Linear Algebra Examples

[2-11]×[5-31]211×531
Step 1
The cross product of two vectors a⃗a⃗ and b⃗b⃗ can be written as a determinant with the standard unit vectors from 3 and the elements of the given vectors.
a⃗×b⃗=|a1a2a3b1b2b3|
Step 2
Set up the determinant with the given values.
|2-115-31|
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.
|+-+-+-+-+|
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 a11 is the determinant with row 1 and column 1 deleted.
|-11-31|
Step 3.4
Multiply element a11 by its cofactor.
|-11-31|
Step 3.5
The minor for a12 is the determinant with row 1 and column 2 deleted.
|2151|
Step 3.6
Multiply element a12 by its cofactor.
-|2151|
Step 3.7
The minor for a13 is the determinant with row 1 and column 3 deleted.
|2-15-3|
Step 3.8
Multiply element a13 by its cofactor.
|2-15-3|
Step 3.9
Add the terms together.
|-11-31|-|2151|+|2-15-3|
|-11-31|-|2151|+|2-15-3|
Step 4
Evaluate |-11-31|.
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Step 4.1
The determinant of a 2×2 matrix can be found using the formula |abcd|=ad-cb.
(-11-(-31))-|2151|+|2-15-3|
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 -1 by 1.
(-1-(-31))-|2151|+|2-15-3|
Step 4.2.1.2
Multiply -(-31).
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Step 4.2.1.2.1
Multiply -3 by 1.
(-1--3)-|2151|+|2-15-3|
Step 4.2.1.2.2
Multiply -1 by -3.
(-1+3)-|2151|+|2-15-3|
(-1+3)-|2151|+|2-15-3|
(-1+3)-|2151|+|2-15-3|
Step 4.2.2
Add -1 and 3.
2-|2151|+|2-15-3|
2-|2151|+|2-15-3|
2-|2151|+|2-15-3|
Step 5
Evaluate |2151|.
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Step 5.1
The determinant of a 2×2 matrix can be found using the formula |abcd|=ad-cb.
2-(21-51)+|2-15-3|
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 2 by 1.
2-(2-51)+|2-15-3|
Step 5.2.1.2
Multiply -5 by 1.
2-(2-5)+|2-15-3|
2-(2-5)+|2-15-3|
Step 5.2.2
Subtract 5 from 2.
2--3+|2-15-3|
2--3+|2-15-3|
2--3+|2-15-3|
Step 6
Evaluate |2-15-3|.
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Step 6.1
The determinant of a 2×2 matrix can be found using the formula |abcd|=ad-cb.
2--3+(2-3-5-1)
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 -3.
2--3+(-6-5-1)
Step 6.2.1.2
Multiply -5 by -1.
2--3+(-6+5)
2--3+(-6+5)
Step 6.2.2
Add -6 and 5.
2--3-
2--3-
2--3-
Step 7
Multiply -1 by -3.
2+3-
Step 8
Rewrite the answer.
[23-1]
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 [x2  12  π  xdx ] 
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