Linear Algebra Examples

x - 6 y = 3

$x - 6 y = 3$ ,

x - y = - 1

$x - y = - 1$

Step 1

Represent the system of equations in matrix format.

[\begin{array}{c} 1 & - 6 \\ 1 & - 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 3 \\ - 1 \end{array}]

$[\begin{array}{c} 1 & - 6 \\ 1 & - 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 3 \\ - 1 \end{array}]$

Step 2

Find the determinant of the coefficient matrix

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

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

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Step 2.1

Write

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

$[\begin{array}{c} 1 & - 6 \\ 1 & - 1 \end{array}]$ in determinant notation.

| \begin{array}{c} 1 & - 6 \\ 1 & - 1 \end{array} |

$| \begin{array}{c} 1 & - 6 \\ 1 & - 1 \end{array} |$

Step 2.2

The determinant of a

2 \times 2

$2 \times 2$ matrix can be found using the formula

| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

1 \cdot - 1 - 1 \cdot - 6

$1 \cdot - 1 - 1 \cdot - 6$

Step 2.3

Simplify the determinant.

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Step 2.3.1

Simplify each term.

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Step 2.3.1.1

Multiply

- 1

$- 1$ by

1

$1$ .

- 1 - 1 \cdot - 6

$- 1 - 1 \cdot - 6$

Step 2.3.1.2

Multiply

- 1

$- 1$ by

- 6

$- 6$ .

- 1 + 6

$- 1 + 6$

- 1 + 6

$- 1 + 6$

Step 2.3.2

Add

- 1

$- 1$ and

6

$6$ .

5

$5$

5

$5$

D = 5

$D = 5$

Step 3

Since the determinant is not

0

$0$ , the system can be solved using Cramer's Rule.

Step 4

Find the value of

x

$x$ by Cramer's Rule, which states that

x = \frac{D_{x}}{D}

$x = \frac{D_{x}}{D}$ .

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Step 4.1

Replace column

1

$1$ of the coefficient matrix that corresponds to the

x

$x$ -coefficients of the system with

[\begin{array}{c} 3 \\ - 1 \end{array}]

$[\begin{array}{c} 3 \\ - 1 \end{array}]$ .

| \begin{array}{c} 3 & - 6 \\ - 1 & - 1 \end{array} |

$| \begin{array}{c} 3 & - 6 \\ - 1 & - 1 \end{array} |$

Step 4.2

Find the determinant.

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Step 4.2.1

The determinant of a

2 \times 2

$2 \times 2$ matrix can be found using the formula

| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

3 \cdot - 1 - - - 6

$3 \cdot - 1 - - - 6$

Step 4.2.2

Simplify the determinant.

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Step 4.2.2.1

Simplify each term.

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Step 4.2.2.1.1

Multiply

3

$3$ by

- 1

$- 1$ .

- 3 - - - 6

$- 3 - - - 6$

Step 4.2.2.1.2

Multiply

- - - 6

$- - - 6$ .

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Step 4.2.2.1.2.1

Multiply

- 1

$- 1$ by

- 6

$- 6$ .

- 3 - 1 \cdot 6

$- 3 - 1 \cdot 6$

Step 4.2.2.1.2.2

Multiply

- 1

$- 1$ by

6

$6$ .

- 3 - 6

$- 3 - 6$

- 3 - 6

$- 3 - 6$

- 3 - 6

$- 3 - 6$

Step 4.2.2.2

Subtract

6

$6$ from

- 3

$- 3$ .

- 9

$- 9$

- 9

$- 9$

D_{x} = - 9

$D_{x} = - 9$

Step 4.3

Use the formula to solve for

x

$x$ .

x = \frac{D_{x}}{D}

$x = \frac{D_{x}}{D}$

Step 4.4

Substitute

5

$5$ for

D

$D$ and

- 9

$- 9$ for

D_{x}

$D_{x}$ in the formula.

x = \frac{- 9}{5}

$x = \frac{- 9}{5}$

Step 4.5

Move the negative in front of the fraction.

x = - \frac{9}{5}

$x = - \frac{9}{5}$

x = - \frac{9}{5}

$x = - \frac{9}{5}$

Step 5

Find the value of

y

$y$ by Cramer's Rule, which states that

y = \frac{D_{y}}{D}

$y = \frac{D_{y}}{D}$ .

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Step 5.1

Replace column

2

$2$ of the coefficient matrix that corresponds to the

y

$y$ -coefficients of the system with

[\begin{array}{c} 3 \\ - 1 \end{array}]

$[\begin{array}{c} 3 \\ - 1 \end{array}]$ .

| \begin{array}{c} 1 & 3 \\ 1 & - 1 \end{array} |

$| \begin{array}{c} 1 & 3 \\ 1 & - 1 \end{array} |$

Step 5.2

Find the determinant.

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Step 5.2.1

The determinant of a

2 \times 2

$2 \times 2$ matrix can be found using the formula

| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

1 \cdot - 1 - 1 \cdot 3

$1 \cdot - 1 - 1 \cdot 3$

Step 5.2.2

Simplify the determinant.

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Step 5.2.2.1

Simplify each term.

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Step 5.2.2.1.1

Multiply

- 1

$- 1$ by

1

$1$ .

- 1 - 1 \cdot 3

$- 1 - 1 \cdot 3$

Step 5.2.2.1.2

Multiply

- 1

$- 1$ by

3

$3$ .

- 1 - 3

$- 1 - 3$

- 1 - 3

$- 1 - 3$

Step 5.2.2.2

Subtract

3

$3$ from

- 1

$- 1$ .

- 4

$- 4$

- 4

$- 4$

D_{y} = - 4

$D_{y} = - 4$

Step 5.3

Use the formula to solve for

y

$y$ .

y = \frac{D_{y}}{D}

$y = \frac{D_{y}}{D}$

Step 5.4

Substitute

5

$5$ for

D

$D$ and

- 4

$- 4$ for

D_{y}

$D_{y}$ in the formula.

y = \frac{- 4}{5}

$y = \frac{- 4}{5}$

Step 5.5

Move the negative in front of the fraction.

y = - \frac{4}{5}

$y = - \frac{4}{5}$

y = - \frac{4}{5}

$y = - \frac{4}{5}$

Step 6

List the solution to the system of equations.

x = - \frac{9}{5}

$x = - \frac{9}{5}$

y = - \frac{4}{5}

$y = - \frac{4}{5}$

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