Linear Algebra Examples

$5 a + 3 b = 35$ , $\frac{a}{b} = \frac{2}{5}$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 5 & 3 & 35 \\ \frac{1}{b} & 0 & \frac{2}{5} \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} \frac{5}{5} & \frac{3}{5} & \frac{35}{5} \\ \frac{1}{b} & 0 & \frac{2}{5} \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Simplify $R_{2}$ .

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

Step 2.3

Multiply each element of $R_{2}$ by $- \frac{5 b}{3}$ to make the entry at $2, 2$ a $1$ .

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Step 2.3.1

Multiply each element of $R_{2}$ by $- \frac{5 b}{3}$ to make the entry at $2, 2$ a $1$ .

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

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{5} & 7 \\ 0 & 1 & - \frac{2 b}{3} + \frac{35}{3} \end{array}]$

Step 2.4

Perform the row operation $R_{1} = R_{1} - \frac{3}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.4.1

Perform the row operation $R_{1} = R_{1} - \frac{3}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{3}{5} \cdot 0 & \frac{3}{5} - \frac{3}{5} \cdot 1 & 7 - \frac{3}{5} (- \frac{2 b}{3} + \frac{35}{3}) \\ 0 & 1 & - \frac{2 b}{3} + \frac{35}{3} \end{array}]$

Step 2.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{2 b}{5} \\ 0 & 1 & - \frac{2 b}{3} + \frac{35}{3} \end{array}]$

Step 3

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

$a = \frac{2 b}{5}$

$b = - \frac{2 b}{3} + \frac{35}{3}$

Step 4

Solve the equation for $b$ .

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Step 4.1

Move all terms containing $b$ to the left side of the equation.

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Step 4.1.1

Add $\frac{2 b}{3}$ to both sides of the equation.

$b + \frac{2 b}{3} = \frac{35}{3}$

$a = \frac{2 b}{5}$

Step 4.1.2

To write $b$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$b \cdot \frac{3}{3} + \frac{2 b}{3} = \frac{35}{3}$

$a = \frac{2 b}{5}$

Step 4.1.3

Combine $b$ and $\frac{3}{3}$ .

$\frac{b \cdot 3}{3} + \frac{2 b}{3} = \frac{35}{3}$

$a = \frac{2 b}{5}$

Step 4.1.4

Combine the numerators over the common denominator.

$\frac{b \cdot 3 + 2 b}{3} = \frac{35}{3}$

$a = \frac{2 b}{5}$

Step 4.1.5

Simplify the numerator.

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Step 4.1.5.1

Move $3$ to the left of $b$ .

$\frac{3 \cdot b + 2 b}{3} = \frac{35}{3}$

$a = \frac{2 b}{5}$

Step 4.1.5.2

Add $3 b$ and $2 b$ .

$\frac{5 b}{3} = \frac{35}{3}$

$a = \frac{2 b}{5}$

$\frac{5 b}{3} = \frac{35}{3}$

$a = \frac{2 b}{5}$

$\frac{5 b}{3} = \frac{35}{3}$

$a = \frac{2 b}{5}$

Step 4.2

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$5 b = 35$

$a = \frac{2 b}{5}$

Step 4.3

Divide each term in $5 b = 35$ by $5$ and simplify.

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Step 4.3.1

Divide each term in $5 b = 35$ by $5$ .

$\frac{5 b}{5} = \frac{35}{5}$

$a = \frac{2 b}{5}$

Step 4.3.2

Simplify the left side.

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Step 4.3.2.1

Cancel the common factor of $5$ .

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Step 4.3.2.1.1

Cancel the common factor.

$\frac{5 b}{5} = \frac{35}{5}$

$a = \frac{2 b}{5}$

Step 4.3.2.1.2

Divide $b$ by $1$ .

$b = \frac{35}{5}$

$a = \frac{2 b}{5}$

$b = \frac{35}{5}$

$a = \frac{2 b}{5}$

$b = \frac{35}{5}$

$a = \frac{2 b}{5}$

Step 4.3.3

Simplify the right side.

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Step 4.3.3.1

Divide $35$ by $5$ .

$b = 7$

$a = \frac{2 b}{5}$

$b = 7$

$a = \frac{2 b}{5}$

$b = 7$

$a = \frac{2 b}{5}$

$b = 7$

$a = \frac{2 b}{5}$

Step 5

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

$(\frac{2 b}{5}, 7)$

Step 6

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.

$X = [\begin{array}{c} a \\ b \end{array}] = [\begin{array}{c} \frac{2 b}{5} \\ 7 \end{array}]$