Linear Algebra Examples

$x + y - 2 z + 4 t = 5$ $2 x + 2 y - 3 z + t = 3$ $3 x + 3 y - 4 z - 2 t = 1$

Step 1

Move variables to the left and constant terms to the right.

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Step 1.1

Move $- 2 z$ .

$x + y + 4 t - 2 z = 5$

$2 x + 2 y - 3 z + t = 3$

$3 x + 3 y - 4 z - 2 t = 1$

Step 1.2

Move $y$ .

$x + 4 t + y - 2 z = 5$

$2 x + 2 y - 3 z + t = 3$

$3 x + 3 y - 4 z - 2 t = 1$

Step 1.3

Reorder $x$ and $4 t$ .

$4 t + x + y - 2 z = 5$

$2 x + 2 y - 3 z + t = 3$

$3 x + 3 y - 4 z - 2 t = 1$

Step 1.4

Move $- 3 z$ .

$4 t + x + y - 2 z = 5$

$2 x + 2 y + t - 3 z = 3$

$3 x + 3 y - 4 z - 2 t = 1$

Step 1.5

Move $2 y$ .

$4 t + x + y - 2 z = 5$

$2 x + t + 2 y - 3 z = 3$

$3 x + 3 y - 4 z - 2 t = 1$

Step 1.6

Reorder $2 x$ and $t$ .

$4 t + x + y - 2 z = 5$

$t + 2 x + 2 y - 3 z = 3$

$3 x + 3 y - 4 z - 2 t = 1$

Step 1.7

Move $- 4 z$ .

$4 t + x + y - 2 z = 5$

$t + 2 x + 2 y - 3 z = 3$

$3 x + 3 y - 2 t - 4 z = 1$

Step 1.8

Move $3 y$ .

$4 t + x + y - 2 z = 5$

$t + 2 x + 2 y - 3 z = 3$

$3 x - 2 t + 3 y - 4 z = 1$

Step 1.9

Reorder $3 x$ and $- 2 t$ .

$4 t + x + y - 2 z = 5$

$t + 2 x + 2 y - 3 z = 3$

$- 2 t + 3 x + 3 y - 4 z = 1$

$4 t + x + y - 2 z = 5$

$t + 2 x + 2 y - 3 z = 3$

$- 2 t + 3 x + 3 y - 4 z = 1$

Step 2

Write the system as a matrix.

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

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Simplify $R_{1}$ .

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{4} & - \frac{1}{2} & \frac{5}{4} \\ 0 & \frac{7}{4} & \frac{7}{4} & - \frac{5}{2} & \frac{7}{4} \\ - 2 & 3 & 3 & - 4 & 1 \end{array}]$

Step 3.3

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 3.3.1

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

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{4} & - \frac{1}{2} & \frac{5}{4} \\ 0 & \frac{7}{4} & \frac{7}{4} & - \frac{5}{2} & \frac{7}{4} \\ - 2 + 2 \cdot 1 & 3 + 2 (\frac{1}{4}) & 3 + 2 (\frac{1}{4}) & - 4 + 2 (- \frac{1}{2}) & 1 + 2 (\frac{5}{4}) \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{4} & - \frac{1}{2} & \frac{5}{4} \\ 0 & \frac{7}{4} & \frac{7}{4} & - \frac{5}{2} & \frac{7}{4} \\ 0 & \frac{7}{2} & \frac{7}{2} & - 5 & \frac{7}{2} \end{array}]$

Step 3.4

Multiply each element of $R_{2}$ by $\frac{4}{7}$ to make the entry at $2, 2$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{2}$ by $\frac{4}{7}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{4} & - \frac{1}{2} & \frac{5}{4} \\ \frac{4}{7} \cdot 0 & \frac{4}{7} \cdot \frac{7}{4} & \frac{4}{7} \cdot \frac{7}{4} & \frac{4}{7} (- \frac{5}{2}) & \frac{4}{7} \cdot \frac{7}{4} \\ 0 & \frac{7}{2} & \frac{7}{2} & - 5 & \frac{7}{2} \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{4} & - \frac{1}{2} & \frac{5}{4} \\ 0 & 1 & 1 & - \frac{10}{7} & 1 \\ 0 & \frac{7}{2} & \frac{7}{2} & - 5 & \frac{7}{2} \end{array}]$

Step 3.5

Perform the row operation $R_{3} = R_{3} - \frac{7}{2} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.5.1

Perform the row operation $R_{3} = R_{3} - \frac{7}{2} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{4} & - \frac{1}{2} & \frac{5}{4} \\ 0 & 1 & 1 & - \frac{10}{7} & 1 \\ 0 - \frac{7}{2} \cdot 0 & \frac{7}{2} - \frac{7}{2} \cdot 1 & \frac{7}{2} - \frac{7}{2} \cdot 1 & - 5 - \frac{7}{2} (- \frac{10}{7}) & \frac{7}{2} - \frac{7}{2} \cdot 1 \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{4} & - \frac{1}{2} & \frac{5}{4} \\ 0 & 1 & 1 & - \frac{10}{7} & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Step 3.6

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

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Step 3.6.1

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

$[\begin{array}{c} 1 - \frac{1}{4} \cdot 0 & \frac{1}{4} - \frac{1}{4} \cdot 1 & \frac{1}{4} - \frac{1}{4} \cdot 1 & - \frac{1}{2} - \frac{1}{4} (- \frac{10}{7}) & \frac{5}{4} - \frac{1}{4} \cdot 1 \\ 0 & 1 & 1 & - \frac{10}{7} & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Step 3.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{1}{7} & 1 \\ 0 & 1 & 1 & - \frac{10}{7} & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Step 4

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

$t - \frac{1}{7} z = 1$

$x + y - \frac{10}{7} z = 1$

$0 = 0$

Step 5

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

$(1 + \frac{z}{7}, 1 - y + \frac{10 z}{7}, y, z)$