Linear Algebra

A vector is a group of numbers in one dimensions; if you have such arrangements of numbers in more than one dimension, you get a tensor. Actually, a vector is simply a special case of a tensor (a 1st-order tensor).

It is: 9*8*7*6*5*4*3*2*1 = 362,880

5

x+3y=6

3y=6+x

y=2+x/3

So the best way is to plot the intercepts (easiest) which are:

if

x=0

therefore y=2 (0,2) is an intercept on the y-axis

y=0

0=2+x/3

-2=x/3

-6=x

so (-6,0) is the x-axis intercept

so plot those points and join a straight line through there. All you have to do is get two points since it is a straight line and will not turn and just draw a line through that.

The question contains an expression, not an equation nor inequality. An expression cannot have a solution.

Check out http://en.wikipedia.org/wiki/Boussinesq_approximation.

If you are asking if x=0, y=5 is a solution to 5x-3y=15, then no. 0, -5 would be as if you sub in 0 for x you get 5(0) -3y=15, i.e. -3y=15, i.e. y=-5

If you have a system, which can be expressed as a set of linear equations, then you can utilize matrices to help solve it. One example is an electrical circuit which uses linear devices (example are constant voltage sources and resistive loads). To find the current through each device, a set of linear equations is derived.

Yes, y = -4x + 3 is a linear equation.

No. Every closed orbit (around and around and around) is an ellipse.

Every open orbit (swish by one time and never return) is a hyperbola.

The one that's exactly precisely on the dividing line between closed and

open is a parabola.

An eigenvalue is a scalar that indicates how much an eigenvector is stretched or compressed during a linear transformation represented by a matrix. In contrast, an eigenvector is a non-zero vector that remains in the same direction after the transformation, only scaled by the eigenvalue. Mathematically, for a square matrix (A), if (A\mathbf{v} = \lambda \mathbf{v}), then (\lambda) is the eigenvalue and (\mathbf{v}) is the corresponding eigenvector.

The singular values of an orthogonal matrix are all equal to 1. This is because an orthogonal matrix ( Q ) satisfies the property ( Q^T Q = I ), where ( I ) is the identity matrix. Consequently, the singular value decomposition of ( Q ) reveals that the singular values, which are the square roots of the eigenvalues of ( Q^T Q ), are all 1. Thus, for an orthogonal matrix, the singular values indicate that the matrix preserves lengths and angles in Euclidean space.

Graphs of equations typically represent a relationship between two variables, showing how one variable changes with respect to the other, often resulting in a curve or line. In contrast, expressions do not have an equality sign and do not define a relationship between variables; thus, they cannot be graphed as a function. Instead, expressions can represent values or calculations, which might be evaluated at specific points rather than graphed. Therefore, while equations can produce visual representations of relationships, expressions remain abstract without a graphical form.

A banana is a non-example of radiation.

To find the best least squares solution to the matrix equation ( AX = b ), where ( A = \begin{pmatrix} 1 & 0 \ 2 & 1 \end{pmatrix} ) and ( b = \begin{pmatrix} 3 \ 6 \end{pmatrix} ), we can use the formula ( X = (A^T A)^{-1} A^T b ). First, we compute ( A^T A = \begin{pmatrix} 1 & 2 \ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 2 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 2 \ 2 & 1 \end{pmatrix} ). After finding ( (A^T A)^{-1} ) and multiplying by ( A^T b ), the least squares solution will yield the values of ( X ). The final values for ( X ) will be the best fit for the original equation.

Phase 1 of linear programming aims to find a feasible solution to the problem by minimizing a "penalty" function, often involving artificial variables. If the feasible region is unbounded or if multiple ways exist to achieve the same minimum value for the penalty function, there can be alternative optimal solutions. This occurs when the objective function is parallel to a constraint boundary, allowing for multiple feasible points that yield the same objective value. Hence, the presence of alternative optimal solutions is tied to the geometry of the feasible region and the nature of the objective function.

To determine whether a real-world situation should be represented by an equation or an inequality, assess the nature of the relationship between the variables involved. If the relationship requires equality—where one quantity is exactly equal to another—you would use an equation. Conversely, if the situation involves a range of possible values, constraints, or conditions where one quantity is greater than, less than, or not equal to another, an inequality is more appropriate. For instance, budgeting scenarios often involve inequalities, while exact measurements may be represented with equations.

To invert a complex matrix using the Armadillo linear algebra library, you can utilize the inv() function, which computes the inverse of a matrix. First, ensure you include the Armadillo header and link against the library. Here's a simple example:

#include <armadillo>

arma::cx_mat A = {{1.0, 2.0}, {3.0, 4.0}}; // Define a complex matrix
arma::cx_mat A_inv = inv(A); // Invert the matrix

Make sure the matrix is square and non-singular for the inversion to be valid.

A math dictionary (or a mathematics dictionary) is usually a dictionary of the definitions of common mathematical terms, formulae and examples of common diagrams and pictures for the uninitiated or those learning math.

It is usually represented by a lower case letter with a horizontal arrow above it. In print, the letter will usually be bold.

it is physically the projection or shadow of a line on a plane...

straight line