Linear Algebra Examples

$2 x - 3 y + z = 4$ $y - 2 z + x - 5 = 0$ $3 - 2 x = 4 y - z$

Step 1

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

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Step 1.1

Add $5$ to both sides of the equation.

$2 x - 3 y + z = 4$

$y - 2 z + x = 5$

$3 - 2 x = 4 y - z$

Step 1.2

Move $- 2 z$ .

$2 x - 3 y + z = 4$

$y + x - 2 z = 5$

$3 - 2 x = 4 y - z$

Step 1.3

Reorder $y$ and $x$ .

$2 x - 3 y + z = 4$

$x + y - 2 z = 5$

$3 - 2 x = 4 y - z$

Step 1.4

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

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Step 1.4.1

Subtract $4 y$ from both sides of the equation.

$2 x - 3 y + z = 4$

$x + y - 2 z = 5$

$3 - 2 x - 4 y = - z$

Step 1.4.2

Add $z$ to both sides of the equation.

$2 x - 3 y + z = 4$

$x + y - 2 z = 5$

$3 - 2 x - 4 y + z = 0$

$2 x - 3 y + z = 4$

$x + y - 2 z = 5$

$3 - 2 x - 4 y + z = 0$

Step 1.5

Subtract $3$ from both sides of the equation.

$2 x - 3 y + z = 4$

$x + y - 2 z = 5$

$- 2 x - 4 y + z = - 3$

$2 x - 3 y + z = 4$

$x + y - 2 z = 5$

$- 2 x - 4 y + z = - 3$

Step 2

Represent the system of equations in matrix format.

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

Step 3

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

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Step 3.1

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

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

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

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

Step 3.2.4

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

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

Step 3.2.5

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

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

Step 3.2.6

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

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

Step 3.2.7

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

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

Step 3.2.8

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

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

Step 3.2.9

Add the terms together.

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

Simplify the determinant.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Multiply $1$ by $1$ .

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

Step 3.3.2.1.2

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

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Step 3.3.2.1.2.1

Multiply $- 4$ by $- 2$ .

$2 (1 - 1 \cdot 8) + 3 | \begin{array}{c} 1 & - 2 \\ - 2 & 1 \end{array} | + 1 | \begin{array}{c} 1 & 1 \\ - 2 & - 4 \end{array} |$

Step 3.3.2.1.2.2

Multiply $- 1$ by $8$ .

$2 (1 - 8) + 3 | \begin{array}{c} 1 & - 2 \\ - 2 & 1 \end{array} | + 1 | \begin{array}{c} 1 & 1 \\ - 2 & - 4 \end{array} |$

Step 3.3.2.2

Subtract $8$ from $1$ .

$2 \cdot - 7 + 3 | \begin{array}{c} 1 & - 2 \\ - 2 & 1 \end{array} | + 1 | \begin{array}{c} 1 & 1 \\ - 2 & - 4 \end{array} |$

Step 3.4

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

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Step 3.4.1

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

$2 \cdot - 7 + 3 (1 \cdot 1 - (- 2 \cdot - 2)) + 1 | \begin{array}{c} 1 & 1 \\ - 2 & - 4 \end{array} |$

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 $1$ by $1$ .

$2 \cdot - 7 + 3 (1 - (- 2 \cdot - 2)) + 1 | \begin{array}{c} 1 & 1 \\ - 2 & - 4 \end{array} |$

Step 3.4.2.1.2

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

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Step 3.4.2.1.2.1

Multiply $- 2$ by $- 2$ .

$2 \cdot - 7 + 3 (1 - 1 \cdot 4) + 1 | \begin{array}{c} 1 & 1 \\ - 2 & - 4 \end{array} |$

Step 3.4.2.1.2.2

Multiply $- 1$ by $4$ .

$2 \cdot - 7 + 3 (1 - 4) + 1 | \begin{array}{c} 1 & 1 \\ - 2 & - 4 \end{array} |$

Step 3.4.2.2

Subtract $4$ from $1$ .

$2 \cdot - 7 + 3 \cdot - 3 + 1 | \begin{array}{c} 1 & 1 \\ - 2 & - 4 \end{array} |$

Step 3.5

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

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Step 3.5.1

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

$2 \cdot - 7 + 3 \cdot - 3 + 1 (1 \cdot - 4 - (- 2 \cdot 1))$

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 $- 4$ by $1$ .

$2 \cdot - 7 + 3 \cdot - 3 + 1 (- 4 - (- 2 \cdot 1))$

Step 3.5.2.1.2

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

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Step 3.5.2.1.2.1

Multiply $- 2$ by $1$ .

$2 \cdot - 7 + 3 \cdot - 3 + 1 (- 4 - - 2)$

Step 3.5.2.1.2.2

Multiply $- 1$ by $- 2$ .

$2 \cdot - 7 + 3 \cdot - 3 + 1 (- 4 + 2)$

Step 3.5.2.2

Add $- 4$ and $2$ .

$2 \cdot - 7 + 3 \cdot - 3 + 1 \cdot - 2$

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 $2$ by $- 7$ .

$- 14 + 3 \cdot - 3 + 1 \cdot - 2$

Step 3.6.1.2

Multiply $3$ by $- 3$ .

$- 14 - 9 + 1 \cdot - 2$

Step 3.6.1.3

Multiply $- 2$ by $1$ .

$- 14 - 9 - 2$

Step 3.6.2

Subtract $9$ from $- 14$ .

$- 23 - 2$

Step 3.6.3

Subtract $2$ from $- 23$ .

$- 25$

$D = - 25$

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 $[\begin{array}{c} 4 \\ 5 \\ - 3 \end{array}]$ .

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

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

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

Step 5.2.1.4

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

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

Step 5.2.1.5

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

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

Step 5.2.1.6

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

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

Step 5.2.1.7

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

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

Step 5.2.1.8

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

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

Step 5.2.1.9

Add the terms together.

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

Step 5.2.2

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

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Step 5.2.2.1

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

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

Step 5.2.2.2

Simplify the determinant.

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Step 5.2.2.2.1

Simplify each term.

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Step 5.2.2.2.1.1

Multiply $1$ by $1$ .

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

Step 5.2.2.2.1.2

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

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Step 5.2.2.2.1.2.1

Multiply $- 4$ by $- 2$ .

$4 (1 - 1 \cdot 8) + 3 | \begin{array}{c} 5 & - 2 \\ - 3 & 1 \end{array} | + 1 | \begin{array}{c} 5 & 1 \\ - 3 & - 4 \end{array} |$

Step 5.2.2.2.1.2.2

Multiply $- 1$ by $8$ .

$4 (1 - 8) + 3 | \begin{array}{c} 5 & - 2 \\ - 3 & 1 \end{array} | + 1 | \begin{array}{c} 5 & 1 \\ - 3 & - 4 \end{array} |$

Step 5.2.2.2.2

Subtract $8$ from $1$ .

$4 \cdot - 7 + 3 | \begin{array}{c} 5 & - 2 \\ - 3 & 1 \end{array} | + 1 | \begin{array}{c} 5 & 1 \\ - 3 & - 4 \end{array} |$

Step 5.2.3

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

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Step 5.2.3.1

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

$4 \cdot - 7 + 3 (5 \cdot 1 - (- 3 \cdot - 2)) + 1 | \begin{array}{c} 5 & 1 \\ - 3 & - 4 \end{array} |$

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 $5$ by $1$ .

$4 \cdot - 7 + 3 (5 - (- 3 \cdot - 2)) + 1 | \begin{array}{c} 5 & 1 \\ - 3 & - 4 \end{array} |$

Step 5.2.3.2.1.2

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

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Step 5.2.3.2.1.2.1

Multiply $- 3$ by $- 2$ .

$4 \cdot - 7 + 3 (5 - 1 \cdot 6) + 1 | \begin{array}{c} 5 & 1 \\ - 3 & - 4 \end{array} |$

Step 5.2.3.2.1.2.2

Multiply $- 1$ by $6$ .

$4 \cdot - 7 + 3 (5 - 6) + 1 | \begin{array}{c} 5 & 1 \\ - 3 & - 4 \end{array} |$

Step 5.2.3.2.2

Subtract $6$ from $5$ .

$4 \cdot - 7 + 3 \cdot - 1 + 1 | \begin{array}{c} 5 & 1 \\ - 3 & - 4 \end{array} |$

Step 5.2.4

Evaluate $| \begin{array}{c} 5 & 1 \\ - 3 & - 4 \end{array} |$ .

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Step 5.2.4.1

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

$4 \cdot - 7 + 3 \cdot - 1 + 1 (5 \cdot - 4 - (- 3 \cdot 1))$

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 $5$ by $- 4$ .

$4 \cdot - 7 + 3 \cdot - 1 + 1 (- 20 - (- 3 \cdot 1))$

Step 5.2.4.2.1.2

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

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Step 5.2.4.2.1.2.1

Multiply $- 3$ by $1$ .

$4 \cdot - 7 + 3 \cdot - 1 + 1 (- 20 - - 3)$

Step 5.2.4.2.1.2.2

Multiply $- 1$ by $- 3$ .

$4 \cdot - 7 + 3 \cdot - 1 + 1 (- 20 + 3)$

Step 5.2.4.2.2

Add $- 20$ and $3$ .

$4 \cdot - 7 + 3 \cdot - 1 + 1 \cdot - 17$

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 $4$ by $- 7$ .

$- 28 + 3 \cdot - 1 + 1 \cdot - 17$

Step 5.2.5.1.2

Multiply $3$ by $- 1$ .

$- 28 - 3 + 1 \cdot - 17$

Step 5.2.5.1.3

Multiply $- 17$ by $1$ .

$- 28 - 3 - 17$

Step 5.2.5.2

Subtract $3$ from $- 28$ .

$- 31 - 17$

Step 5.2.5.3

Subtract $17$ from $- 31$ .

$- 48$

$D_{x} = - 48$

Step 5.3

Use the formula to solve for $x$ .

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

Step 5.4

Substitute $- 25$ for $D$ and $- 48$ for $D_{x}$ in the formula.

$x = \frac{- 48}{- 25}$

Step 5.5

Dividing two negative values results in a positive value.

$x = \frac{48}{25}$

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 $[\begin{array}{c} 4 \\ 5 \\ - 3 \end{array}]$ .

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

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

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

Step 6.2.1.4

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

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

Step 6.2.1.5

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

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

Step 6.2.1.6

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

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

Step 6.2.1.7

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

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

Step 6.2.1.8

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

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

Step 6.2.1.9

Add the terms together.

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

Step 6.2.2

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

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Step 6.2.2.1

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

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

Step 6.2.2.2

Simplify the determinant.

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Step 6.2.2.2.1

Simplify each term.

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Step 6.2.2.2.1.1

Multiply $5$ by $1$ .

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

Step 6.2.2.2.1.2

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

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Step 6.2.2.2.1.2.1

Multiply $- 3$ by $- 2$ .

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

Step 6.2.2.2.1.2.2

Multiply $- 1$ by $6$ .

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

Step 6.2.2.2.2

Subtract $6$ from $5$ .

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

Step 6.2.3

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

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Step 6.2.3.1

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

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

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 $1$ by $1$ .

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

Step 6.2.3.2.1.2

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

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Step 6.2.3.2.1.2.1

Multiply $- 2$ by $- 2$ .

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

Step 6.2.3.2.1.2.2

Multiply $- 1$ by $4$ .

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

Step 6.2.3.2.2

Subtract $4$ from $1$ .

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

Step 6.2.4

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

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Step 6.2.4.1

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

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

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 $1$ .

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

Step 6.2.4.2.1.2

Multiply $- (- 2 \cdot 5)$ .

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Step 6.2.4.2.1.2.1

Multiply $- 2$ by $5$ .

$2 \cdot - 1 - 4 \cdot - 3 + 1 (- 3 - - 10)$

Step 6.2.4.2.1.2.2

Multiply $- 1$ by $- 10$ .

$2 \cdot - 1 - 4 \cdot - 3 + 1 (- 3 + 10)$

Step 6.2.4.2.2

Add $- 3$ and $10$ .

$2 \cdot - 1 - 4 \cdot - 3 + 1 \cdot 7$

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 $2$ by $- 1$ .

$- 2 - 4 \cdot - 3 + 1 \cdot 7$

Step 6.2.5.1.2

Multiply $- 4$ by $- 3$ .

$- 2 + 12 + 1 \cdot 7$

Step 6.2.5.1.3

Multiply $7$ by $1$ .

$- 2 + 12 + 7$

Step 6.2.5.2

Add $- 2$ and $12$ .

$10 + 7$

Step 6.2.5.3

Add $10$ and $7$ .

$17$

$D_{y} = 17$

Step 6.3

Use the formula to solve for $y$ .

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

Step 6.4

Substitute $- 25$ for $D$ and $17$ for $D_{y}$ in the formula.

$y = \frac{17}{- 25}$

Step 6.5

Move the negative in front of the fraction.

$y = - \frac{17}{25}$

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 $[\begin{array}{c} 4 \\ 5 \\ - 3 \end{array}]$ .

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

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.

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

Step 7.2.1.4

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

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

Step 7.2.1.5

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

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

Step 7.2.1.6

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

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

Step 7.2.1.7

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

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

Step 7.2.1.8

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

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

Step 7.2.1.9

Add the terms together.

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

Step 7.2.2

Evaluate $| \begin{array}{c} 1 & 5 \\ - 4 & - 3 \end{array} |$ .

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Step 7.2.2.1

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

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

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 $- 3$ by $1$ .

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

Step 7.2.2.2.1.2

Multiply $- (- 4 \cdot 5)$ .

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Step 7.2.2.2.1.2.1

Multiply $- 4$ by $5$ .

$2 (- 3 - - 20) + 3 | \begin{array}{c} 1 & 5 \\ - 2 & - 3 \end{array} | + 4 | \begin{array}{c} 1 & 1 \\ - 2 & - 4 \end{array} |$

Step 7.2.2.2.1.2.2

Multiply $- 1$ by $- 20$ .

$2 (- 3 + 20) + 3 | \begin{array}{c} 1 & 5 \\ - 2 & - 3 \end{array} | + 4 | \begin{array}{c} 1 & 1 \\ - 2 & - 4 \end{array} |$

Step 7.2.2.2.2

Add $- 3$ and $20$ .

$2 \cdot 17 + 3 | \begin{array}{c} 1 & 5 \\ - 2 & - 3 \end{array} | + 4 | \begin{array}{c} 1 & 1 \\ - 2 & - 4 \end{array} |$

Step 7.2.3

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

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Step 7.2.3.1

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

$2 \cdot 17 + 3 (1 \cdot - 3 - (- 2 \cdot 5)) + 4 | \begin{array}{c} 1 & 1 \\ - 2 & - 4 \end{array} |$

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 $- 3$ by $1$ .

$2 \cdot 17 + 3 (- 3 - (- 2 \cdot 5)) + 4 | \begin{array}{c} 1 & 1 \\ - 2 & - 4 \end{array} |$

Step 7.2.3.2.1.2

Multiply $- (- 2 \cdot 5)$ .

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Step 7.2.3.2.1.2.1

Multiply $- 2$ by $5$ .

$2 \cdot 17 + 3 (- 3 - - 10) + 4 | \begin{array}{c} 1 & 1 \\ - 2 & - 4 \end{array} |$

Step 7.2.3.2.1.2.2

Multiply $- 1$ by $- 10$ .

$2 \cdot 17 + 3 (- 3 + 10) + 4 | \begin{array}{c} 1 & 1 \\ - 2 & - 4 \end{array} |$

Step 7.2.3.2.2

Add $- 3$ and $10$ .

$2 \cdot 17 + 3 \cdot 7 + 4 | \begin{array}{c} 1 & 1 \\ - 2 & - 4 \end{array} |$

Step 7.2.4

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

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Step 7.2.4.1

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

$2 \cdot 17 + 3 \cdot 7 + 4 (1 \cdot - 4 - (- 2 \cdot 1))$

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 $- 4$ by $1$ .

$2 \cdot 17 + 3 \cdot 7 + 4 (- 4 - (- 2 \cdot 1))$

Step 7.2.4.2.1.2

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

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Step 7.2.4.2.1.2.1

Multiply $- 2$ by $1$ .

$2 \cdot 17 + 3 \cdot 7 + 4 (- 4 - - 2)$

Step 7.2.4.2.1.2.2

Multiply $- 1$ by $- 2$ .

$2 \cdot 17 + 3 \cdot 7 + 4 (- 4 + 2)$

Step 7.2.4.2.2

Add $- 4$ and $2$ .

$2 \cdot 17 + 3 \cdot 7 + 4 \cdot - 2$

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 $2$ by $17$ .

$34 + 3 \cdot 7 + 4 \cdot - 2$

Step 7.2.5.1.2

Multiply $3$ by $7$ .

$34 + 21 + 4 \cdot - 2$

Step 7.2.5.1.3

Multiply $4$ by $- 2$ .

$34 + 21 - 8$

Step 7.2.5.2

Add $34$ and $21$ .

$55 - 8$

Step 7.2.5.3

Subtract $8$ from $55$ .

$47$

$D_{z} = 47$

Step 7.3

Use the formula to solve for $z$ .

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

Step 7.4

Substitute $- 25$ for $D$ and $47$ for $D_{z}$ in the formula.

$z = \frac{47}{- 25}$

Step 7.5

Move the negative in front of the fraction.

$z = - \frac{47}{25}$

Step 8

List the solution to the system of equations.

$x = \frac{48}{25}$

$y = - \frac{17}{25}$

$z = - \frac{47}{25}$