Linear Algebra Examples

$3 z + 3 x + 3 y = 19$ , $x + 3 = y$ , $z = y - 4 x + 1$

Step 1

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

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Step 1.1

Move $3 z$ .

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

$x + 3 = y$

$z = y - 4 x + 1$

Step 1.2

Subtract $y$ from both sides of the equation.

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

$x + 3 - y = 0$

$z = y - 4 x + 1$

Step 1.3

Subtract $3$ from both sides of the equation.

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

$x - y = - 3$

$z = y - 4 x + 1$

Step 1.4

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

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Step 1.4.1

Subtract $y$ from both sides of the equation.

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

$x - y = - 3$

$z - y = - 4 x + 1$

Step 1.4.2

Add $4 x$ to both sides of the equation.

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

$x - y = - 3$

$z - y + 4 x = 1$

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

$x - y = - 3$

$z - y + 4 x = 1$

Step 1.5

Move $z$ .

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

$x - y = - 3$

$- y + 4 x + z = 1$

Step 1.6

Reorder $- y$ and $4 x$ .

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

$x - y = - 3$

$4 x - y + z = 1$

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

$x - y = - 3$

$4 x - y + z = 1$

Step 2

Represent the system of equations in matrix format.

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

Step 3

Find the determinant of the coefficient matrix $[\begin{array}{c} 3 & 3 & 3 \\ 1 & - 1 & 0 \\ 4 & - 1 & 1 \end{array}]$ .

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Step 3.1

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

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

Step 3.2

Choose the row or column with the most $0$ 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 3.2.1

Consider the corresponding sign chart.

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

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

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

Step 3.2.4

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

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

Step 3.2.5

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 3.2.6

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

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

Step 3.2.7

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

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

Step 3.2.8

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

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

Step 3.2.9

Add the terms together.

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

Step 3.3

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

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

Step 3.4

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

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Step 3.4.1

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

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

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 $3$ by $1$ .

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

Step 3.4.2.1.2

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

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Step 3.4.2.1.2.1

Multiply $- 1$ by $3$ .

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

Step 3.4.2.1.2.2

Multiply $- 1$ by $- 3$ .

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

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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