Linear Algebra Examples

5 x + 3 = 4 y

$5 x + 3 = 4 y$ ,

y = 8 x - 2

$y = 8 x - 2$

Step 1

Move all of the variables to the left side of each equation.

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Step 1.1

Subtract

4 y

$4 y$ from both sides of the equation.

5 x + 3 - 4 y = 0

$5 x + 3 - 4 y = 0$

y = 8 x - 2

$y = 8 x - 2$

Step 1.2

Subtract

3

$3$ from both sides of the equation.

5 x - 4 y = - 3

$5 x - 4 y = - 3$

y = 8 x - 2

$y = 8 x - 2$

Step 1.3

Subtract

8 x

$8 x$ from both sides of the equation.

5 x - 4 y = - 3

$5 x - 4 y = - 3$

y - 8 x = - 2

$y - 8 x = - 2$

Step 1.4

Reorder

y

$y$ and

- 8 x

$- 8 x$ .

5 x - 4 y = - 3

$5 x - 4 y = - 3$

- 8 x + y = - 2

$- 8 x + y = - 2$

5 x - 4 y = - 3

$5 x - 4 y = - 3$

- 8 x + y = - 2

$- 8 x + y = - 2$

Step 2

Represent the system of equations in matrix format.

[\begin{matrix} 5 & - 4 - 8 & 1 \end{matrix}] [\begin{matrix} x y \end{matrix}] = [\begin{matrix} - 3 - 2 \end{matrix}]

$[\begin{array}{c} 5 & - 4 \\ - 8 & 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 3 \\ - 2 \end{array}]$

Step 3

Find the determinant of the coefficient matrix

[\begin{matrix} 5 & - 4 - 8 & 1 \end{matrix}]

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

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Step 3.1

Write

[\begin{matrix} 5 & - 4 - 8 & 1 \end{matrix}]

$[\begin{array}{c} 5 & - 4 \\ - 8 & 1 \end{array}]$ in determinant notation.

∣ ∣ ∣ \begin{matrix} 5 & - 4 - 8 & 1 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 5 & - 4 \\ - 8 & 1 \end{array} |$

Step 3.2

The determinant of a

2 \times 2

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

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

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

5 \cdot 1 - (- 8 \cdot - 4)

$5 \cdot 1 - (- 8 \cdot - 4)$

Step 3.3

Simplify the determinant.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Multiply

5

$5$ by

1

$1$ .

5 - (- 8 \cdot - 4)

$5 - (- 8 \cdot - 4)$

Step 3.3.1.2

Multiply

- (- 8 \cdot - 4)

$- (- 8 \cdot - 4)$ .

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Step 3.3.1.2.1

Multiply

- 8

$- 8$ by

- 4

$- 4$ .

5 - 1 \cdot 32

$5 - 1 \cdot 32$

Step 3.3.1.2.2

Multiply

- 1

$- 1$ by

32

$32$ .

5 - 32

$5 - 32$

5 - 32

$5 - 32$

5 - 32

$5 - 32$

Step 3.3.2

Subtract

32

$32$ from

5

$5$ .

- 27

$- 27$

- 27

$- 27$

D = - 27

$D = - 27$

Step 4

Since the determinant is not

0

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

Step 5

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 5.1

Replace column

1

$1$ of the coefficient matrix that corresponds to the

x

$x$ -coefficients of the system with

[\begin{matrix} - 3 - 2 \end{matrix}]

$[\begin{array}{c} - 3 \\ - 2 \end{array}]$ .

∣ ∣ ∣ \begin{matrix} - 3 & - 4 - 2 & 1 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} - 3 & - 4 \\ - 2 & 1 \end{array} |$

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{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

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

- 3 \cdot 1 - (- 2 \cdot - 4)

$- 3 \cdot 1 - (- 2 \cdot - 4)$

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

- 3

$- 3$ by

1

$1$ .

- 3 - (- 2 \cdot - 4)

$- 3 - (- 2 \cdot - 4)$

Step 5.2.2.1.2

Multiply

- (- 2 \cdot - 4)

$- (- 2 \cdot - 4)$ .

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Step 5.2.2.1.2.1

Multiply

- 2

$- 2$ by

- 4

$- 4$ .

- 3 - 1 \cdot 8

$- 3 - 1 \cdot 8$

Step 5.2.2.1.2.2

Multiply

- 1

$- 1$ by

8

$8$ .

- 3 - 8

$- 3 - 8$

- 3 - 8

$- 3 - 8$

- 3 - 8

$- 3 - 8$

Step 5.2.2.2

Subtract

8

$8$ from

- 3

$- 3$ .

- 11

$- 11$

- 11

$- 11$

D_{x} = - 11

$D_{x} = - 11$

Step 5.3

Use the formula to solve for

x

$x$ .

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

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

Step 5.4

Substitute

- 27

$- 27$ for

D

$D$ and

- 11

$- 11$ for

D_{x}

$D_{x}$ in the formula.

x = \frac{- 11}{- 27}

$x = \frac{- 11}{- 27}$

Step 5.5

Dividing two negative values results in a positive value.

x = \frac{11}{27}

$x = \frac{11}{27}$

x = \frac{11}{27}

$x = \frac{11}{27}$

Step 6

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 6.1

Replace column

2

$2$ of the coefficient matrix that corresponds to the

y

$y$ -coefficients of the system with

[\begin{matrix} - 3 - 2 \end{matrix}]

$[\begin{array}{c} - 3 \\ - 2 \end{array}]$ .

∣ ∣ ∣ \begin{matrix} 5 & - 3 - 8 & - 2 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 5 & - 3 \\ - 8 & - 2 \end{array} |$

Step 6.2

Find the determinant.

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Step 6.2.1

The determinant of a

2 \times 2

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

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

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

5 \cdot - 2 - (- 8 \cdot - 3)

$5 \cdot - 2 - (- 8 \cdot - 3)$

Step 6.2.2

Simplify the determinant.

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Step 6.2.2.1

Simplify each term.

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Step 6.2.2.1.1

Multiply

5

$5$ by

- 2

$- 2$ .

- 10 - (- 8 \cdot - 3)

$- 10 - (- 8 \cdot - 3)$

Step 6.2.2.1.2

Multiply

- (- 8 \cdot - 3)

$- (- 8 \cdot - 3)$ .

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Step 6.2.2.1.2.1

Multiply

- 8

$- 8$ by

- 3

$- 3$ .

- 10 - 1 \cdot 24

$- 10 - 1 \cdot 24$

Step 6.2.2.1.2.2

Multiply

- 1

$- 1$ by

24

$24$ .

- 10 - 24

$- 10 - 24$

- 10 - 24

$- 10 - 24$

- 10 - 24

$- 10 - 24$

Step 6.2.2.2

Subtract

24

$24$ from

- 10

$- 10$ .

- 34

$- 34$

- 34

$- 34$

D_{y} = - 34

$D_{y} = - 34$

Step 6.3

Use the formula to solve for

y

$y$ .

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

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

Step 6.4

Substitute

- 27

$- 27$ for

D

$D$ and

- 34

$- 34$ for

D_{y}

$D_{y}$ in the formula.

y = \frac{- 34}{- 27}

$y = \frac{- 34}{- 27}$

Step 6.5

Dividing two negative values results in a positive value.

y = \frac{34}{27}

$y = \frac{34}{27}$

y = \frac{34}{27}

$y = \frac{34}{27}$

Step 7

List the solution to the system of equations.

x = \frac{11}{27}

$x = \frac{11}{27}$

y = \frac{34}{27}

$y = \frac{34}{27}$