Linear Algebra Examples

3 x - 2 y = 8

$3 x - 2 y = 8$ ,

6 y = 15 x + 12

$6 y = 15 x + 12$

Step 1

Subtract

15 x

$15 x$ from both sides of the equation.

3 x - 2 y = 8, 6 y - 15 x = 12

$3 x - 2 y = 8, 6 y - 15 x = 12$

Step 2

Write the system of equations in matrix form.

[\begin{array}{c} 3 & - 2 & 8 \\ - 15 & 6 & 12 \end{array}]

$[\begin{array}{c} 3 & - 2 & 8 \\ - 15 & 6 & 12 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{3}

$\frac{1}{3}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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Step 3.1.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{3}

$\frac{1}{3}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

[\begin{array}{c} \frac{3}{3} & \frac{- 2}{3} & \frac{8}{3} \\ - 15 & 6 & 12 \end{array}]

$[\begin{array}{c} \frac{3}{3} & \frac{- 2}{3} & \frac{8}{3} \\ - 15 & 6 & 12 \end{array}]$

Step 3.1.2

Simplify

R_{1}

$R_{1}$ .

[\begin{array}{c} 1 & - \frac{2}{3} & \frac{8}{3} \\ - 15 & 6 & 12 \end{array}]

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{8}{3} \\ - 15 & 6 & 12 \end{array}]$

[\begin{array}{c} 1 & - \frac{2}{3} & \frac{8}{3} \\ - 15 & 6 & 12 \end{array}]

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{8}{3} \\ - 15 & 6 & 12 \end{array}]$

Step 3.2

Perform the row operation

R_{2} = R_{2} + 15 R_{1}

$R_{2} = R_{2} + 15 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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Step 3.2.1

Perform the row operation

R_{2} = R_{2} + 15 R_{1}

$R_{2} = R_{2} + 15 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

[\begin{array}{c} 1 & - \frac{2}{3} & \frac{8}{3} \\ - 15 + 15 \cdot 1 & 6 + 15 (- \frac{2}{3}) & 12 + 15 (\frac{8}{3}) \end{array}]

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{8}{3} \\ - 15 + 15 \cdot 1 & 6 + 15 (- \frac{2}{3}) & 12 + 15 (\frac{8}{3}) \end{array}]$

Step 3.2.2

Simplify

R_{2}

$R_{2}$ .

[\begin{array}{c} 1 & - \frac{2}{3} & \frac{8}{3} \\ 0 & - 4 & 52 \end{array}]

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{8}{3} \\ 0 & - 4 & 52 \end{array}]$

[\begin{array}{c} 1 & - \frac{2}{3} & \frac{8}{3} \\ 0 & - 4 & 52 \end{array}]

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{8}{3} \\ 0 & - 4 & 52 \end{array}]$

Step 3.3

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{1}{4}

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

2, 2

$2, 2$ a

1

$1$ .

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Step 3.3.1

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{1}{4}

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

2, 2

$2, 2$ a

1

$1$ .

[\begin{array}{c} 1 & - \frac{2}{3} & \frac{8}{3} \\ - \frac{1}{4} \cdot 0 & - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot 52 \end{array}]

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{8}{3} \\ - \frac{1}{4} \cdot 0 & - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot 52 \end{array}]$

Step 3.3.2

Simplify

R_{2}

$R_{2}$ .

[\begin{array}{c} 1 & - \frac{2}{3} & \frac{8}{3} \\ 0 & 1 & - 13 \end{array}]

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{8}{3} \\ 0 & 1 & - 13 \end{array}]$

[\begin{array}{c} 1 & - \frac{2}{3} & \frac{8}{3} \\ 0 & 1 & - 13 \end{array}]

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{8}{3} \\ 0 & 1 & - 13 \end{array}]$

Step 3.4

Perform the row operation

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

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

1, 2

$1, 2$ a

0

$0$ .

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Step 3.4.1

Perform the row operation

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

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

1, 2

$1, 2$ a

0

$0$ .

[\begin{array}{c} 1 + \frac{2}{3} \cdot 0 & - \frac{2}{3} + \frac{2}{3} \cdot 1 & \frac{8}{3} + \frac{2}{3} \cdot - 13 \\ 0 & 1 & - 13 \end{array}]

$[\begin{array}{c} 1 + \frac{2}{3} \cdot 0 & - \frac{2}{3} + \frac{2}{3} \cdot 1 & \frac{8}{3} + \frac{2}{3} \cdot - 13 \\ 0 & 1 & - 13 \end{array}]$

Step 3.4.2

Simplify

R_{1}

$R_{1}$ .

[\begin{array}{c} 1 & 0 & - 6 \\ 0 & 1 & - 13 \end{array}]

$[\begin{array}{c} 1 & 0 & - 6 \\ 0 & 1 & - 13 \end{array}]$

[\begin{array}{c} 1 & 0 & - 6 \\ 0 & 1 & - 13 \end{array}]

$[\begin{array}{c} 1 & 0 & - 6 \\ 0 & 1 & - 13 \end{array}]$

[\begin{array}{c} 1 & 0 & - 6 \\ 0 & 1 & - 13 \end{array}]

$[\begin{array}{c} 1 & 0 & - 6 \\ 0 & 1 & - 13 \end{array}]$

Step 4

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

x = - 6

$x = - 6$

y = - 13

$y = - 13$

Step 5

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

(- 6, - 13)

$(- 6, - 13)$

Step 6

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} - 6 \\ - 13 \end{array}]

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 6 \\ - 13 \end{array}]$