Linear Algebra Examples

$x + y + z = 53$ , $x - y - z = - 3$ , $2 y - x = - 11$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 1 & 1 & 53 \\ 1 & - 1 & - 1 & - 3 \\ - 1 & 2 & 0 & - 11 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} 1 & 1 & 1 & 53 \\ 1 - 1 & - 1 - 1 & - 1 - 1 & - 3 - 53 \\ - 1 & 2 & 0 & - 11 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 53 \\ 0 & - 2 & - 2 & - 56 \\ - 1 & 2 & 0 & - 11 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & 1 & 1 & 53 \\ 0 & - 2 & - 2 & - 56 \\ - 1 + 1 \cdot 1 & 2 + 1 \cdot 1 & 0 + 1 \cdot 1 & - 11 + 1 \cdot 53 \end{array}]$

Step 2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 53 \\ 0 & - 2 & - 2 & - 56 \\ 0 & 3 & 1 & 42 \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 & 1 & 1 & 53 \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot - 56 \\ 0 & 3 & 1 & 42 \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 53 \\ 0 & 1 & 1 & 28 \\ 0 & 3 & 1 & 42 \end{array}]$

Step 2.4

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

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Step 2.4.1

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

$[\begin{array}{c} 1 & 1 & 1 & 53 \\ 0 & 1 & 1 & 28 \\ 0 - 3 \cdot 0 & 3 - 3 \cdot 1 & 1 - 3 \cdot 1 & 42 - 3 \cdot 28 \end{array}]$

Step 2.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 53 \\ 0 & 1 & 1 & 28 \\ 0 & 0 & - 2 & - 42 \end{array}]$

Step 2.5

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

$[\begin{array}{c} 1 & 1 & 1 & 53 \\ 0 & 1 & 1 & 28 \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot - 42 \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 53 \\ 0 & 1 & 1 & 28 \\ 0 & 0 & 1 & 21 \end{array}]$

Step 2.6

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

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Step 2.6.1

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

$[\begin{array}{c} 1 & 1 & 1 & 53 \\ 0 - 0 & 1 - 0 & 1 - 1 & 28 - 21 \\ 0 & 0 & 1 & 21 \end{array}]$

Step 2.6.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 53 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & 21 \end{array}]$

Step 2.7

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

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Step 2.7.1

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

$[\begin{array}{c} 1 - 0 & 1 - 0 & 1 - 1 & 53 - 21 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & 21 \end{array}]$

Step 2.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 1 & 0 & 32 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & 21 \end{array}]$

Step 2.8

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

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Step 2.8.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 0 - 0 & 32 - 7 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & 21 \end{array}]$

Step 2.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 25 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & 21 \end{array}]$

Step 3

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

$x = 25$

$y = 7$

$z = 21$

Step 4

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

$(25, 7, 21)$

Step 5

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 25 \\ 7 \\ 21 \end{array}]$