Finite Math Examples

$x + y + z = 7$ , $x - y + 2 z = 7$ , $2 x + 3 z = 15$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 1 & 1 & 1 \\ 1 & - 1 & 2 \\ 2 & 0 & 3 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 7 \\ 7 \\ 15 \end{array}]$

Step 2

Find the determinant of the coefficient matrix $[\begin{array}{c} 1 & 1 & 1 \\ 1 & - 1 & 2 \\ 2 & 0 & 3 \end{array}]$ .

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Step 2.1

Write $[\begin{array}{c} 1 & 1 & 1 \\ 1 & - 1 & 2 \\ 2 & 0 & 3 \end{array}]$ in determinant notation.

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

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

The minor for $a_{32}$ is the determinant with row $3$ and column $2$ deleted.

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

Step 2.2.8

Multiply element $a_{32}$ by its cofactor.

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

Step 2.2.9

Add the terms together.

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

Step 2.3

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

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

Step 2.4

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

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Step 2.4.1

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

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

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 $3$ by $1$ .

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

Step 2.4.2.1.2

Multiply $- 2$ by $2$ .

$- 1 (3 - 4) - 1 | \begin{array}{c} 1 & 1 \\ 2 & 3 \end{array} | + 0$

Step 2.4.2.2

Subtract $4$ from $3$ .

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

Step 2.5

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

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Step 2.5.1

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

$- 1 \cdot - 1 - 1 (1 \cdot 3 - 2 \cdot 1) + 0$

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 $3$ by $1$ .

$- 1 \cdot - 1 - 1 (3 - 2 \cdot 1) + 0$

Step 2.5.2.1.2

Multiply $- 2$ by $1$ .

$- 1 \cdot - 1 - 1 (3 - 2) + 0$

Step 2.5.2.2

Subtract $2$ from $3$ .

$- 1 \cdot - 1 - 1 \cdot 1 + 0$

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 $- 1$ by $- 1$ .

$1 - 1 \cdot 1 + 0$

Step 2.6.1.2

Multiply $- 1$ by $1$ .

$1 - 1 + 0$

Step 2.6.2

Subtract $1$ from $1$ .

$0 + 0$

Step 2.6.3

Add $0$ and $0$ .

$0$

$D = 0$

Step 3

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