Finite Math Examples

$t = [\begin{array}{c} x & 0.4 & 0.3 \\ 0 & y & 0.5 \\ 0.2 & 0.5 & Z \end{array}]$

Step 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 1.1

Consider the corresponding sign chart.

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

Step 1.2

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

Step 1.3

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

$| \begin{array}{c} y & 0.5 \\ 0.5 & Z \end{array} |$

Step 1.4

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

$x | \begin{array}{c} y & 0.5 \\ 0.5 & Z \end{array} |$

Step 1.5

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

$| \begin{array}{c} 0.4 & 0.3 \\ 0.5 & Z \end{array} |$

Step 1.6

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

$0 | \begin{array}{c} 0.4 & 0.3 \\ 0.5 & Z \end{array} |$

Step 1.7

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

$| \begin{array}{c} 0.4 & 0.3 \\ y & 0.5 \end{array} |$

Step 1.8

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

$0.2 | \begin{array}{c} 0.4 & 0.3 \\ y & 0.5 \end{array} |$

Step 1.9

Add the terms together.

$x | \begin{array}{c} y & 0.5 \\ 0.5 & Z \end{array} | + 0 | \begin{array}{c} 0.4 & 0.3 \\ 0.5 & Z \end{array} | + 0.2 | \begin{array}{c} 0.4 & 0.3 \\ y & 0.5 \end{array} |$

Step 2

Multiply $0$ by $| \begin{array}{c} 0.4 & 0.3 \\ 0.5 & Z \end{array} |$ .

$x | \begin{array}{c} y & 0.5 \\ 0.5 & Z \end{array} | + 0 + 0.2 | \begin{array}{c} 0.4 & 0.3 \\ y & 0.5 \end{array} |$

Step 3

Evaluate $| \begin{array}{c} y & 0.5 \\ 0.5 & Z \end{array} |$ .

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Step 3.1

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

$x (y Z - 0.5 \cdot 0.5) + 0 + 0.2 | \begin{array}{c} 0.4 & 0.3 \\ y & 0.5 \end{array} |$

Step 3.2

Multiply $- 0.5$ by $0.5$ .

$x (y Z - 0.25) + 0 + 0.2 | \begin{array}{c} 0.4 & 0.3 \\ y & 0.5 \end{array} |$

Step 4

Evaluate $| \begin{array}{c} 0.4 & 0.3 \\ y & 0.5 \end{array} |$ .

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Step 4.1

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

$x (y Z - 0.25) + 0 + 0.2 (0.4 \cdot 0.5 - y \cdot 0.3)$

Step 4.2

Simplify each term.

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Step 4.2.1

Multiply $0.4$ by $0.5$ .

$x (y Z - 0.25) + 0 + 0.2 (0.2 - y \cdot 0.3)$

Step 4.2.2

Multiply $0.3$ by $- 1$ .

$x (y Z - 0.25) + 0 + 0.2 (0.2 - 0.3 y)$

Step 5

Simplify the determinant.

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Step 5.1

Add $x (y Z - 0.25)$ and $0$ .

$x (y Z - 0.25) + 0.2 (0.2 - 0.3 y)$

Step 5.2

Simplify each term.

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Step 5.2.1

Apply the distributive property.

$x (y Z) + x \cdot - 0.25 + 0.2 (0.2 - 0.3 y)$

Step 5.2.2

Move $- 0.25$ to the left of $x$ .

$x y Z - 0.25 \cdot x + 0.2 (0.2 - 0.3 y)$

Step 5.2.3

Apply the distributive property.

$x y Z - 0.25 x + 0.2 \cdot 0.2 + 0.2 (- 0.3 y)$

Step 5.2.4

Multiply $0.2$ by $0.2$ .

$x y Z - 0.25 x + 0.04 + 0.2 (- 0.3 y)$

Step 5.2.5

Multiply $- 0.3$ by $0.2$ .

$x y Z - 0.25 x + 0.04 - 0.06 y$