Linear Algebra Examples

4 x - y = - 4

$4 x - y = - 4$ ,

6 x - y = - 5

$6 x - y = - 5$

Step 1

Write the system as a matrix.

[\begin{array}{c} 4 & - 1 & - 4 \\ 6 & - 1 & - 5 \end{array}]

$[\begin{array}{c} 4 & - 1 & - 4 \\ 6 & - 1 & - 5 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{4}

$\frac{1}{4}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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Step 2.1.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{4}

$\frac{1}{4}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

[\begin{array}{c} \frac{4}{4} & \frac{- 1}{4} & \frac{- 4}{4} \\ 6 & - 1 & - 5 \end{array}]

$[\begin{array}{c} \frac{4}{4} & \frac{- 1}{4} & \frac{- 4}{4} \\ 6 & - 1 & - 5 \end{array}]$

Step 2.1.2

Simplify

R_{1}

$R_{1}$ .

[\begin{array}{c} 1 & - \frac{1}{4} & - 1 \\ 6 & - 1 & - 5 \end{array}]

$[\begin{array}{c} 1 & - \frac{1}{4} & - 1 \\ 6 & - 1 & - 5 \end{array}]$

[\begin{array}{c} 1 & - \frac{1}{4} & - 1 \\ 6 & - 1 & - 5 \end{array}]

$[\begin{array}{c} 1 & - \frac{1}{4} & - 1 \\ 6 & - 1 & - 5 \end{array}]$

Step 2.2

Perform the row operation

R_{2} = R_{2} - 6 R_{1}

$R_{2} = R_{2} - 6 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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Step 2.2.1

Perform the row operation

R_{2} = R_{2} - 6 R_{1}

$R_{2} = R_{2} - 6 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

[\begin{array}{c} 1 & - \frac{1}{4} & - 1 \\ 6 - 6 \cdot 1 & - 1 - 6 (- \frac{1}{4}) & - 5 - 6 \cdot - 1 \end{array}]

$[\begin{array}{c} 1 & - \frac{1}{4} & - 1 \\ 6 - 6 \cdot 1 & - 1 - 6 (- \frac{1}{4}) & - 5 - 6 \cdot - 1 \end{array}]$

Step 2.2.2

Simplify

R_{2}

$R_{2}$ .

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

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

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

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

Step 2.3

Multiply each element of

R_{2}

$R_{2}$ by

2

$2$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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Step 2.3.1

Multiply each element of

R_{2}

$R_{2}$ by

2

$2$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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

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

Step 2.3.2

Simplify

R_{2}

$R_{2}$ .

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

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

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

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

Step 2.4

Perform the row operation

R_{1} = R_{1} + \frac{1}{4} R_{2}

$R_{1} = R_{1} + \frac{1}{4} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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Step 2.4.1

Perform the row operation

R_{1} = R_{1} + \frac{1}{4} R_{2}

$R_{1} = R_{1} + \frac{1}{4} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

[\begin{array}{c} 1 + \frac{1}{4} \cdot 0 & - \frac{1}{4} + \frac{1}{4} \cdot 1 & - 1 + \frac{1}{4} \cdot 2 \\ 0 & 1 & 2 \end{array}]

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

Step 2.4.2

Simplify

R_{1}

$R_{1}$ .

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

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

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

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

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

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

Step 3

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

x = - \frac{1}{2}

$x = - \frac{1}{2}$

y = 2

$y = 2$

Step 4

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

(- \frac{1}{2}, 2)

$(- \frac{1}{2}, 2)$

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