Linear Algebra Examples

$[\begin{array}{c} 1 \\ - 7 \\ 1 \end{array}] \times [\begin{array}{c} 5 \\ 2 \\ 4 \end{array}]$

Step 1

The cross product of two vectors $a⃗$ and $b⃗$ can be written as a determinant with the standard unit vectors from $ℝ^{3}$ and the elements of the given vectors.

$a⃗ \times b⃗ = | \begin{array}{c} î & ĵ & k̂ \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{array} |$

Step 2

Set up the determinant with the given values.

$| \begin{array}{c} î & ĵ & k̂ \\ 1 & - 7 & 1 \\ 5 & 2 & 4 \end{array} |$

Step 3

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 3.1

Consider the corresponding sign chart.

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

Step 3.2

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

Step 3.3

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

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

Step 3.4

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

$| \begin{array}{c} - 7 & 1 \\ 2 & 4 \end{array} | î$

Step 3.5

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

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

Step 3.6

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

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

Step 3.7

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

$| \begin{array}{c} 1 & - 7 \\ 5 & 2 \end{array} |$

Step 3.8

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

$| \begin{array}{c} 1 & - 7 \\ 5 & 2 \end{array} | k̂$

Step 3.9

Add the terms together.

$| \begin{array}{c} - 7 & 1 \\ 2 & 4 \end{array} | î - | \begin{array}{c} 1 & 1 \\ 5 & 4 \end{array} | ĵ + | \begin{array}{c} 1 & - 7 \\ 5 & 2 \end{array} | k̂$

Step 4

Evaluate $| \begin{array}{c} - 7 & 1 \\ 2 & 4 \end{array} |$ .

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Step 4.1

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

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

Step 4.2

Simplify the determinant.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Multiply $- 7$ by $4$ .

$(- 28 - 2 \cdot 1) î - | \begin{array}{c} 1 & 1 \\ 5 & 4 \end{array} | ĵ + | \begin{array}{c} 1 & - 7 \\ 5 & 2 \end{array} | k̂$

Step 4.2.1.2

Multiply $- 2$ by $1$ .

$(- 28 - 2) î - | \begin{array}{c} 1 & 1 \\ 5 & 4 \end{array} | ĵ + | \begin{array}{c} 1 & - 7 \\ 5 & 2 \end{array} | k̂$

Step 4.2.2

Subtract $2$ from $- 28$ .

$- 30 î - | \begin{array}{c} 1 & 1 \\ 5 & 4 \end{array} | ĵ + | \begin{array}{c} 1 & - 7 \\ 5 & 2 \end{array} | k̂$

Step 5

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

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Step 5.1

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

$- 30 î - (1 \cdot 4 - 5 \cdot 1) ĵ + | \begin{array}{c} 1 & - 7 \\ 5 & 2 \end{array} | k̂$

Step 5.2

Simplify the determinant.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Multiply $4$ by $1$ .

$- 30 î - (4 - 5 \cdot 1) ĵ + | \begin{array}{c} 1 & - 7 \\ 5 & 2 \end{array} | k̂$

Step 5.2.1.2

Multiply $- 5$ by $1$ .

$- 30 î - (4 - 5) ĵ + | \begin{array}{c} 1 & - 7 \\ 5 & 2 \end{array} | k̂$

Step 5.2.2

Subtract $5$ from $4$ .

$- 30 î - - 1 ĵ + | \begin{array}{c} 1 & - 7 \\ 5 & 2 \end{array} | k̂$

Step 6

Evaluate $| \begin{array}{c} 1 & - 7 \\ 5 & 2 \end{array} |$ .

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Step 6.1

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

$- 30 î - - 1 ĵ + (1 \cdot 2 - 5 \cdot - 7) k̂$

Step 6.2

Simplify the determinant.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Multiply $2$ by $1$ .

$- 30 î - - 1 ĵ + (2 - 5 \cdot - 7) k̂$

Step 6.2.1.2

Multiply $- 5$ by $- 7$ .

$- 30 î - - 1 ĵ + (2 + 35) k̂$

Step 6.2.2

Add $2$ and $35$ .

$- 30 î - - 1 ĵ + 37 k̂$

Step 7

Multiply $- 1$ by $- 1$ .

$- 30 î + 1 ĵ + 37 k̂$

Step 8

Rewrite the answer.

$[\begin{array}{c} - 30 \\ 1 \\ 37 \end{array}]$

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