Linear Algebra Examples

$y = 2 x + 3$ , $y = - (\frac{1}{2} x) + 5.5$ , $y = 2 x + 4$ , $y = - (\frac{1}{2} x) + 3.5$

Step 1

Subtract $2 x$ from both sides of the equation.

$y - 2 x = 3, y = - (\frac{1}{2} x) + 5.5, y = 2 x + 4, y = - (\frac{1}{2} x) + 3.5$

Step 2

Combine $\frac{1}{2}$ and $x$ .

$y - 2 x = 3, y = - \frac{x}{2} + 5.5, y = 2 x + 4, y = - (\frac{1}{2} x) + 3.5$

Step 3

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

$y - 2 x = 3, y + \frac{x}{2} = 5.5, y = 2 x + 4, y = - (\frac{1}{2} x) + 3.5$

Step 4

Subtract $2 x$ from both sides of the equation.

$y - 2 x = 3, y + \frac{x}{2} = 5.5, y - 2 x = 4, y = - (\frac{1}{2} x) + 3.5$

Step 5

Combine $\frac{1}{2}$ and $x$ .

$y - 2 x = 3, y + \frac{x}{2} = 5.5, y - 2 x = 4, y = - \frac{x}{2} + 3.5$

Step 6

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

$y - 2 x = 3, y + \frac{x}{2} = 5.5, y - 2 x = 4, y + \frac{x}{2} = 3.5$

Step 7

Write the system of equations in matrix form.

$[\begin{array}{c} - 2 & 1 & 3 \\ \frac{1}{2} & 1 & 5.5 \\ - 2 & 1 & 4 \\ \frac{1}{2} & 1 & 3.5 \end{array}]$

Step 8

Find the reduced row echelon form.

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Step 8.1

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

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Step 8.1.1

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

$[\begin{array}{c} - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot 1 & - \frac{1}{2} \cdot 3 \\ \frac{1}{2} & 1 & 5.5 \\ - 2 & 1 & 4 \\ \frac{1}{2} & 1 & 3.5 \end{array}]$

Step 8.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} \\ \frac{1}{2} & 1 & 5.5 \\ - 2 & 1 & 4 \\ \frac{1}{2} & 1 & 3.5 \end{array}]$

Step 8.2

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

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Step 8.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} \\ \frac{1}{2} - \frac{1}{2} \cdot 1 & 1 - \frac{1}{2} (- \frac{1}{2}) & 5.5 - \frac{1}{2} (- \frac{3}{2}) \\ - 2 & 1 & 4 \\ \frac{1}{2} & 1 & 3.5 \end{array}]$

Step 8.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} \\ 0 & \frac{5}{4} & \frac{25}{4} \\ - 2 & 1 & 4 \\ \frac{1}{2} & 1 & 3.5 \end{array}]$

Step 8.3

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

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Step 8.3.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} \\ 0 & \frac{5}{4} & \frac{25}{4} \\ - 2 + 2 \cdot 1 & 1 + 2 (- \frac{1}{2}) & 4 + 2 (- \frac{3}{2}) \\ \frac{1}{2} & 1 & 3.5 \end{array}]$

Step 8.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} \\ 0 & \frac{5}{4} & \frac{25}{4} \\ 0 & 0 & 1 \\ \frac{1}{2} & 1 & 3.5 \end{array}]$

Step 8.4

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

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Step 8.4.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} \\ 0 & \frac{5}{4} & \frac{25}{4} \\ 0 & 0 & 1 \\ \frac{1}{2} - \frac{1}{2} \cdot 1 & 1 - \frac{1}{2} (- \frac{1}{2}) & 3.5 - \frac{1}{2} (- \frac{3}{2}) \end{array}]$

Step 8.4.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} \\ 0 & \frac{5}{4} & \frac{25}{4} \\ 0 & 0 & 1 \\ 0 & \frac{5}{4} & \frac{17}{4} \end{array}]$

Step 8.5

Multiply each element of $R_{2}$ by $\frac{4}{5}$ to make the entry at $2, 2$ a $1$ .

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Step 8.5.1

Multiply each element of $R_{2}$ by $\frac{4}{5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} \\ \frac{4}{5} \cdot 0 & \frac{4}{5} \cdot \frac{5}{4} & \frac{4}{5} \cdot \frac{25}{4} \\ 0 & 0 & 1 \\ 0 & \frac{5}{4} & \frac{17}{4} \end{array}]$

Step 8.5.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} \\ 0 & 1 & 5 \\ 0 & 0 & 1 \\ 0 & \frac{5}{4} & \frac{17}{4} \end{array}]$

Step 8.6

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

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Step 8.6.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} \\ 0 & 1 & 5 \\ 0 & 0 & 1 \\ 0 - \frac{5}{4} \cdot 0 & \frac{5}{4} - \frac{5}{4} \cdot 1 & \frac{17}{4} - \frac{5}{4} \cdot 5 \end{array}]$

Step 8.6.2

Simplify $R_{4}$ .

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

Step 8.7

Perform the row operation $R_{4} = R_{4} + 2 R_{3}$ to make the entry at $4, 3$ a $0$ .

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Step 8.7.1

Perform the row operation $R_{4} = R_{4} + 2 R_{3}$ to make the entry at $4, 3$ a $0$ .

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

Step 8.7.2

Simplify $R_{4}$ .

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

Step 8.8

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

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Step 8.8.1

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

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

Step 8.8.2

Simplify $R_{2}$ .

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

Step 8.9

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

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Step 8.9.1

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

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

Step 8.9.2

Simplify $R_{1}$ .

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

Step 8.10

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

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Step 8.10.1

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

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & 0 + \frac{1}{2} \cdot 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}]$

Step 8.10.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}]$

Step 9

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

$x = 0$

$y = 0$

$0 = 1$

Step 10

Since $0 \neq 1$ , there are no solutions.

No solution