Finite Math Examples

$2 x + 5 y - 2 z = 14$ , $5 x - 6 y + 2 z = 0$ , $4 x - y + 3 z = - 7$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 2 & 5 & - 2 \\ 5 & - 6 & 2 \\ 4 & - 1 & 3 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 14 \\ 0 \\ - 7 \end{array}]$

Step 2

Find the determinant of the coefficient matrix $[\begin{array}{c} 2 & 5 & - 2 \\ 5 & - 6 & 2 \\ 4 & - 1 & 3 \end{array}]$ .

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Step 2.1

Write $[\begin{array}{c} 2 & 5 & - 2 \\ 5 & - 6 & 2 \\ 4 & - 1 & 3 \end{array}]$ in determinant notation.

$| \begin{array}{c} 2 & 5 & - 2 \\ 5 & - 6 & 2 \\ 4 & - 1 & 3 \end{array} |$

Step 2.2

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 2.2.1

Consider the corresponding sign chart.

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

Step 2.2.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 2.2.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

$| \begin{array}{c} - 6 & 2 \\ - 1 & 3 \end{array} |$

Step 2.2.4

Multiply element $a_{11}$ by its cofactor.

$2 | \begin{array}{c} - 6 & 2 \\ - 1 & 3 \end{array} |$

Step 2.2.5

The minor for $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

$| \begin{array}{c} 5 & 2 \\ 4 & 3 \end{array} |$

Step 2.2.6

Multiply element $a_{12}$ by its cofactor.

$- 5 | \begin{array}{c} 5 & 2 \\ 4 & 3 \end{array} |$

Step 2.2.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

$| \begin{array}{c} 5 & - 6 \\ 4 & - 1 \end{array} |$

Step 2.2.8

Multiply element $a_{13}$ by its cofactor.

$- 2 | \begin{array}{c} 5 & - 6 \\ 4 & - 1 \end{array} |$

Step 2.2.9

Add the terms together.

$2 | \begin{array}{c} - 6 & 2 \\ - 1 & 3 \end{array} | - 5 | \begin{array}{c} 5 & 2 \\ 4 & 3 \end{array} | - 2 | \begin{array}{c} 5 & - 6 \\ 4 & - 1 \end{array} |$

Step 2.3

Evaluate $| \begin{array}{c} - 6 & 2 \\ - 1 & 3 \end{array} |$ .

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Step 2.3.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$2 (- 6 \cdot 3 - (- 1 \cdot 2)) - 5 | \begin{array}{c} 5 & 2 \\ 4 & 3 \end{array} | - 2 | \begin{array}{c} 5 & - 6 \\ 4 & - 1 \end{array} |$

Step 2.3.2

Simplify the determinant.

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Step 2.3.2.1

Simplify each term.

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Step 2.3.2.1.1

Multiply $- 6$ by $3$ .

$2 (- 18 - (- 1 \cdot 2)) - 5 | \begin{array}{c} 5 & 2 \\ 4 & 3 \end{array} | - 2 | \begin{array}{c} 5 & - 6 \\ 4 & - 1 \end{array} |$

Step 2.3.2.1.2

Multiply $- (- 1 \cdot 2)$ .

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Step 2.3.2.1.2.1

Multiply $- 1$ by $2$ .

$2 (- 18 - - 2) - 5 | \begin{array}{c} 5 & 2 \\ 4 & 3 \end{array} | - 2 | \begin{array}{c} 5 & - 6 \\ 4 & - 1 \end{array} |$

Step 2.3.2.1.2.2

Multiply $- 1$ by $- 2$ .

$2 (- 18 + 2) - 5 | \begin{array}{c} 5 & 2 \\ 4 & 3 \end{array} | - 2 | \begin{array}{c} 5 & - 6 \\ 4 & - 1 \end{array} |$

Step 2.3.2.2

Add $- 18$ and $2$ .

$2 \cdot - 16 - 5 | \begin{array}{c} 5 & 2 \\ 4 & 3 \end{array} | - 2 | \begin{array}{c} 5 & - 6 \\ 4 & - 1 \end{array} |$

Step 2.4

Evaluate $| \begin{array}{c} 5 & 2 \\ 4 & 3 \end{array} |$ .

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Step 2.4.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$2 \cdot - 16 - 5 (5 \cdot 3 - 4 \cdot 2) - 2 | \begin{array}{c} 5 & - 6 \\ 4 & - 1 \end{array} |$

Step 2.4.2

Simplify the determinant.

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Step 2.4.2.1

Simplify each term.

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Step 2.4.2.1.1

Multiply $5$ by $3$ .

$2 \cdot - 16 - 5 (15 - 4 \cdot 2) - 2 | \begin{array}{c} 5 & - 6 \\ 4 & - 1 \end{array} |$

Step 2.4.2.1.2

Multiply $- 4$ by $2$ .

$2 \cdot - 16 - 5 (15 - 8) - 2 | \begin{array}{c} 5 & - 6 \\ 4 & - 1 \end{array} |$

Step 2.4.2.2

Subtract $8$ from $15$ .

$2 \cdot - 16 - 5 \cdot 7 - 2 | \begin{array}{c} 5 & - 6 \\ 4 & - 1 \end{array} |$

Step 2.5

Evaluate $| \begin{array}{c} 5 & - 6 \\ 4 & - 1 \end{array} |$ .

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Step 2.5.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$2 \cdot - 16 - 5 \cdot 7 - 2 (5 \cdot - 1 - 4 \cdot - 6)$

Step 2.5.2

Simplify the determinant.

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Step 2.5.2.1

Simplify each term.

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Step 2.5.2.1.1

Multiply $5$ by $- 1$ .

$2 \cdot - 16 - 5 \cdot 7 - 2 (- 5 - 4 \cdot - 6)$

Step 2.5.2.1.2

Multiply $- 4$ by $- 6$ .

$2 \cdot - 16 - 5 \cdot 7 - 2 (- 5 + 24)$

Step 2.5.2.2

Add $- 5$ and $24$ .

$2 \cdot - 16 - 5 \cdot 7 - 2 \cdot 19$

Step 2.6

Simplify the determinant.

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Step 2.6.1

Simplify each term.

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Step 2.6.1.1

Multiply $2$ by $- 16$ .

$- 32 - 5 \cdot 7 - 2 \cdot 19$

Step 2.6.1.2

Multiply $- 5$ by $7$ .

$- 32 - 35 - 2 \cdot 19$

Step 2.6.1.3

Multiply $- 2$ by $19$ .

$- 32 - 35 - 38$

Step 2.6.2

Subtract $35$ from $- 32$ .

$- 67 - 38$

Step 2.6.3

Subtract $38$ from $- 67$ .

$- 105$

$D = - 105$

Step 3

Since the determinant is not $0$ , the system can be solved using Cramer's Rule.

Step 4

Find the value of $x$ by Cramer's Rule, which states that $x = \frac{D_{x}}{D}$ .

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Step 4.1

Replace column $1$ of the coefficient matrix that corresponds to the $x$ -coefficients of the system with $[\begin{array}{c} 14 \\ 0 \\ - 7 \end{array}]$ .

$| \begin{array}{c} 14 & 5 & - 2 \\ 0 & - 6 & 2 \\ - 7 & - 1 & 3 \end{array} |$

Step 4.2

Find the determinant.

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Step 4.2.1

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 column $1$ by its cofactor and add.

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Step 4.2.1.1

Consider the corresponding sign chart.

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

Step 4.2.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 4.2.1.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

$| \begin{array}{c} - 6 & 2 \\ - 1 & 3 \end{array} |$

Step 4.2.1.4

Multiply element $a_{11}$ by its cofactor.

$14 | \begin{array}{c} - 6 & 2 \\ - 1 & 3 \end{array} |$

Step 4.2.1.5

The minor for $a_{21}$ is the determinant with row $2$ and column $1$ deleted.

$| \begin{array}{c} 5 & - 2 \\ - 1 & 3 \end{array} |$

Step 4.2.1.6

Multiply element $a_{21}$ by its cofactor.

$0 | \begin{array}{c} 5 & - 2 \\ - 1 & 3 \end{array} |$

Step 4.2.1.7

The minor for $a_{31}$ is the determinant with row $3$ and column $1$ deleted.

$| \begin{array}{c} 5 & - 2 \\ - 6 & 2 \end{array} |$

Step 4.2.1.8

Multiply element $a_{31}$ by its cofactor.

$- 7 | \begin{array}{c} 5 & - 2 \\ - 6 & 2 \end{array} |$

Step 4.2.1.9

Add the terms together.

$14 | \begin{array}{c} - 6 & 2 \\ - 1 & 3 \end{array} | + 0 | \begin{array}{c} 5 & - 2 \\ - 1 & 3 \end{array} | - 7 | \begin{array}{c} 5 & - 2 \\ - 6 & 2 \end{array} |$

Step 4.2.2

Multiply $0$ by $| \begin{array}{c} 5 & - 2 \\ - 1 & 3 \end{array} |$ .

$14 | \begin{array}{c} - 6 & 2 \\ - 1 & 3 \end{array} | + 0 - 7 | \begin{array}{c} 5 & - 2 \\ - 6 & 2 \end{array} |$

Step 4.2.3

Evaluate $| \begin{array}{c} - 6 & 2 \\ - 1 & 3 \end{array} |$ .

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Step 4.2.3.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$14 (- 6 \cdot 3 - (- 1 \cdot 2)) + 0 - 7 | \begin{array}{c} 5 & - 2 \\ - 6 & 2 \end{array} |$

Step 4.2.3.2

Simplify the determinant.

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Step 4.2.3.2.1

Simplify each term.

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Step 4.2.3.2.1.1

Multiply $- 6$ by $3$ .

$14 (- 18 - (- 1 \cdot 2)) + 0 - 7 | \begin{array}{c} 5 & - 2 \\ - 6 & 2 \end{array} |$

Step 4.2.3.2.1.2

Multiply $- (- 1 \cdot 2)$ .

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Step 4.2.3.2.1.2.1

Multiply $- 1$ by $2$ .

$14 (- 18 - - 2) + 0 - 7 | \begin{array}{c} 5 & - 2 \\ - 6 & 2 \end{array} |$

Step 4.2.3.2.1.2.2

Multiply $- 1$ by $- 2$ .

$14 (- 18 + 2) + 0 - 7 | \begin{array}{c} 5 & - 2 \\ - 6 & 2 \end{array} |$

Step 4.2.3.2.2

Add $- 18$ and $2$ .

$14 \cdot - 16 + 0 - 7 | \begin{array}{c} 5 & - 2 \\ - 6 & 2 \end{array} |$

Step 4.2.4

Evaluate $| \begin{array}{c} 5 & - 2 \\ - 6 & 2 \end{array} |$ .

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Step 4.2.4.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$14 \cdot - 16 + 0 - 7 (5 \cdot 2 - (- 6 \cdot - 2))$

Step 4.2.4.2

Simplify the determinant.

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Step 4.2.4.2.1

Simplify each term.

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Step 4.2.4.2.1.1

Multiply $5$ by $2$ .

$14 \cdot - 16 + 0 - 7 (10 - (- 6 \cdot - 2))$

Step 4.2.4.2.1.2

Multiply $- (- 6 \cdot - 2)$ .

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Step 4.2.4.2.1.2.1

Multiply $- 6$ by $- 2$ .

$14 \cdot - 16 + 0 - 7 (10 - 1 \cdot 12)$

Step 4.2.4.2.1.2.2

Multiply $- 1$ by $12$ .

$14 \cdot - 16 + 0 - 7 (10 - 12)$

Step 4.2.4.2.2

Subtract $12$ from $10$ .

$14 \cdot - 16 + 0 - 7 \cdot - 2$

Step 4.2.5

Simplify the determinant.

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Step 4.2.5.1

Simplify each term.

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Step 4.2.5.1.1

Multiply $14$ by $- 16$ .

$- 224 + 0 - 7 \cdot - 2$

Step 4.2.5.1.2

Multiply $- 7$ by $- 2$ .

$- 224 + 0 + 14$

Step 4.2.5.2

Add $- 224$ and $0$ .

$- 224 + 14$

Step 4.2.5.3

Add $- 224$ and $14$ .

$- 210$

$D_{x} = - 210$

Step 4.3

Use the formula to solve for $x$ .

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

Step 4.4

Substitute $- 105$ for $D$ and $- 210$ for $D_{x}$ in the formula.

$x = \frac{- 210}{- 105}$

Step 4.5

Divide $- 210$ by $- 105$ .

$x = 2$

Step 5

Find the value of $y$ by Cramer's Rule, which states that $y = \frac{D_{y}}{D}$ .

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Step 5.1

Replace column $2$ of the coefficient matrix that corresponds to the $y$ -coefficients of the system with $[\begin{array}{c} 14 \\ 0 \\ - 7 \end{array}]$ .

$| \begin{array}{c} 2 & 14 & - 2 \\ 5 & 0 & 2 \\ 4 & - 7 & 3 \end{array} |$

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.

$| \begin{array}{c} 14 & - 2 \\ - 7 & 3 \end{array} |$

Step 5.2.1.4

Multiply element $a_{21}$ by its cofactor.

$- 5 | \begin{array}{c} 14 & - 2 \\ - 7 & 3 \end{array} |$

Step 5.2.1.5

The minor for $a_{22}$ is the determinant with row $2$ and column $2$ deleted.

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

Step 5.2.1.6

Multiply element $a_{22}$ by its cofactor.

$0 | \begin{array}{c} 2 & - 2 \\ 4 & 3 \end{array} |$

Step 5.2.1.7

The minor for $a_{23}$ is the determinant with row $2$ and column $3$ deleted.

$| \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} |$

Step 5.2.1.8

Multiply element $a_{23}$ by its cofactor.

$- 2 | \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} |$

Step 5.2.1.9

Add the terms together.

$- 5 | \begin{array}{c} 14 & - 2 \\ - 7 & 3 \end{array} | + 0 | \begin{array}{c} 2 & - 2 \\ 4 & 3 \end{array} | - 2 | \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} |$

Step 5.2.2

Multiply $0$ by $| \begin{array}{c} 2 & - 2 \\ 4 & 3 \end{array} |$ .

$- 5 | \begin{array}{c} 14 & - 2 \\ - 7 & 3 \end{array} | + 0 - 2 | \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} |$

Step 5.2.3

Evaluate $| \begin{array}{c} 14 & - 2 \\ - 7 & 3 \end{array} |$ .

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Step 5.2.3.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$- 5 (14 \cdot 3 - (- 7 \cdot - 2)) + 0 - 2 | \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} |$

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 $14$ by $3$ .

$- 5 (42 - (- 7 \cdot - 2)) + 0 - 2 | \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} |$

Step 5.2.3.2.1.2

Multiply $- (- 7 \cdot - 2)$ .

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Step 5.2.3.2.1.2.1

Multiply $- 7$ by $- 2$ .

$- 5 (42 - 1 \cdot 14) + 0 - 2 | \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} |$

Step 5.2.3.2.1.2.2

Multiply $- 1$ by $14$ .

$- 5 (42 - 14) + 0 - 2 | \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} |$

Step 5.2.3.2.2

Subtract $14$ from $42$ .

$- 5 \cdot 28 + 0 - 2 | \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} |$

Step 5.2.4

Evaluate $| \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} |$ .

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Step 5.2.4.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$- 5 \cdot 28 + 0 - 2 (2 \cdot - 7 - 4 \cdot 14)$

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 $2$ by $- 7$ .

$- 5 \cdot 28 + 0 - 2 (- 14 - 4 \cdot 14)$

Step 5.2.4.2.1.2

Multiply $- 4$ by $14$ .

$- 5 \cdot 28 + 0 - 2 (- 14 - 56)$

Step 5.2.4.2.2

Subtract $56$ from $- 14$ .

$- 5 \cdot 28 + 0 - 2 \cdot - 70$

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 $- 5$ by $28$ .

$- 140 + 0 - 2 \cdot - 70$

Step 5.2.5.1.2

Multiply $- 2$ by $- 70$ .

$- 140 + 0 + 140$

Step 5.2.5.2

Add $- 140$ and $0$ .

$- 140 + 140$

Step 5.2.5.3

Add $- 140$ and $140$ .

$0$

$D_{y} = 0$

Step 5.3

Use the formula to solve for $y$ .

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

Step 5.4

Substitute $- 105$ for $D$ and $0$ for $D_{y}$ in the formula.

$y = \frac{0}{- 105}$

Step 5.5

Divide $0$ by $- 105$ .

$y = 0$

Step 6

Find the value of $z$ by Cramer's Rule, which states that $z = \frac{D_{z}}{D}$ .

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Step 6.1

Replace column $3$ of the coefficient matrix that corresponds to the $z$ -coefficients of the system with $[\begin{array}{c} 14 \\ 0 \\ - 7 \end{array}]$ .

$| \begin{array}{c} 2 & 5 & 14 \\ 5 & - 6 & 0 \\ 4 & - 1 & - 7 \end{array} |$

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.

$| \begin{array}{c} 5 & 14 \\ - 1 & - 7 \end{array} |$

Step 6.2.1.4

Multiply element $a_{21}$ by its cofactor.

$- 5 | \begin{array}{c} 5 & 14 \\ - 1 & - 7 \end{array} |$

Step 6.2.1.5

The minor for $a_{22}$ is the determinant with row $2$ and column $2$ deleted.

$| \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} |$

Step 6.2.1.6

Multiply element $a_{22}$ by its cofactor.

$- 6 | \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} |$

Step 6.2.1.7

The minor for $a_{23}$ is the determinant with row $2$ and column $3$ deleted.

$| \begin{array}{c} 2 & 5 \\ 4 & - 1 \end{array} |$

Step 6.2.1.8

Multiply element $a_{23}$ by its cofactor.

$0 | \begin{array}{c} 2 & 5 \\ 4 & - 1 \end{array} |$

Step 6.2.1.9

Add the terms together.

$- 5 | \begin{array}{c} 5 & 14 \\ - 1 & - 7 \end{array} | - 6 | \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} | + 0 | \begin{array}{c} 2 & 5 \\ 4 & - 1 \end{array} |$

Step 6.2.2

Multiply $0$ by $| \begin{array}{c} 2 & 5 \\ 4 & - 1 \end{array} |$ .

$- 5 | \begin{array}{c} 5 & 14 \\ - 1 & - 7 \end{array} | - 6 | \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} | + 0$

Step 6.2.3

Evaluate $| \begin{array}{c} 5 & 14 \\ - 1 & - 7 \end{array} |$ .

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Step 6.2.3.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$- 5 (5 \cdot - 7 - (- 1 \cdot 14)) - 6 | \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} | + 0$

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 $5$ by $- 7$ .

$- 5 (- 35 - (- 1 \cdot 14)) - 6 | \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} | + 0$

Step 6.2.3.2.1.2

Multiply $- (- 1 \cdot 14)$ .

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Step 6.2.3.2.1.2.1

Multiply $- 1$ by $14$ .

$- 5 (- 35 - - 14) - 6 | \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} | + 0$

Step 6.2.3.2.1.2.2

Multiply $- 1$ by $- 14$ .

$- 5 (- 35 + 14) - 6 | \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} | + 0$

Step 6.2.3.2.2

Add $- 35$ and $14$ .

$- 5 \cdot - 21 - 6 | \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} | + 0$

Step 6.2.4

Evaluate $| \begin{array}{c} 2 & 14 \\ 4 & - 7 \end{array} |$ .

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Step 6.2.4.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$- 5 \cdot - 21 - 6 (2 \cdot - 7 - 4 \cdot 14) + 0$

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 $2$ by $- 7$ .

$- 5 \cdot - 21 - 6 (- 14 - 4 \cdot 14) + 0$

Step 6.2.4.2.1.2

Multiply $- 4$ by $14$ .

$- 5 \cdot - 21 - 6 (- 14 - 56) + 0$

Step 6.2.4.2.2

Subtract $56$ from $- 14$ .

$- 5 \cdot - 21 - 6 \cdot - 70 + 0$

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 $- 5$ by $- 21$ .

$105 - 6 \cdot - 70 + 0$

Step 6.2.5.1.2

Multiply $- 6$ by $- 70$ .

$105 + 420 + 0$

Step 6.2.5.2

Add $105$ and $420$ .

$525 + 0$

Step 6.2.5.3

Add $525$ and $0$ .

$525$

$D_{z} = 525$

Step 6.3

Use the formula to solve for $z$ .

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

Step 6.4

Substitute $- 105$ for $D$ and $525$ for $D_{z}$ in the formula.

$z = \frac{525}{- 105}$

Step 6.5

Divide $525$ by $- 105$ .

$z = - 5$

Step 7

List the solution to the system of equations.

$x = 2$

$y = 0$

$z = - 5$