 & 11 - 11 \cdot 1 & - 15 - 11 \cdot - 3 & 0 - 11 \cdot 2 & - 14 - 11 \cdot 12 \end{array}]$

Step 2.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 & 1 & - 3 & 2 & 12 \\ 0 & 0 & 18 & - 22 & - 146 \end{array}]$

Step 2.5

Multiply each element of $R_{3}$ by $\frac{1}{18}$ to make the entry at $3, 3$ a $1$ .

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Step 2.5.1

Multiply each element of $R_{3}$ by $\frac{1}{18}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 & 1 & - 3 & 2 & 12 \\ \frac{0}{18} & \frac{0}{18} & \frac{18}{18} & \frac{- 22}{18} & \frac{- 146}{18} \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 & 1 & - 3 & 2 & 12 \\ 0 & 0 & 1 & - \frac{11}{9} & - \frac{73}{9} \end{array}]$

Step 2.6

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

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Step 2.6.1

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

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 + 3 \cdot 0 & 1 + 3 \cdot 0 & - 3 + 3 \cdot 1 & 2 + 3 (- \frac{11}{9}) & 12 + 3 (- \frac{73}{9}) \\ 0 & 0 & 1 & - \frac{11}{9} & - \frac{73}{9} \end{array}]$

Step 2.6.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 & 1 & 0 & - \frac{5}{3} & - \frac{37}{3} \\ 0 & 0 & 1 & - \frac{11}{9} & - \frac{73}{9} \end{array}]$

Step 2.7

Perform the row operation $R_{1} = R_{1} - \frac{7}{5} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 2.7.1

Perform the row operation $R_{1} = R_{1} - \frac{7}{5} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{7}{5} \cdot 0 & - \frac{4}{5} - \frac{7}{5} \cdot 0 & \frac{7}{5} - \frac{7}{5} \cdot 1 & 0 - \frac{7}{5} (- \frac{11}{9}) & \frac{27}{5} - \frac{7}{5} (- \frac{73}{9}) \\ 0 & 1 & 0 & - \frac{5}{3} & - \frac{37}{3} \\ 0 & 0 & 1 & - \frac{11}{9} & - \frac{73}{9} \end{array}]$

Step 2.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & 0 & \frac{77}{45} & \frac{754}{45} \\ 0 & 1 & 0 & - \frac{5}{3} & - \frac{37}{3} \\ 0 & 0 & 1 & - \frac{11}{9} & - \frac{73}{9} \end{array}]$

Step 2.8

Perform the row operation $R_{1} = R_{1} + \frac{4}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.8.1

Perform the row operation $R_{1} = R_{1} + \frac{4}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{4}{5} \cdot 0 & - \frac{4}{5} + \frac{4}{5} \cdot 1 & 0 + \frac{4}{5} \cdot 0 & \frac{77}{45} + \frac{4}{5} (- \frac{5}{3}) & \frac{754}{45} + \frac{4}{5} (- \frac{37}{3}) \\ 0 & 1 & 0 & - \frac{5}{3} & - \frac{37}{3} \\ 0 & 0 & 1 & - \frac{11}{9} & - \frac{73}{9} \end{array}]$

Step 2.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{17}{45} & \frac{62}{9} \\ 0 & 1 & 0 & - \frac{5}{3} & - \frac{37}{3} \\ 0 & 0 & 1 & - \frac{11}{9} & - \frac{73}{9} \end{array}]$

Step 3

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

$a + \frac{17 x}{45} = \frac{62}{9}$

$b - \frac{5 x}{3} = - \frac{37}{3}$

$c - \frac{11 x}{9} = - \frac{73}{9}$

Step 4

Subtract $\frac{17 x}{45}$ from both sides of the equation.

$a = \frac{62}{9} - \frac{17 x}{45}$

$b - \frac{5 x}{3} = - \frac{37}{3}$

$c - \frac{11 x}{9} = - \frac{73}{9}$

Step 5

Add $\frac{5 x}{3}$ to both sides of the equation.

$b = - \frac{37}{3} + \frac{5 x}{3}$

$a = \frac{62}{9} - \frac{17 x}{45}$

$c - \frac{11 x}{9} = - \frac{73}{9}$

Step 6

Add $\frac{11 x}{9}$ to both sides of the equation.

$c = - \frac{73}{9} + \frac{11 x}{9}$

$a = \frac{62}{9} - \frac{17 x}{45}$

$b = - \frac{37}{3} + \frac{5 x}{3}$

Step 7

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

$(\frac{62}{9} - \frac{17 x}{45}, - \frac{37}{3} + \frac{5 x}{3}, - \frac{73}{9} + \frac{11 x}{9}, x)$

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.

$X = [\begin{array}{c} a \\ b \\ c \\ x \end{array}] = [\begin{array}{c} \frac{62}{9} - \frac{17 x}{45} \\ - \frac{37}{3} + \frac{5 x}{3} \\ - \frac{73}{9} + \frac{11 x}{9} \\ x \end{array}]$