Finite Math Examples

3 z + 3 x + 3 y = 19

$3 z + 3 x + 3 y = 19$ ,

x + 3 = y

$x + 3 = y$ ,

z = y - 4 x + 1

$z = y - 4 x + 1$

Step 1

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

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Step 1.1

Move

3 z

$3 z$ .

3 x + 3 y + 3 z = 19

$3 x + 3 y + 3 z = 19$

x + 3 = y

$x + 3 = y$

z = y - 4 x + 1

$z = y - 4 x + 1$

Step 1.2

Subtract

y

$y$ from both sides of the equation.

3 x + 3 y + 3 z = 19

$3 x + 3 y + 3 z = 19$

x + 3 - y = 0

$x + 3 - y = 0$

z = y - 4 x + 1

$z = y - 4 x + 1$

Step 1.3

Subtract

3

$3$ from both sides of the equation.

3 x + 3 y + 3 z = 19

$3 x + 3 y + 3 z = 19$

x - y = - 3

$x - y = - 3$

z = y - 4 x + 1

$z = y - 4 x + 1$

Step 1.4

Move all terms containing variables to the left side of the equation.

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Step 1.4.1

Subtract

y

$y$ from both sides of the equation.

3 x + 3 y + 3 z = 19

$3 x + 3 y + 3 z = 19$

x - y = - 3

$x - y = - 3$

z - y = - 4 x + 1

$z - y = - 4 x + 1$

Step 1.4.2

Add

4 x

$4 x$ to both sides of the equation.

3 x + 3 y + 3 z = 19

$3 x + 3 y + 3 z = 19$

x - y = - 3

$x - y = - 3$

z - y + 4 x = 1

$z - y + 4 x = 1$

3 x + 3 y + 3 z = 19

$3 x + 3 y + 3 z = 19$

x - y = - 3

$x - y = - 3$

z - y + 4 x = 1

$z - y + 4 x = 1$

Step 1.5

Move

z

$z$ .

3 x + 3 y + 3 z = 19

$3 x + 3 y + 3 z = 19$

x - y = - 3

$x - y = - 3$

- y + 4 x + z = 1

$- y + 4 x + z = 1$

Step 1.6

Reorder

- y

$- y$ and

4 x

$4 x$ .

3 x + 3 y + 3 z = 19

$3 x + 3 y + 3 z = 19$

x - y = - 3

$x - y = - 3$

4 x - y + z = 1

$4 x - y + z = 1$

3 x + 3 y + 3 z = 19

$3 x + 3 y + 3 z = 19$

x - y = - 3

$x - y = - 3$

4 x - y + z = 1

$4 x - y + z = 1$

Step 2

Represent the system of equations in matrix format.

⎡ ⎢ ⎣ \begin{matrix} 3 & 3 & 3 1 & - 1 & 0 4 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 19 - 3 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 3 & 3 & 3 \\ 1 & - 1 & 0 \\ 4 & - 1 & 1 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 19 \\ - 3 \\ 1 \end{array}]$

Step 3

Find the determinant of the coefficient matrix

⎡ ⎢ ⎣ \begin{matrix} 3 & 3 & 3 1 & - 1 & 0 4 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 3 & 3 & 3 \\ 1 & - 1 & 0 \\ 4 & - 1 & 1 \end{array}]$ .

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Step 3.1

Write

⎡ ⎢ ⎣ \begin{matrix} 3 & 3 & 3 1 & - 1 & 0 4 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 3 & 3 & 3 \\ 1 & - 1 & 0 \\ 4 & - 1 & 1 \end{array}]$ in determinant notation.

∣ ∣ ∣ ∣ \begin{matrix} 3 & 3 & 3 1 & - 1 & 0 4 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣

$| \begin{array}{c} 3 & 3 & 3 \\ 1 & - 1 & 0 \\ 4 & - 1 & 1 \end{array} |$

Step 3.2

Choose the row or column with the most

0

$0$ elements. If there are no

0

$0$ elements choose any row or column. Multiply every element in row

2

$2$ by its cofactor and add.

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Step 3.2.1

Consider the corresponding sign chart.

∣ ∣ ∣ ∣ \begin{matrix} + & - & + - & + & - + & - & + \end{matrix} ∣ ∣ ∣ ∣

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 3.2.2

The cofactor is the minor with the sign changed if the indices match a

-

$-$ position on the sign chart.

Step 3.2.3

The minor for

a_{21}

$a_{21}$ is the determinant with row

2

$2$ and column

1

$1$ deleted.

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

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

Step 3.2.4

Multiply element

a_{21}

$a_{21}$ by its cofactor.

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

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

Step 3.2.5

The minor for

a_{22}

$a_{22}$ is the determinant with row

2

$2$ and column

2

$2$ deleted.

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

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

Step 3.2.6

Multiply element

a_{22}

$a_{22}$ by its cofactor.

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

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

Step 3.2.7

The minor for

a_{23}

$a_{23}$ is the determinant with row

2

$2$ and column

3

$3$ deleted.

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

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

Step 3.2.8

Multiply element

a_{23}

$a_{23}$ by its cofactor.

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

$0 | \begin{array}{c} 3 & 3 \\ 4 & - 1 \end{array} |$

Step 3.2.9

Add the terms together.

- 1 ∣ ∣ ∣ \begin{matrix} 3 & 3 - 1 & 1 \end{matrix} ∣ ∣ ∣ - 1 ∣ ∣ ∣ \begin{matrix} 3 & 3 4 & 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 3 & 3 4 & - 1 \end{matrix} ∣ ∣ ∣

$- 1 | \begin{array}{c} 3 & 3 \\ - 1 & 1 \end{array} | - 1 | \begin{array}{c} 3 & 3 \\ 4 & 1 \end{array} | + 0 | \begin{array}{c} 3 & 3 \\ 4 & - 1 \end{array} |$

- 1 ∣ ∣ ∣ \begin{matrix} 3 & 3 - 1 & 1 \end{matrix} ∣ ∣ ∣ - 1 ∣ ∣ ∣ \begin{matrix} 3 & 3 4 & 1 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 3 & 3 4 & - 1 \end{matrix} ∣ ∣ ∣

$- 1 | \begin{array}{c} 3 & 3 \\ - 1 & 1 \end{array} | - 1 | \begin{array}{c} 3 & 3 \\ 4 & 1 \end{array} | + 0 | \begin{array}{c} 3 & 3 \\ 4 & - 1 \end{array} |$

Step 3.3

Multiply

0

$0$ by

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

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

- 1 ∣ ∣ ∣ \begin{matrix} 3 & 3 - 1 & 1 \end{matrix} ∣ ∣ ∣ - 1 ∣ ∣ ∣ \begin{matrix} 3 & 3 4 & 1 \end{matrix} ∣ ∣ ∣ + 0

$- 1 | \begin{array}{c} 3 & 3 \\ - 1 & 1 \end{array} | - 1 | \begin{array}{c} 3 & 3 \\ 4 & 1 \end{array} | + 0$

Step 3.4

Evaluate

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

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

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Step 3.4.1

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

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

- 1 (3 \cdot 1 - (- 1 \cdot 3)) - 1 ∣ ∣ ∣ \begin{matrix} 3 & 3 4 & 1 \end{matrix} ∣ ∣ ∣ + 0

$- 1 (3 \cdot 1 - (- 1 \cdot 3)) - 1 | \begin{array}{c} 3 & 3 \\ 4 & 1 \end{array} | + 0$

Step 3.4.2

Simplify the determinant.

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Step 3.4.2.1

Simplify each term.

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Step 3.4.2.1.1

Multiply

3

$3$ by

1

$1$ .

- 1 (3 - (- 1 \cdot 3)) - 1 ∣ ∣ ∣ \begin{matrix} 3 & 3 4 & 1 \end{matrix} ∣ ∣ ∣ + 0

$- 1 (3 - (- 1 \cdot 3)) - 1 | \begin{array}{c} 3 & 3 \\ 4 & 1 \end{array} | + 0$

Step 3.4.2.1.2

Multiply

- (- 1 \cdot 3)

$- (- 1 \cdot 3)$ .

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Step 3.4.2.1.2.1

Multiply

- 1

$- 1$ by

3

$3$ .

- 1 (3 - - 3) - 1 ∣ ∣ ∣ \begin{matrix} 3 & 3 4 & 1 \end{matrix} ∣ ∣ ∣ + 0

$- 1 (3 - - 3) - 1 | \begin{array}{c} 3 & 3 \\ 4 & 1 \end{array} | + 0$

Step 3.4.2.1.2.2

Multiply

- 1

$- 1$ by

- 3

$- 3$ .

- 1 (3 + 3) - 1 ∣ ∣ ∣ \begin{matrix} 3 & 3 4 & 1 \end{matrix} ∣ ∣ ∣ + 0

$- 1 (3 + 3) - 1 | \begin{array}{c} 3 & 3 \\ 4 & 1 \end{array} | + 0$

- 1 (3 + 3) - 1 ∣ ∣ ∣ \begin{matrix} 3 & 3 4 & 1 \end{matrix} ∣ ∣ ∣ + 0

$- 1 (3 + 3) - 1 | \begin{array}{c} 3 & 3 \\ 4 & 1 \end{array} | + 0$

Step 3.4.2.2

Add

$3$ and

$3$ .

$- 1 \cdot 6 - 1 | \begin{array}{c} 3 & 3 \\ 4 & 1 \end{array} | + 0$

Step 3.5

Evaluate

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

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Step 3.5.1

The determinant of a

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

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

$- 1 \cdot 6 - 1 (3 \cdot 1 - 4 \cdot 3) + 0$

Step 3.5.2

Simplify the determinant.

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Step 3.5.2.1

Simplify each term.

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Step 3.5.2.1.1

Multiply

$3$ by

$1$ .

$- 1 \cdot 6 - 1 (3 - 4 \cdot 3) + 0$

Step 3.5.2.1.2

Multiply

$- 4$ by

$3$ .

$- 1 \cdot 6 - 1 (3 - 12) + 0$

Step 3.5.2.2

Subtract

$12$ from

$3$ .

$- 1 \cdot 6 - 1 \cdot - 9 + 0$

Step 3.6

Simplify the determinant.

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Step 3.6.1

Simplify each term.

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Step 3.6.1.1

Multiply

$- 1$ by

$6$ .

$- 6 - 1 \cdot - 9 + 0$

Step 3.6.1.2

Multiply

$- 1$ by

$- 9$ .

$- 6 + 9 + 0$

Step 3.6.2

Add

$- 6$ and

$9$ .

$3 + 0$

Step 3.6.3

Add

$3$ and

$0$ .

$3$

$D = 3$

Step 4

Since the determinant is not

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

Step 5

Find the value of

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

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

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Step 5.1

Replace column

$1$ of the coefficient matrix that corresponds to the

$x$ -coefficients of the system with

$[\begin{array}{c} 19 \\ - 3 \\ 1 \end{array}]$ .

$| \begin{array}{c} 19 & 3 & 3 \\ - 3 & - 1 & 0 \\ 1 & - 1 & 1 \end{array} |$

Step 5.2

Find the determinant.

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Step 5.2.1

Choose the row or column with the most

$0$ elements. If there are no

$0$ elements choose any row or column. Multiply every element in row

$2$ by its cofactor and add.

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Step 5.2.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 5.2.1.2

The cofactor is the minor with the sign changed if the indices match a

$-$ position on the sign chart.

Step 5.2.1.3

The minor for

$a_{21}$ is the determinant with row

$2$ and column

$1$ deleted.

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

Step 5.2.1.4

Multiply element

$a_{21}$ by its cofactor.

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

Step 5.2.1.5

The minor for

$a_{22}$ is the determinant with row

$2$ and column

$2$ deleted.

$| \begin{array}{c} 19 & 3 \\ 1 & 1 \end{array} |$

Step 5.2.1.6

Multiply element

$a_{22}$ by its cofactor.

$- 1 | \begin{array}{c} 19 & 3 \\ 1 & 1 \end{array} |$

Step 5.2.1.7

The minor for

$a_{23}$ is the determinant with row

$2$ and column

$3$ deleted.

$| \begin{array}{c} 19 & 3 \\ 1 & - 1 \end{array} |$

Step 5.2.1.8

Multiply element

$a_{23}$ by its cofactor.

$0 | \begin{array}{c} 19 & 3 \\ 1 & - 1 \end{array} |$

Step 5.2.1.9

Add the terms together.

$3 | \begin{array}{c} 3 & 3 \\ - 1 & 1 \end{array} | - 1 | \begin{array}{c} 19 & 3 \\ 1 & 1 \end{array} | + 0 | \begin{array}{c} 19 & 3 \\ 1 & - 1 \end{array} |$

Step 5.2.2

Multiply

$0$ by

$| \begin{array}{c} 19 & 3 \\ 1 & - 1 \end{array} |$ .

$3 | \begin{array}{c} 3 & 3 \\ - 1 & 1 \end{array} | - 1 | \begin{array}{c} 19 & 3 \\ 1 & 1 \end{array} | + 0$

Step 5.2.3

Evaluate

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

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Step 5.2.3.1

The determinant of a

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

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

$3 (3 \cdot 1 - (- 1 \cdot 3)) - 1 | \begin{array}{c} 19 & 3 \\ 1 & 1 \end{array} | + 0$

Step 5.2.3.2

Simplify the determinant.

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Step 5.2.3.2.1

Simplify each term.

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Step 5.2.3.2.1.1

Multiply

$3$ by

$1$ .

$3 (3 - (- 1 \cdot 3)) - 1 | \begin{array}{c} 19 & 3 \\ 1 & 1 \end{array} | + 0$

Step 5.2.3.2.1.2

Multiply

$- (- 1 \cdot 3)$ .

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Step 5.2.3.2.1.2.1

Multiply

$- 1$ by

$3$ .

$3 (3 - - 3) - 1 | \begin{array}{c} 19 & 3 \\ 1 & 1 \end{array} | + 0$

Step 5.2.3.2.1.2.2

Multiply

$- 1$ by

$- 3$ .

$3 (3 + 3) - 1 | \begin{array}{c} 19 & 3 \\ 1 & 1 \end{array} | + 0$

Step 5.2.3.2.2

Add

$3$ and

$3$ .

$3 \cdot 6 - 1 | \begin{array}{c} 19 & 3 \\ 1 & 1 \end{array} | + 0$

Step 5.2.4

Evaluate

$| \begin{array}{c} 19 & 3 \\ 1 & 1 \end{array} |$ .

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Step 5.2.4.1

The determinant of a

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

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

$3 \cdot 6 - 1 (19 \cdot 1 - 1 \cdot 3) + 0$

Step 5.2.4.2

Simplify the determinant.

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Step 5.2.4.2.1

Simplify each term.

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Step 5.2.4.2.1.1

Multiply

$19$ by

$1$ .

$3 \cdot 6 - 1 (19 - 1 \cdot 3) + 0$

Step 5.2.4.2.1.2

Multiply

$- 1$ by

$3$ .

$3 \cdot 6 - 1 (19 - 3) + 0$

Step 5.2.4.2.2

Subtract

$3$ from

$19$ .

$3 \cdot 6 - 1 \cdot 16 + 0$

Step 5.2.5

Simplify the determinant.

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Step 5.2.5.1

Simplify each term.

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Step 5.2.5.1.1

Multiply

$3$ by

$6$ .

$18 - 1 \cdot 16 + 0$

Step 5.2.5.1.2

Multiply

$- 1$ by

$16$ .

$18 - 16 + 0$

Step 5.2.5.2

Subtract

$16$ from

$18$ .

$2 + 0$

Step 5.2.5.3

Add

$2$ and

$0$ .

$2$

$D_{x} = 2$

Step 5.3

Use the formula to solve for

$x$ .

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

Step 5.4

Substitute

$3$ for

$D$ and

$2$ for

$D_{x}$ in the formula.

$x = \frac{2}{3}$

Step 6

Find the value of

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

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

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Step 6.1

Replace column

$2$ of the coefficient matrix that corresponds to the

$y$ -coefficients of the system with

$[\begin{array}{c} 19 \\ - 3 \\ 1 \end{array}]$ .

$| \begin{array}{c} 3 & 19 & 3 \\ 1 & - 3 & 0 \\ 4 & 1 & 1 \end{array} |$

Step 6.2

Find the determinant.

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Step 6.2.1

Choose the row or column with the most

$0$ elements. If there are no

$0$ elements choose any row or column. Multiply every element in row

$2$ by its cofactor and add.

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Step 6.2.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 6.2.1.2

The cofactor is the minor with the sign changed if the indices match a

$-$ position on the sign chart.

Step 6.2.1.3

The minor for

$a_{21}$ is the determinant with row

$2$ and column

$1$ deleted.

$| \begin{array}{c} 19 & 3 \\ 1 & 1 \end{array} |$

Step 6.2.1.4

Multiply element

$a_{21}$ by its cofactor.

$- 1 | \begin{array}{c} 19 & 3 \\ 1 & 1 \end{array} |$

Step 6.2.1.5

The minor for

$a_{22}$ is the determinant with row

$2$ and column

$2$ deleted.

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

Step 6.2.1.6

Multiply element

$a_{22}$ by its cofactor.

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

Step 6.2.1.7

The minor for

$a_{23}$ is the determinant with row

$2$ and column

$3$ deleted.

$| \begin{array}{c} 3 & 19 \\ 4 & 1 \end{array} |$

Step 6.2.1.8

Multiply element

$a_{23}$ by its cofactor.

$0 | \begin{array}{c} 3 & 19 \\ 4 & 1 \end{array} |$

Step 6.2.1.9

Add the terms together.

$- 1 | \begin{array}{c} 19 & 3 \\ 1 & 1 \end{array} | - 3 | \begin{array}{c} 3 & 3 \\ 4 & 1 \end{array} | + 0 | \begin{array}{c} 3 & 19 \\ 4 & 1 \end{array} |$

Step 6.2.2

Multiply

$0$ by

$| \begin{array}{c} 3 & 19 \\ 4 & 1 \end{array} |$ .

$- 1 | \begin{array}{c} 19 & 3 \\ 1 & 1 \end{array} | - 3 | \begin{array}{c} 3 & 3 \\ 4 & 1 \end{array} | + 0$

Step 6.2.3

Evaluate

$| \begin{array}{c} 19 & 3 \\ 1 & 1 \end{array} |$ .

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Step 6.2.3.1

The determinant of a

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

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

$- 1 (19 \cdot 1 - 1 \cdot 3) - 3 | \begin{array}{c} 3 & 3 \\ 4 & 1 \end{array} | + 0$

Step 6.2.3.2

Simplify the determinant.

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Step 6.2.3.2.1

Simplify each term.

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Step 6.2.3.2.1.1

Multiply

$19$ by

$1$ .

$- 1 (19 - 1 \cdot 3) - 3 | \begin{array}{c} 3 & 3 \\ 4 & 1 \end{array} | + 0$

Step 6.2.3.2.1.2

Multiply

$- 1$ by

$3$ .

$- 1 (19 - 3) - 3 | \begin{array}{c} 3 & 3 \\ 4 & 1 \end{array} | + 0$

Step 6.2.3.2.2

Subtract

$3$ from

$19$ .

$- 1 \cdot 16 - 3 | \begin{array}{c} 3 & 3 \\ 4 & 1 \end{array} | + 0$

Step 6.2.4

Evaluate

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

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Step 6.2.4.1

The determinant of a

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

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

$- 1 \cdot 16 - 3 (3 \cdot 1 - 4 \cdot 3) + 0$

Step 6.2.4.2

Simplify the determinant.

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Step 6.2.4.2.1

Simplify each term.

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Step 6.2.4.2.1.1

Multiply

$3$ by

$1$ .

$- 1 \cdot 16 - 3 (3 - 4 \cdot 3) + 0$

Step 6.2.4.2.1.2

Multiply

$- 4$ by

$3$ .

$- 1 \cdot 16 - 3 (3 - 12) + 0$

Step 6.2.4.2.2

Subtract

$12$ from

$3$ .

$- 1 \cdot 16 - 3 \cdot - 9 + 0$

Step 6.2.5

Simplify the determinant.

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Step 6.2.5.1

Simplify each term.

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Step 6.2.5.1.1

Multiply

$- 1$ by

$16$ .

$- 16 - 3 \cdot - 9 + 0$

Step 6.2.5.1.2

Multiply

$- 3$ by

$- 9$ .

$- 16 + 27 + 0$

Step 6.2.5.2

Add

$- 16$ and

$27$ .

$11 + 0$

Step 6.2.5.3

Add

$11$ and

$0$ .

$11$

$D_{y} = 11$

Step 6.3

Use the formula to solve for

$y$ .

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

Step 6.4

Substitute

$3$ for

$D$ and

$11$ for

$D_{y}$ in the formula.

$y = \frac{11}{3}$

Step 7

Find the value of

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

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

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Step 7.1

Replace column

$3$ of the coefficient matrix that corresponds to the

$z$ -coefficients of the system with

$[\begin{array}{c} 19 \\ - 3 \\ 1 \end{array}]$ .

$| \begin{array}{c} 3 & 3 & 19 \\ 1 & - 1 & - 3 \\ 4 & - 1 & 1 \end{array} |$

Step 7.2

Find the determinant.

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Step 7.2.1

Choose the row or column with the most

$0$ elements. If there are no

$0$ elements choose any row or column. Multiply every element in row

$1$ by its cofactor and add.

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Step 7.2.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 7.2.1.2

The cofactor is the minor with the sign changed if the indices match a

$-$ position on the sign chart.

Step 7.2.1.3

The minor for

$a_{11}$ is the determinant with row

$1$ and column

$1$ deleted.

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

Step 7.2.1.4

Multiply element

$a_{11}$ by its cofactor.

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

Step 7.2.1.5

The minor for

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

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

Step 7.2.1.6

Multiply element

$a_{12}$ by its cofactor.

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

Step 7.2.1.7

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

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

Step 7.2.1.8

Multiply element

$a_{13}$ by its cofactor.

$19 | \begin{array}{c} 1 & - 1 \\ 4 & - 1 \end{array} |$

Step 7.2.1.9

Add the terms together.

$3 | \begin{array}{c} - 1 & - 3 \\ - 1 & 1 \end{array} | - 3 | \begin{array}{c} 1 & - 3 \\ 4 & 1 \end{array} | + 19 | \begin{array}{c} 1 & - 1 \\ 4 & - 1 \end{array} |$

Step 7.2.2

Evaluate

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

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Step 7.2.2.1

The determinant of a

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

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

$3 (- 1 \cdot 1 - - - 3) - 3 | \begin{array}{c} 1 & - 3 \\ 4 & 1 \end{array} | + 19 | \begin{array}{c} 1 & - 1 \\ 4 & - 1 \end{array} |$

Step 7.2.2.2

Simplify the determinant.

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Step 7.2.2.2.1

Simplify each term.

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Step 7.2.2.2.1.1

Multiply

$- 1$ by

$1$ .

$3 (- 1 - - - 3) - 3 | \begin{array}{c} 1 & - 3 \\ 4 & 1 \end{array} | + 19 | \begin{array}{c} 1 & - 1 \\ 4 & - 1 \end{array} |$

Step 7.2.2.2.1.2

Multiply

$- - - 3$ .

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Step 7.2.2.2.1.2.1

Multiply

$- 1$ by

$- 3$ .

$3 (- 1 - 1 \cdot 3) - 3 | \begin{array}{c} 1 & - 3 \\ 4 & 1 \end{array} | + 19 | \begin{array}{c} 1 & - 1 \\ 4 & - 1 \end{array} |$

Step 7.2.2.2.1.2.2

Multiply

$- 1$ by

$3$ .

$3 (- 1 - 3) - 3 | \begin{array}{c} 1 & - 3 \\ 4 & 1 \end{array} | + 19 | \begin{array}{c} 1 & - 1 \\ 4 & - 1 \end{array} |$

Step 7.2.2.2.2

Subtract

$3$ from

$- 1$ .

$3 \cdot - 4 - 3 | \begin{array}{c} 1 & - 3 \\ 4 & 1 \end{array} | + 19 | \begin{array}{c} 1 & - 1 \\ 4 & - 1 \end{array} |$

Step 7.2.3

Evaluate

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

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Step 7.2.3.1

The determinant of a

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

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

$3 \cdot - 4 - 3 (1 \cdot 1 - 4 \cdot - 3) + 19 | \begin{array}{c} 1 & - 1 \\ 4 & - 1 \end{array} |$

Step 7.2.3.2

Simplify the determinant.

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Step 7.2.3.2.1

Simplify each term.

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Step 7.2.3.2.1.1

Multiply

$1$ by

$1$ .

$3 \cdot - 4 - 3 (1 - 4 \cdot - 3) + 19 | \begin{array}{c} 1 & - 1 \\ 4 & - 1 \end{array} |$

Step 7.2.3.2.1.2

Multiply

$- 4$ by

$- 3$ .

$3 \cdot - 4 - 3 (1 + 12) + 19 | \begin{array}{c} 1 & - 1 \\ 4 & - 1 \end{array} |$

Step 7.2.3.2.2

Add

$1$ and

$12$ .

$3 \cdot - 4 - 3 \cdot 13 + 19 | \begin{array}{c} 1 & - 1 \\ 4 & - 1 \end{array} |$

Step 7.2.4

Evaluate

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

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Step 7.2.4.1

The determinant of a

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

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

$3 \cdot - 4 - 3 \cdot 13 + 19 (1 \cdot - 1 - 4 \cdot - 1)$

Step 7.2.4.2

Simplify the determinant.

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Step 7.2.4.2.1

Simplify each term.

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Step 7.2.4.2.1.1

Multiply

$- 1$ by

$1$ .

$3 \cdot - 4 - 3 \cdot 13 + 19 (- 1 - 4 \cdot - 1)$

Step 7.2.4.2.1.2

Multiply

$- 4$ by

$- 1$ .

$3 \cdot - 4 - 3 \cdot 13 + 19 (- 1 + 4)$

Step 7.2.4.2.2

Add

$- 1$ and

$4$ .

$3 \cdot - 4 - 3 \cdot 13 + 19 \cdot 3$

Step 7.2.5

Simplify the determinant.

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Step 7.2.5.1

Simplify each term.

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Step 7.2.5.1.1

Multiply

$3$ by

$- 4$ .

$- 12 - 3 \cdot 13 + 19 \cdot 3$

Step 7.2.5.1.2

Multiply

$- 3$ by

$13$ .

$- 12 - 39 + 19 \cdot 3$

Step 7.2.5.1.3

Multiply

$19$ by

$3$ .

$- 12 - 39 + 57$

Step 7.2.5.2

Subtract

$39$ from

$- 12$ .

$- 51 + 57$

Step 7.2.5.3

Add

$- 51$ and

$57$ .

$6$

$D_{z} = 6$

Step 7.3

Use the formula to solve for

$z$ .

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

Step 7.4

Substitute

$3$ for

$D$ and

$6$ for

$D_{z}$ in the formula.

$z = \frac{6}{3}$

Step 7.5

Divide

$6$ by

$3$ .

$z = 2$

Step 8

List the solution to the system of equations.

$x = \frac{2}{3}$

$y = \frac{11}{3}$

$z = 2$

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