Linear Algebra Questions

For any matrix A, there exists a matrix B so that A+B=0.

Asking whether the linear system corresponding to an augmented matrix [a1a2a3|b] has a solution amounts to asking whether b is in Span{a1,a2,a3}.

If S is a set of linearly dependent vectors, then every vector in S can be written as a linear combination of the other vectors in S.

If a set S of vectors contains fewer vectors than there are entries in the vectors, then the set must be linearly independent.

The homogeneous system Ax=0 has the trivial solution if and only if the system has at least one free variable.

There are exactly three vectors in Span{a1,a2,a3}.

The vector b is a linear combination of the columns of matrix A if and only if Ax=b has at least one solution.

Suppose A and B are invertible matrices. A+B is invertible.

The equation Ax=b is homogenous if the zero vector is a solution.

For any matrix A, we have the equality 2A+3A=5A.

If the columns of an m×n matrix A span Rm, then the equation Ax=b is consistent for each b in Rm.

The columns of matrix A are linearly independent if equation Ax=0 has the trivial solution.

If A is an m×n matrix and if the equation Ax=b is inconsistent for some b in Rm, then A cannot have a pivot position in every row.

There are exactly three vectors in the set {a1,a2,a3}.

A set of three vectors in R4 can span all of R4.

A| is diagonalizable if and only if A| has n eigenvalues, counting multiplicity.

If A is an m×n matrix whose columns do not span Rm, then the equation Ax=b is inconsistent for some b in Rm.

The solution set of Ax=b is obtained by translating the solution set of Ax=0.

The equation Ax=b is consistent if the augmented matrix [A|b] has a pivot position in every row.

If the equation Ax=b is consistent, then b is in the span of the columns of A.