The calculator solves linear algebra problems. It is used for answering questions related to vectors and matrices.
-------
add matrices
add vectors
adjoint (adjugate) matrix
angle between vectors
basis
characteristic polynomial
cofactor matrix
column space
Cramer's rule
cross product
diagonalize matrix
divide matrices
dot (inner) product
eigenvalues and eigenvectors
Gauss-Jordan elimination
Gram-Schmidt process
linear independence
LU decomposition
matrix determinant
matrix exponential
matrix inverse
matrix of minors
matrix power
matrix rank
matrix scalar multiplication
matrix trace
matrix transpose
Moore-Penrose inverse (pseudoinverse)
multiply matrices
null space (kernel), nullity
operations on matrices
operations on vectors
orthogonal complement
QR factorization
reduced row echelon form (rref)
row space
scalar projection
singular value decomposition
solve system of linear equations
subtract matrices
subtract vectors
transition matrix
triple product
unit vector
vector magnitude
vector projection
vector scalar multiplication
add matrices
add vectors
adjoint (adjugate) matrix
angle between vectors
basis
characteristic polynomial
cofactor matrix
column space
Cramer's rule
cross product
diagonalize matrix
divide matrices
dot (inner) product
eigenvalues and eigenvectors
Gauss-Jordan elimination
Gram-Schmidt process
linear independence
LU decomposition
matrix determinant
matrix exponential
matrix inverse
matrix of minors
matrix power
matrix rank
matrix scalar multiplication
matrix trace
matrix transpose
Moore-Penrose inverse (pseudoinverse)
multiply matrices
null space (kernel), nullity
operations on matrices
operations on vectors
orthogonal complement
QR factorization
reduced row echelon form (rref)
row space
scalar projection
singular value decomposition
solve system of linear equations
subtract matrices
subtract vectors
transition matrix
triple product
unit vector
vector magnitude
vector projection
vector scalar multiplication
Solution To find the cross product, we form a formal determinant the first row of which consists of unit vectors, the second row is our first vector, and the third row is our second vector: $$$ \left|\begin{array}{ccc}\mathbf{\vec{i}} & \mathbf{\vec{j}} & \mathbf{\vec{k}}\\3 & 1 & 4\\-2 & 0 & 5\end{array}\right|$$$ .
Now, just expand along the first row (for steps in finding a determinant, see determinant calculator ):
$$$ \left|\begin{array}{ccc}\mathbf{\vec{i}} & \mathbf{\vec{j}} & \mathbf{\vec{k}}\\3 & 1 & 4\\-2 & 0 & 5\end{array}\right| = \left|\begin{array}{cc}1 & 4\\0 & 5\end{array}\right| \mathbf{\vec{i}} - \left|\begin{array}{cc}3 & 4\\-2 & 5\end{array}\right| \mathbf{\vec{j}} + \left|\begin{array}{cc}3 & 1\\-2 & 0\end{array}\right| \mathbf{\vec{k}} = \left(\left(1\right)\cdot \left(5\right) - \left(4\right)\cdot \left(0\right)\right) \mathbf{\vec{i}} - \left(\left(3\right)\cdot \left(5\right) - \left(4\right)\cdot \left(-2\right)\right) \mathbf{\vec{j}} + \left(\left(3\right)\cdot \left(0\right) - \left(1\right)\cdot \left(-2\right)\right) \mathbf{\vec{k}} = 5 \mathbf{\vec{i}} - 23 \mathbf{\vec{j}} + 2 \mathbf{\vec{k}}$$$
Thus, $$$ \left\langle 3, 1, 4\right\rangle\times \left\langle -2, 0, 5\right\rangle = \left\langle 5, -23, 2\right\rangle.$$$
Answer $$$ \left\langle 3, 1, 4\right\rangle\times \left\langle -2, 0, 5\right\rangle = \left\langle 5, -23, 2\right\rangle$$$ A
