3x1 matrix
7 x 6A+
No.Two matrices A and B can be added or subtracted if and only if they have the same number of rows and columns. That is a 3 x 2 matrix can be added or subtracted only with another 3 x 2 matrix.
2 x 5 matrix
The answer is yes, and here's why: Remember that for the eigenvalues (k) and eigenvectors (v) of a matrix (M) the following holds: M.v = k*v, where "." denotes matrix multiplication. This operation is only defined if the number of columns in the first matrix is equal to the number of rows in the second, and the resulting matrix/vector will have as many rows as the first matrix, and as many columns as the second matrix. For example, if you have a 3 x 2 matrix and multiply with a 2 x 4 matrix, the result will be a 3 x 4 matrix. Applying this to the eigenvalue problem, where the second matrix is a vector, we see that if the matrix M is m x n and the vector is n x 1, the result will be an m x 1 vector. Clearly, this can never be a scalar multiple of the original vector.
It will be a 2 x 5 matrix.
3x1 matrix
7 x 6A+
3 x 3 matrix
Yes it is possible. The resulting matrix would be of the 2x3 order.
no
No.Two matrices A and B can be added or subtracted if and only if they have the same number of rows and columns. That is a 3 x 2 matrix can be added or subtracted only with another 3 x 2 matrix.
2 x 5 matrix
Yes.
The eigen values of a matirx are the values L such that Ax = Lxwhere A is a matrix, x is a vector, and L is a constant.The vector x is known as the eigenvector.
Consider the linear system of equations AX = YwhereX is a n x 1 matrix of variables,Y is a n x 1 matrix of constants, andA is an n x n matrix of coefficients.Provided A is not a singular matrix, A has an inverse, A-1, an n x n matrix.Premultiplying by A-1 gives A-1AX = A-1Y or X = A-1Y, the solution to the linear system.
The answer is yes, and here's why: Remember that for the eigenvalues (k) and eigenvectors (v) of a matrix (M) the following holds: M.v = k*v, where "." denotes matrix multiplication. This operation is only defined if the number of columns in the first matrix is equal to the number of rows in the second, and the resulting matrix/vector will have as many rows as the first matrix, and as many columns as the second matrix. For example, if you have a 3 x 2 matrix and multiply with a 2 x 4 matrix, the result will be a 3 x 4 matrix. Applying this to the eigenvalue problem, where the second matrix is a vector, we see that if the matrix M is m x n and the vector is n x 1, the result will be an m x 1 vector. Clearly, this can never be a scalar multiple of the original vector.