Rank of a matrix is used to find consistency of linear system of equations.As we know most of the engineering problems land up with the problem of finding solution of a linear system of equations ,at that point rank of matrix is useful.
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First, You have to reduce the matrix to echelon form . The number of nonzero rows in the reduced echelon form matrix (number of linearly independent rows) indicates the rank of the matrix. Go to any search engine and type "Rank of a matrix, Cliffnotes" for an example.
Then it has (not have!) a unique solution.
A scalar, which is a tensor of rank 0, is just a number, e.g. 6 A vector, which is a tensor of rank 1, is a group of scalars, e.g. [1, 6, 3] A matrix, which is a tensor of rank 2, is a group of vectors, e.g. 1 6 3 9 4 2 0 1 3 A tensor of rank 3 would be a group of matrix and would look like a 3d matrix. A tensor is the general term for all of these, and the generalization into high dimensions.
Matrix inverses and determinants, square and nonsingular, the equations AX = I and XA = I have the same solution, X. This solution is called the inverse of A.
No.A 3x3 matrix A is a representation of a linear map \alpha : \mathbb{R}^3 \longrightarrow \mathbb{R}^3 .For any linear map T : U \longrightarrow V ,we have the rank-nullity theorum:rank(T)+nullity(T) = dim(U)where the rank and nullity are the dimensions of the image and kernal of T respectively.Im(T) = ker(T) \Rightarrow rank(T) = nullity(T) = m, sayfor some non-negative integer m. Then the rank-nullity theorum implies that dim(U)=2m.The image and kernal of a matrix A are the same as those for the corresponding basis-free linear map \alpha .For a 3x3 matrix, dim(U) = 3, so there are no such matrices (since 3 is odd).