Linear Algebra Examples

- 4 x + y = - 8

$- 4 x + y = - 8$ ,

- 3 x + 5 y = 3

$- 3 x + 5 y = 3$

Step 1

Write the system as a matrix.

[\begin{array}{c} - 4 & 1 & - 8 \\ - 3 & 5 & 3 \end{array}]

$[\begin{array}{c} - 4 & 1 & - 8 \\ - 3 & 5 & 3 \end{array}]$

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{1}{4} \cdot - 4 & - \frac{1}{4} \cdot 1 & - \frac{1}{4} \cdot - 8 \\ - 3 & 5 & 3 \end{array}]

$[\begin{array}{c} - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot 1 & - \frac{1}{4} \cdot - 8 \\ - 3 & 5 & 3 \end{array}]$

Step 2.1.2

Simplify

R_{1}

$R_{1}$ .

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

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

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

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

Step 2.2

Perform the row operation

R_{2} = R_{2} + 3 R_{1}

$R_{2} = R_{2} + 3 R_{1}$ 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} + 3 R_{1}

$R_{2} = R_{2} + 3 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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

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

Step 2.2.2

Simplify

R_{2}

$R_{2}$ .

[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ 0 & \frac{17}{4} & 9 \end{array}]

$[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ 0 & \frac{17}{4} & 9 \end{array}]$

[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ 0 & \frac{17}{4} & 9 \end{array}]

$[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ 0 & \frac{17}{4} & 9 \end{array}]$

Step 2.3

Multiply each element of

R_{2}

$R_{2}$ by

\frac{4}{17}

$\frac{4}{17}$ 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

\frac{4}{17}

$\frac{4}{17}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ \frac{4}{17} \cdot 0 & \frac{4}{17} \cdot \frac{17}{4} & \frac{4}{17} \cdot 9 \end{array}]

$[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ \frac{4}{17} \cdot 0 & \frac{4}{17} \cdot \frac{17}{4} & \frac{4}{17} \cdot 9 \end{array}]$

Step 2.3.2

Simplify

R_{2}

$R_{2}$ .

[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ 0 & 1 & \frac{36}{17} \end{array}]

$[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ 0 & 1 & \frac{36}{17} \end{array}]$

[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ 0 & 1 & \frac{36}{17} \end{array}]

$[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ 0 & 1 & \frac{36}{17} \end{array}]$

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 & 2 + \frac{1}{4} \cdot \frac{36}{17} \\ 0 & 1 & \frac{36}{17} \end{array}]

$[\begin{array}{c} 1 + \frac{1}{4} \cdot 0 & - \frac{1}{4} + \frac{1}{4} \cdot 1 & 2 + \frac{1}{4} \cdot \frac{36}{17} \\ 0 & 1 & \frac{36}{17} \end{array}]$

Step 2.4.2

Simplify

R_{1}

$R_{1}$ .

[\begin{array}{c} 1 & 0 & \frac{43}{17} \\ 0 & 1 & \frac{36}{17} \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{43}{17} \\ 0 & 1 & \frac{36}{17} \end{array}]$

[\begin{array}{c} 1 & 0 & \frac{43}{17} \\ 0 & 1 & \frac{36}{17} \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{43}{17} \\ 0 & 1 & \frac{36}{17} \end{array}]$

[\begin{array}{c} 1 & 0 & \frac{43}{17} \\ 0 & 1 & \frac{36}{17} \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{43}{17} \\ 0 & 1 & \frac{36}{17} \end{array}]$

Step 3

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

x = \frac{43}{17}

$x = \frac{43}{17}$

y = \frac{36}{17}

$y = \frac{36}{17}$

Step 4

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

(\frac{43}{17}, \frac{36}{17})

$(\frac{43}{17}, \frac{36}{17})$

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