Linear Algebra Examples

$45 a - 30 b + 2 d = - 1$ , $18 x + 17 b - 13 d = - 18$ , $9 a - 24 b + 42 d = 23$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 45 & - 30 & 2 & 0 & - 1 \\ 0 & 17 & - 13 & 18 & - 18 \\ 9 & - 24 & 42 & 0 & 23 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of $R_{1}$ by $\frac{1}{45}$ to make the entry at $1, 1$ a $1$ .

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Step 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{45}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{45}{45} & \frac{- 30}{45} & \frac{2}{45} & \frac{0}{45} & \frac{- 1}{45} \\ 0 & 17 & - 13 & 18 & - 18 \\ 9 & - 24 & 42 & 0 & 23 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{2}{45} & 0 & - \frac{1}{45} \\ 0 & 17 & - 13 & 18 & - 18 \\ 9 & - 24 & 42 & 0 & 23 \end{array}]$

Step 2.2

Perform the row operation $R_{3} = R_{3} - 9 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 2.2.1

Perform the row operation $R_{3} = R_{3} - 9 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{2}{45} & 0 & - \frac{1}{45} \\ 0 & 17 & - 13 & 18 & - 18 \\ 9 - 9 \cdot 1 & - 24 - 9 (- \frac{2}{3}) & 42 - 9 (\frac{2}{45}) & 0 - 9 \cdot 0 & 23 - 9 (- \frac{1}{45}) \end{array}]$

Step 2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{2}{45} & 0 & - \frac{1}{45} \\ 0 & 17 & - 13 & 18 & - 18 \\ 0 & - 18 & \frac{208}{5} & 0 & \frac{116}{5} \end{array}]$

Step 2.3

Multiply each element of $R_{2}$ by $\frac{1}{17}$ to make the entry at $2, 2$ a $1$ .

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Step 2.3.1

Multiply each element of $R_{2}$ by $\frac{1}{17}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{2}{45} & 0 & - \frac{1}{45} \\ \frac{0}{17} & \frac{17}{17} & \frac{- 13}{17} & \frac{18}{17} & \frac{- 18}{17} \\ 0 & - 18 & \frac{208}{5} & 0 & \frac{116}{5} \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{2}{45} & 0 & - \frac{1}{45} \\ 0 & 1 & - \frac{13}{17} & \frac{18}{17} & - \frac{18}{17} \\ 0 & - 18 & \frac{208}{5} & 0 & \frac{116}{5} \end{array}]$

Step 2.4

Perform the row operation $R_{3} = R_{3} + 18 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.4.1

Perform the row operation $R_{3} = R_{3} + 18 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{2}{45} & 0 & - \frac{1}{45} \\ 0 & 1 & - \frac{13}{17} & \frac{18}{17} & - \frac{18}{17} \\ 0 + 18 \cdot 0 & - 18 + 18 \cdot 1 & \frac{208}{5} + 18 (- \frac{13}{17}) & 0 + 18 (\frac{18}{17}) & \frac{116}{5} + 18 (- \frac{18}{17}) \end{array}]$

Step 2.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{2}{45} & 0 & - \frac{1}{45} \\ 0 & 1 & - \frac{13}{17} & \frac{18}{17} & - \frac{18}{17} \\ 0 & 0 & \frac{2366}{85} & \frac{324}{17} & \frac{352}{85} \end{array}]$

Step 2.5

Multiply each element of $R_{3}$ by $\frac{85}{2366}$ to make the entry at $3, 3$ a $1$ .

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Step 2.5.1

Multiply each element of $R_{3}$ by $\frac{85}{2366}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{2}{45} & 0 & - \frac{1}{45} \\ 0 & 1 & - \frac{13}{17} & \frac{18}{17} & - \frac{18}{17} \\ \frac{85}{2366} \cdot 0 & \frac{85}{2366} \cdot 0 & \frac{85}{2366} \cdot \frac{2366}{85} & \frac{85}{2366} \cdot \frac{324}{17} & \frac{85}{2366} \cdot \frac{352}{85} \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{2}{45} & 0 & - \frac{1}{45} \\ 0 & 1 & - \frac{13}{17} & \frac{18}{17} & - \frac{18}{17} \\ 0 & 0 & 1 & \frac{810}{1183} & \frac{176}{1183} \end{array}]$

Step 2.6

Perform the row operation $R_{2} = R_{2} + \frac{13}{17} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 2.6.1

Perform the row operation $R_{2} = R_{2} + \frac{13}{17} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{2}{45} & 0 & - \frac{1}{45} \\ 0 + \frac{13}{17} \cdot 0 & 1 + \frac{13}{17} \cdot 0 & - \frac{13}{17} + \frac{13}{17} \cdot 1 & \frac{18}{17} + \frac{13}{17} \cdot \frac{810}{1183} & - \frac{18}{17} + \frac{13}{17} \cdot \frac{176}{1183} \\ 0 & 0 & 1 & \frac{810}{1183} & \frac{176}{1183} \end{array}]$

Step 2.6.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{2}{45} & 0 & - \frac{1}{45} \\ 0 & 1 & 0 & \frac{144}{91} & - \frac{86}{91} \\ 0 & 0 & 1 & \frac{810}{1183} & \frac{176}{1183} \end{array}]$

Step 2.7

Perform the row operation $R_{1} = R_{1} - \frac{2}{45} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 2.7.1

Perform the row operation $R_{1} = R_{1} - \frac{2}{45} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{2}{45} \cdot 0 & - \frac{2}{3} - \frac{2}{45} \cdot 0 & \frac{2}{45} - \frac{2}{45} \cdot 1 & 0 - \frac{2}{45} \cdot \frac{810}{1183} & - \frac{1}{45} - \frac{2}{45} \cdot \frac{176}{1183} \\ 0 & 1 & 0 & \frac{144}{91} & - \frac{86}{91} \\ 0 & 0 & 1 & \frac{810}{1183} & \frac{176}{1183} \end{array}]$

Step 2.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & 0 & - \frac{36}{1183} & - \frac{307}{10647} \\ 0 & 1 & 0 & \frac{144}{91} & - \frac{86}{91} \\ 0 & 0 & 1 & \frac{810}{1183} & \frac{176}{1183} \end{array}]$

Step 2.8

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

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Step 2.8.1

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

$[\begin{array}{c} 1 + \frac{2}{3} \cdot 0 & - \frac{2}{3} + \frac{2}{3} \cdot 1 & 0 + \frac{2}{3} \cdot 0 & - \frac{36}{1183} + \frac{2}{3} \cdot \frac{144}{91} & - \frac{307}{10647} + \frac{2}{3} (- \frac{86}{91}) \\ 0 & 1 & 0 & \frac{144}{91} & - \frac{86}{91} \\ 0 & 0 & 1 & \frac{810}{1183} & \frac{176}{1183} \end{array}]$

Step 2.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{1212}{1183} & - \frac{7015}{10647} \\ 0 & 1 & 0 & \frac{144}{91} & - \frac{86}{91} \\ 0 & 0 & 1 & \frac{810}{1183} & \frac{176}{1183} \end{array}]$

Step 3

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

$a + \frac{1212 x}{1183} = - \frac{7015}{10647}$

$b + \frac{144 x}{91} = - \frac{86}{91}$

$d + \frac{810 x}{1183} = \frac{176}{1183}$

Step 4

Subtract $\frac{1212 x}{1183}$ from both sides of the equation.

$a = - \frac{7015}{10647} - \frac{1212 x}{1183}$

$b + \frac{144 x}{91} = - \frac{86}{91}$

$d + \frac{810 x}{1183} = \frac{176}{1183}$

Step 5

Subtract $\frac{144 x}{91}$ from both sides of the equation.

$b = - \frac{86}{91} - \frac{144 x}{91}$

$a = - \frac{7015}{10647} - \frac{1212 x}{1183}$

$d + \frac{810 x}{1183} = \frac{176}{1183}$

Step 6

Subtract $\frac{810 x}{1183}$ from both sides of the equation.

$d = \frac{176}{1183} - \frac{810 x}{1183}$

$a = - \frac{7015}{10647} - \frac{1212 x}{1183}$

$b = - \frac{86}{91} - \frac{144 x}{91}$

Step 7

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

$(- \frac{7015}{10647} - \frac{1212 x}{1183}, - \frac{86}{91} - \frac{144 x}{91}, \frac{176}{1183} - \frac{810 x}{1183}, x)$

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.

$X = [\begin{array}{c} a \\ b \\ d \\ x \end{array}] = [\begin{array}{c} - \frac{7015}{10647} - \frac{1212 x}{1183} \\ - \frac{86}{91} - \frac{144 x}{91} \\ \frac{176}{1183} - \frac{810 x}{1183} \\ x \end{array}]$