Finite Math Examples

$x - 2 y + 3 z = - 1$ , $- 2 x + y - z = 2$ , $3 x - 3 y + 2 z = - 1$

Step 1

Represent the system of equations in matrix format.

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

Step 2

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

Tap for more steps...

Step 2.1

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

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

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.

Tap for more steps...

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.

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

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

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

Step 2.2.8

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

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

Step 2.2.9

Add the terms together.

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

Step 2.3

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

Tap for more steps...

Step 2.3.1

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

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

Step 2.3.2

Simplify the determinant.

Tap for more steps...

Step 2.3.2.1

Simplify each term.

Tap for more steps...

Step 2.3.2.1.1

Multiply $2$ by $1$ .

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

Step 2.3.2.1.2

Multiply $- (- 3 \cdot - 1)$ .

Tap for more steps...

Step 2.3.2.1.2.1

Multiply $- 3$ by $- 1$ .

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

Step 2.3.2.1.2.2

Multiply $- 1$ by $3$ .

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

Step 2.3.2.2

Subtract $3$ from $2$ .

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

Step 2.4

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

Tap for more steps...

Step 2.4.1

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

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

Step 2.4.2

Simplify the determinant.

Tap for more steps...

Step 2.4.2.1

Simplify each term.

Tap for more steps...

Step 2.4.2.1.1

Multiply $- 2$ by $2$ .

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

Step 2.4.2.1.2

Multiply $- 3$ by $- 1$ .

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

Step 2.4.2.2

Add $- 4$ and $3$ .

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

Step 2.5

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

Tap for more steps...

Step 2.5.1

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

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

Step 2.5.2

Simplify the determinant.

Tap for more steps...

Step 2.5.2.1

Simplify each term.

Tap for more steps...

Step 2.5.2.1.1

Multiply $- 2$ by $- 3$ .

$1 \cdot - 1 + 2 \cdot - 1 + 3 (6 - 3 \cdot 1)$

Step 2.5.2.1.2

Multiply $- 3$ by $1$ .

$1 \cdot - 1 + 2 \cdot - 1 + 3 (6 - 3)$

Step 2.5.2.2

Subtract $3$ from $6$ .

$1 \cdot - 1 + 2 \cdot - 1 + 3 \cdot 3$

Step 2.6

Simplify the determinant.

Tap for more steps...

Step 2.6.1

Simplify each term.

Tap for more steps...

Step 2.6.1.1

Multiply $- 1$ by $1$ .

$- 1 + 2 \cdot - 1 + 3 \cdot 3$

Step 2.6.1.2

Multiply $2$ by $- 1$ .

$- 1 - 2 + 3 \cdot 3$

Step 2.6.1.3

Multiply $3$ by $3$ .

$- 1 - 2 + 9$

Step 2.6.2

Subtract $2$ from $- 1$ .

$- 3 + 9$

Step 2.6.3

Add $- 3$ and $9$ .

$6$

$D = 6$

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

Tap for more steps...

Step 4.1

Replace column $1$ of the coefficient matrix that corresponds to the $x$ -coefficients of the system with $[\begin{array}{c} - 1 \\ 2 \\ - 1 \end{array}]$ .

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

Step 4.2

Find the determinant.

Tap for more steps...

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 row $1$ by its cofactor and add.

Tap for more steps...

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.

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

Step 4.2.1.4

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

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

Step 4.2.1.5

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

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

Step 4.2.1.6

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

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

Step 4.2.1.7

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

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

Step 4.2.1.8

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

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

Step 4.2.1.9

Add the terms together.

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

Step 4.2.2

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

Tap for more steps...

Step 4.2.2.1

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

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

Step 4.2.2.2

Simplify the determinant.

Tap for more steps...

Step 4.2.2.2.1

Simplify each term.

Tap for more steps...

Step 4.2.2.2.1.1

Multiply $2$ by $1$ .

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

Step 4.2.2.2.1.2

Multiply $- (- 3 \cdot - 1)$ .

Tap for more steps...

Step 4.2.2.2.1.2.1

Multiply $- 3$ by $- 1$ .

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

Step 4.2.2.2.1.2.2

Multiply $- 1$ by $3$ .

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

Step 4.2.2.2.2

Subtract $3$ from $2$ .

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

Step 4.2.3

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

Tap for more steps...

Step 4.2.3.1

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

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

Step 4.2.3.2

Simplify the determinant.

Tap for more steps...

Step 4.2.3.2.1

Simplify each term.

Tap for more steps...

Step 4.2.3.2.1.1

Multiply $2$ by $2$ .

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

Step 4.2.3.2.1.2

Multiply $- - - 1$ .

Tap for more steps...

Step 4.2.3.2.1.2.1

Multiply $- 1$ by $- 1$ .

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

Step 4.2.3.2.1.2.2

Multiply $- 1$ by $1$ .

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

Step 4.2.3.2.2

Subtract $1$ from $4$ .

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

Step 4.2.4

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

Tap for more steps...

Step 4.2.4.1

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

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

Step 4.2.4.2

Simplify the determinant.

Tap for more steps...

Step 4.2.4.2.1

Simplify each term.

Tap for more steps...

Step 4.2.4.2.1.1

Multiply $2$ by $- 3$ .

$- 1 \cdot - 1 + 2 \cdot 3 + 3 (- 6 - (- 1 \cdot 1))$

Step 4.2.4.2.1.2

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

Tap for more steps...

Step 4.2.4.2.1.2.1

Multiply $- 1$ by $1$ .

$- 1 \cdot - 1 + 2 \cdot 3 + 3 (- 6 - - 1)$

Step 4.2.4.2.1.2.2

Multiply $- 1$ by $- 1$ .

$- 1 \cdot - 1 + 2 \cdot 3 + 3 (- 6 + 1)$

Step 4.2.4.2.2

Add $- 6$ and $1$ .

$- 1 \cdot - 1 + 2 \cdot 3 + 3 \cdot - 5$

Step 4.2.5

Simplify the determinant.

Tap for more steps...

Step 4.2.5.1

Simplify each term.

Tap for more steps...

Step 4.2.5.1.1

Multiply $- 1$ by $- 1$ .

$1 + 2 \cdot 3 + 3 \cdot - 5$

Step 4.2.5.1.2

Multiply $2$ by $3$ .

$1 + 6 + 3 \cdot - 5$

Step 4.2.5.1.3

Multiply $3$ by $- 5$ .

$1 + 6 - 15$

Step 4.2.5.2

Add $1$ and $6$ .

$7 - 15$

Step 4.2.5.3

Subtract $15$ from $7$ .

$- 8$

$D_{x} = - 8$

Step 4.3

Use the formula to solve for $x$ .

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

Step 4.4

Substitute $6$ for $D$ and $- 8$ for $D_{x}$ in the formula.

$x = \frac{- 8}{6}$

Step 4.5

Cancel the common factor of $- 8$ and $6$ .

Tap for more steps...

Step 4.5.1

Factor $2$ out of $- 8$ .

$x = \frac{2 (- 4)}{6}$

Step 4.5.2

Cancel the common factors.

Tap for more steps...

Step 4.5.2.1

Factor $2$ out of $6$ .

$x = \frac{2 \cdot - 4}{2 \cdot 3}$

Step 4.5.2.2

Cancel the common factor.

$x = \frac{2 \cdot - 4}{2 \cdot 3}$

Step 4.5.2.3

Rewrite the expression.

$x = \frac{- 4}{3}$

Step 4.6

Move the negative in front of the fraction.

$x = - \frac{4}{3}$

Step 5

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

Tap for more steps...

Step 5.1

Replace column $2$ of the coefficient matrix that corresponds to the $y$ -coefficients of the system with $[\begin{array}{c} - 1 \\ 2 \\ - 1 \end{array}]$ .

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

Step 5.2

Find the determinant.

Tap for more steps...

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 $1$ by its cofactor and add.

Tap for more steps...

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

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

Step 5.2.1.4

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

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

Step 5.2.1.5

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

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

Step 5.2.1.6

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

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

Step 5.2.1.7

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

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

Step 5.2.1.8

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

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

Step 5.2.1.9

Add the terms together.

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

Step 5.2.2

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

Tap for more steps...

Step 5.2.2.1

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

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

Step 5.2.2.2

Simplify the determinant.

Tap for more steps...

Step 5.2.2.2.1

Simplify each term.

Tap for more steps...

Step 5.2.2.2.1.1

Multiply $2$ by $2$ .

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

Step 5.2.2.2.1.2

Multiply $- - - 1$ .

Tap for more steps...

Step 5.2.2.2.1.2.1

Multiply $- 1$ by $- 1$ .

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

Step 5.2.2.2.1.2.2

Multiply $- 1$ by $1$ .

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

Step 5.2.2.2.2

Subtract $1$ from $4$ .

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

Step 5.2.3

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

Tap for more steps...

Step 5.2.3.1

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

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

Step 5.2.3.2

Simplify the determinant.

Tap for more steps...

Step 5.2.3.2.1

Simplify each term.

Tap for more steps...

Step 5.2.3.2.1.1

Multiply $- 2$ by $2$ .

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

Step 5.2.3.2.1.2

Multiply $- 3$ by $- 1$ .

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

Step 5.2.3.2.2

Add $- 4$ and $3$ .

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

Step 5.2.4

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

Tap for more steps...

Step 5.2.4.1

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

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

Step 5.2.4.2

Simplify the determinant.

Tap for more steps...

Step 5.2.4.2.1

Simplify each term.

Tap for more steps...

Step 5.2.4.2.1.1

Multiply $- 2$ by $- 1$ .

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

Step 5.2.4.2.1.2

Multiply $- 3$ by $2$ .

$1 \cdot 3 + 1 \cdot - 1 + 3 (2 - 6)$

Step 5.2.4.2.2

Subtract $6$ from $2$ .

$1 \cdot 3 + 1 \cdot - 1 + 3 \cdot - 4$

Step 5.2.5

Simplify the determinant.

Tap for more steps...

Step 5.2.5.1

Simplify each term.

Tap for more steps...

Step 5.2.5.1.1

Multiply $3$ by $1$ .

$3 + 1 \cdot - 1 + 3 \cdot - 4$

Step 5.2.5.1.2

Multiply $- 1$ by $1$ .

$3 - 1 + 3 \cdot - 4$

Step 5.2.5.1.3

Multiply $3$ by $- 4$ .

$3 - 1 - 12$

Step 5.2.5.2

Subtract $1$ from $3$ .

$2 - 12$

Step 5.2.5.3

Subtract $12$ from $2$ .

$- 10$

$D_{y} = - 10$

Step 5.3

Use the formula to solve for $y$ .

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

Step 5.4

Substitute $6$ for $D$ and $- 10$ for $D_{y}$ in the formula.

$y = \frac{- 10}{6}$

Step 5.5

Cancel the common factor of $- 10$ and $6$ .

Tap for more steps...

Step 5.5.1

Factor $2$ out of $- 10$ .

$y = \frac{2 (- 5)}{6}$

Step 5.5.2

Cancel the common factors.

Tap for more steps...

Step 5.5.2.1

Factor $2$ out of $6$ .

$y = \frac{2 \cdot - 5}{2 \cdot 3}$

Step 5.5.2.2

Cancel the common factor.

$y = \frac{2 \cdot - 5}{2 \cdot 3}$

Step 5.5.2.3

Rewrite the expression.

$y = \frac{- 5}{3}$

Step 5.6

Move the negative in front of the fraction.

$y = - \frac{5}{3}$

Step 6

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

Tap for more steps...

Step 6.1

Replace column $3$ of the coefficient matrix that corresponds to the $z$ -coefficients of the system with $[\begin{array}{c} - 1 \\ 2 \\ - 1 \end{array}]$ .

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

Step 6.2

Find the determinant.

Tap for more steps...

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 $1$ by its cofactor and add.

Tap for more steps...

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

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

Step 6.2.1.4

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

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

Step 6.2.1.5

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

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

Step 6.2.1.6

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

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

Step 6.2.1.7

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

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

Step 6.2.1.8

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

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

Step 6.2.1.9

Add the terms together.

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

Step 6.2.2

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

Tap for more steps...

Step 6.2.2.1

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

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

Step 6.2.2.2

Simplify the determinant.

Tap for more steps...

Step 6.2.2.2.1

Simplify each term.

Tap for more steps...

Step 6.2.2.2.1.1

Multiply $- 1$ by $1$ .

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

Step 6.2.2.2.1.2

Multiply $- (- 3 \cdot 2)$ .

Tap for more steps...

Step 6.2.2.2.1.2.1

Multiply $- 3$ by $2$ .

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

Step 6.2.2.2.1.2.2

Multiply $- 1$ by $- 6$ .

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

Step 6.2.2.2.2

Add $- 1$ and $6$ .

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

Step 6.2.3

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

Tap for more steps...

Step 6.2.3.1

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

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

Step 6.2.3.2

Simplify the determinant.

Tap for more steps...

Step 6.2.3.2.1

Simplify each term.

Tap for more steps...

Step 6.2.3.2.1.1

Multiply $- 2$ by $- 1$ .

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

Step 6.2.3.2.1.2

Multiply $- 3$ by $2$ .

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

Step 6.2.3.2.2

Subtract $6$ from $2$ .

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

Step 6.2.4

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

Tap for more steps...

Step 6.2.4.1

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

$1 \cdot 5 + 2 \cdot - 4 - 1 (- 2 \cdot - 3 - 3 \cdot 1)$

Step 6.2.4.2

Simplify the determinant.

Tap for more steps...

Step 6.2.4.2.1

Simplify each term.

Tap for more steps...

Step 6.2.4.2.1.1

Multiply $- 2$ by $- 3$ .

$1 \cdot 5 + 2 \cdot - 4 - 1 (6 - 3 \cdot 1)$

Step 6.2.4.2.1.2

Multiply $- 3$ by $1$ .

$1 \cdot 5 + 2 \cdot - 4 - 1 (6 - 3)$

Step 6.2.4.2.2

Subtract $3$ from $6$ .

$1 \cdot 5 + 2 \cdot - 4 - 1 \cdot 3$

Step 6.2.5

Simplify the determinant.

Tap for more steps...

Step 6.2.5.1

Simplify each term.

Tap for more steps...

Step 6.2.5.1.1

Multiply $5$ by $1$ .

$5 + 2 \cdot - 4 - 1 \cdot 3$

Step 6.2.5.1.2

Multiply $2$ by $- 4$ .

$5 - 8 - 1 \cdot 3$

Step 6.2.5.1.3

Multiply $- 1$ by $3$ .

$5 - 8 - 3$

Step 6.2.5.2

Subtract $8$ from $5$ .

$- 3 - 3$

Step 6.2.5.3

Subtract $3$ from $- 3$ .

$- 6$

$D_{z} = - 6$

Step 6.3

Use the formula to solve for $z$ .

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

Step 6.4

Substitute $6$ for $D$ and $- 6$ for $D_{z}$ in the formula.

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

Step 6.5

Divide $- 6$ by $6$ .

$z = - 1$

Step 7

List the solution to the system of equations.

$x = - \frac{4}{3}$

$y = - \frac{5}{3}$

$z = - 1$