Examples

x + y = - 2

$x + y = - 2$ ,

53 x - 8 y = 0

$53 x - 8 y = 0$

Step 1

Write the system of equations in matrix form.

[\begin{matrix} 1 & 1 & - 2 53 & - 8 & 0 \end{matrix}]

$[\begin{array}{c} 1 & 1 & - 2 \\ 53 & - 8 & 0 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Perform the row operation

R_{2} = R_{2} - 53 R_{1}

$R_{2} = R_{2} - 53 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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Step 2.1.1

Perform the row operation

R_{2} = R_{2} - 53 R_{1}

$R_{2} = R_{2} - 53 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

[\begin{matrix} 1 & 1 & - 2 53 - 53 \cdot 1 & - 8 - 53 \cdot 1 & 0 - 53 \cdot - 2 \end{matrix}]

$[\begin{array}{c} 1 & 1 & - 2 \\ 53 - 53 \cdot 1 & - 8 - 53 \cdot 1 & 0 - 53 \cdot - 2 \end{array}]$

Step 2.1.2

Simplify

R_{2}

$R_{2}$ .

[\begin{matrix} 1 & 1 & - 2 0 & - 61 & 106 \end{matrix}]

$[\begin{array}{c} 1 & 1 & - 2 \\ 0 & - 61 & 106 \end{array}]$

[\begin{matrix} 1 & 1 & - 2 0 & - 61 & 106 \end{matrix}]

$[\begin{array}{c} 1 & 1 & - 2 \\ 0 & - 61 & 106 \end{array}]$

Step 2.2

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{1}{61}

$- \frac{1}{61}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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Step 2.2.1

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{1}{61}

$- \frac{1}{61}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

[\begin{matrix} 1 & 1 & - 2 - \frac{1}{61} \cdot 0 & - \frac{1}{61} \cdot - 61 & - \frac{1}{61} \cdot 106 \end{matrix}]

$[\begin{array}{c} 1 & 1 & - 2 \\ - \frac{1}{61} \cdot 0 & - \frac{1}{61} \cdot - 61 & - \frac{1}{61} \cdot 106 \end{array}]$

Step 2.2.2

Simplify

R_{2}

$R_{2}$ .

[\begin{matrix} 1 & 1 & - 2 0 & 1 & - \frac{106}{61} \end{matrix}]

$[\begin{array}{c} 1 & 1 & - 2 \\ 0 & 1 & - \frac{106}{61} \end{array}]$

[\begin{matrix} 1 & 1 & - 2 0 & 1 & - \frac{106}{61} \end{matrix}]

$[\begin{array}{c} 1 & 1 & - 2 \\ 0 & 1 & - \frac{106}{61} \end{array}]$

Step 2.3

Perform the row operation

R_{1} = R_{1} - R_{2}

$R_{1} = R_{1} - R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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Step 2.3.1

Perform the row operation

R_{1} = R_{1} - R_{2}

$R_{1} = R_{1} - R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

[\begin{matrix} 1 - 0 & 1 - 1 & - 2 + \frac{106}{61} 0 & 1 & - \frac{106}{61} \end{matrix}]

$[\begin{array}{c} 1 - 0 & 1 - 1 & - 2 + \frac{106}{61} \\ 0 & 1 & - \frac{106}{61} \end{array}]$

Step 2.3.2

Simplify

R_{1}

$R_{1}$ .

[\begin{matrix} 1 & 0 & - \frac{16}{61} 0 & 1 & - \frac{106}{61} \end{matrix}]

$[\begin{array}{c} 1 & 0 & - \frac{16}{61} \\ 0 & 1 & - \frac{106}{61} \end{array}]$

[\begin{matrix} 1 & 0 & - \frac{16}{61} 0 & 1 & - \frac{106}{61} \end{matrix}]

$[\begin{array}{c} 1 & 0 & - \frac{16}{61} \\ 0 & 1 & - \frac{106}{61} \end{array}]$

Step 3

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

$x = - \frac{16}{61}$

$y = - \frac{106}{61}$

Step 4

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

$(- \frac{16}{61}, - \frac{106}{61})$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - \frac{16}{61} \\ - \frac{106}{61} \end{array}]$

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