Linear Algebra Examples

$x - y - z + 2 s = 1$ , $2 x - 2 y - z + 3 s = 3$ , $- x + y - z = - 5$

Step 1

Write the system of equations in matrix form.

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

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

Simplify $R_{1}$ .

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Simplify $R_{2}$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

Simplify $R_{2}$ .

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

Simplify $R_{3}$ .

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

Step 2.5

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

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Step 2.5.1

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

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

Step 2.5.2

Simplify $R_{3}$ .

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

Step 2.6

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

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

Step 2.6.2

Simplify $R_{2}$ .

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

Step 2.7

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

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Step 2.7.1

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

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

Step 2.7.2

Simplify $R_{1}$ .

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

Step 2.8

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 2.8.1

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

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

Step 2.8.2

Simplify $R_{1}$ .

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

Step 3

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

$s - z = 0$

$x - y + z = 0$

$0 = 1$

Step 4

Add $z$ to both sides of the equation.

$s = z$

$x - y + z = 0$

$0 = 1$

Step 5

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

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Step 5.1

Add $y$ to both sides of the equation.

$x + z = y$

$s = z$

$0 = 1$

Step 5.2

Subtract $z$ from both sides of the equation.

$x = y - z$

$s = z$

$0 = 1$

$x = y - z$

$s = z$

$0 = 1$

Step 6

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

No solution