Linear Algebra Examples

$10 x + 2 y = 6$ , $- 25 x + k y = - 14$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 0 & 10 & 2 & 6 \\ y & - 25 & 0 & - 14 \end{array}]$

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 $1, 1$ .

$[\begin{array}{c} y & - 25 & 0 & - 14 \\ 0 & 10 & 2 & 6 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} \frac{y}{y} & \frac{- 25}{y} & \frac{0}{y} & \frac{- 14}{y} \\ 0 & 10 & 2 & 6 \end{array}]$

Step 2.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{25}{y} & 0 & - \frac{14}{y} \\ 0 & 10 & 2 & 6 \end{array}]$

Step 2.3

Multiply each element of $R_{2}$ by $\frac{1}{10}$ to make the entry at $2, 2$ a $1$ .

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Step 2.3.1

Multiply each element of $R_{2}$ by $\frac{1}{10}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{25}{y} & 0 & - \frac{14}{y} \\ \frac{0}{10} & \frac{10}{10} & \frac{2}{10} & \frac{6}{10} \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{25}{y} & 0 & - \frac{14}{y} \\ 0 & 1 & \frac{1}{5} & \frac{3}{5} \end{array}]$

Step 2.4

Perform the row operation $R_{1} = R_{1} + \frac{25}{y} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.4.1

Perform the row operation $R_{1} = R_{1} + \frac{25}{y} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{25}{y} \cdot 0 & - \frac{25}{y} + \frac{25}{y} \cdot 1 & 0 + \frac{25}{y} \cdot \frac{1}{5} & - \frac{14}{y} + \frac{25}{y} \cdot \frac{3}{5} \\ 0 & 1 & \frac{1}{5} & \frac{3}{5} \end{array}]$

Step 2.4.2

Simplify $R_{1}$ .

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

Step 3

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

$k + \frac{5 y}{y} = \frac{1}{y}$

$x + \frac{y}{5} = \frac{3}{5}$

Step 4

Solve the equation for $k$ .

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Step 4.1

Move all terms containing variables to the left side of the equation.

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Step 4.1.1

Subtract $\frac{1}{y}$ from both sides of the equation.

$k + \frac{5 y}{y} - \frac{1}{y} = 0$

$x + \frac{y}{5} = \frac{3}{5}$

Step 4.1.2

Cancel the common factor of $y$ .

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Step 4.1.2.1

Cancel the common factor.

$k + \frac{5 y}{y} - \frac{1}{y} = 0$

$x + \frac{y}{5} = \frac{3}{5}$

Step 4.1.2.2

Divide $5$ by $1$ .

$k + 5 - \frac{1}{y} = 0$

$x + \frac{y}{5} = \frac{3}{5}$

$k + 5 - \frac{1}{y} = 0$

$x + \frac{y}{5} = \frac{3}{5}$

$k + 5 - \frac{1}{y} = 0$

$x + \frac{y}{5} = \frac{3}{5}$

Step 4.2

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

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Step 4.2.1

Subtract $5$ from both sides of the equation.

$k - \frac{1}{y} = - 5$

$x + \frac{y}{5} = \frac{3}{5}$

Step 4.2.2

Add $\frac{1}{y}$ to both sides of the equation.

$k = - 5 + \frac{1}{y}$

$x + \frac{y}{5} = \frac{3}{5}$

$k = - 5 + \frac{1}{y}$

$x + \frac{y}{5} = \frac{3}{5}$

$k = - 5 + \frac{1}{y}$

$x + \frac{y}{5} = \frac{3}{5}$

Step 5

Subtract $\frac{y}{5}$ from both sides of the equation.

$x = \frac{3}{5} - \frac{y}{5}$

$k = - 5 + \frac{1}{y}$

Step 6

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

$(- 5 + \frac{1}{y}, \frac{3}{5} - \frac{y}{5}, y)$

Step 7

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.

$X = [\begin{array}{c} k \\ x \\ y \end{array}] = [\begin{array}{c} - 5 + \frac{1}{y} \\ \frac{3}{5} - \frac{y}{5} \\ y \end{array}]$