Finite Math Examples

$[\begin{array}{c} b & a & c \\ e & d & f \\ h & g & i \end{array}]$

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 row $1$ by its cofactor and add.

Tap for more steps...

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.

$| \begin{array}{c} d & f \\ g & i \end{array} |$

Step 1.4

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

$b | \begin{array}{c} d & f \\ g & i \end{array} |$

Step 1.5

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

$| \begin{array}{c} e & f \\ h & i \end{array} |$

Step 1.6

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

$- a | \begin{array}{c} e & f \\ h & i \end{array} |$

Step 1.7

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

$| \begin{array}{c} e & d \\ h & g \end{array} |$

Step 1.8

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

$c | \begin{array}{c} e & d \\ h & g \end{array} |$

Step 1.9

Add the terms together.

$b | \begin{array}{c} d & f \\ g & i \end{array} | - a | \begin{array}{c} e & f \\ h & i \end{array} | + c | \begin{array}{c} e & d \\ h & g \end{array} |$

Step 2

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

$b (d i - g f) - a | \begin{array}{c} e & f \\ h & i \end{array} | + c | \begin{array}{c} e & d \\ h & g \end{array} |$

Step 3

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

$b (d i - g f) - a (e i - h f) + c | \begin{array}{c} e & d \\ h & g \end{array} |$

Step 4

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

$b (d i - g f) - a (e i - h f) + c (e g - h d)$

Step 5

Simplify each term.

Tap for more steps...

Step 5.1

Apply the distributive property.

$b (d i) + b (- g f) - a (e i - h f) + c (e g - h d)$

Step 5.2

Rewrite using the commutative property of multiplication.

$i b d + b (- g f) - a (e i - h f) + c (e g - h d)$

Step 5.3

Rewrite using the commutative property of multiplication.

$i b d - b g f - a (e i - h f) + c (e g - h d)$

Step 5.4

Apply the distributive property.

$i b d - b g f - a (e i) - a (- h f) + c (e g - h d)$

Step 5.5

Multiply $- a (- h f)$ .

Tap for more steps...

Step 5.5.1

Multiply $- 1$ by $- 1$ .

$i b d - b g f - a e i + 1 a (h f) + c (e g - h d)$

Step 5.5.2

Multiply $a$ by $1$ .

$i b d - b g f - a e i + a (h f) + c (e g - h d)$

Step 5.6

Apply the distributive property.

$i b d - b g f - a e i + a h f + c (e g) + c (- h d)$

Step 5.7

Rewrite using the commutative property of multiplication.

$i b d - b g f - a e i + a h f + e c g + c (- h d)$

Step 5.8

Rewrite using the commutative property of multiplication.

$i b d - b g f - a e i + a h f + e c g - c h d$