3x1 matrix
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.
No. Matrix addition (or subtraction) is defined only for matrices of the same dimensions.
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.
3x1 matrix
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.
It will be a 2 x 5 matrix.
No. Matrix addition (or subtraction) is defined only for matrices of the same dimensions.
3 x 3 matrix
no
A rectangle containing numbers are called "matrix" (1 0 0 1) (3 4 8 0) is a 2 x 4 matrix a SQUARE containing numbers is a n x n matrix, or square matrix (1 0) (5 6) is a square matrix (1) is a square matrix
Yes.
2 x 5 matrix
A 2x4 matrix has 2 rows (horizontal) and 4 columns (vertical). Ex: [1 2 3 4] [5 6 7 8]
That is called the identity matrix. For example, (3,1,4)t x (1,1,1) = (3,1,4)t In this case (1,1,1) is the identity matrix. Another example is 5 11 1 0 1 11 x = 4 3 0 1 4 3 (You will have to imagine the brackets around the matrices as I did not know how to draw them in.) In this case 1 0 is the identity matrix. 0 1
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.