No. Matrix addition (or subtraction) is defined only for matrices of the same dimensions.
x-2(x)+4/x^2 -4=x-2x+4/x^2 -4=-x-4+4/x^2
2^(6) = 64 2 x 2 x 2 x 2 x 2 x2 = 4 x 4 x 4 = 16 x 4 = 64
2x - 3y = 4 x + 4y = 3 Write the augmented matrix: 2 -3 4 1 4 3 Multiply the first row by 1/2: 1 -3/2 2 1 4 3 Subtract the second row from the first row: 1 - 3/2 2 0 -11/2 -1 Multiply the second row by -2/11: 1 -3/2 2 0 1 2/11 Multiply the second row by 3/2 and add it to the first row: 1 0 25/11 0 1 2/11 Thus x = 25/11 and y = 2/11 Check:
The question is unclear - but the information provided may help. 2 squared (which can be written 22) = 2 x 2 = 4 2 cubed (which can be written 23) = 2 x 2 x 2 = 8 4 squared = 42 = 4 x 4 = 16 4 cubed = 43 = 4 x 4 x 4 = 64
It will be a 2 x 5 matrix.
no
Yes.
2 x 5 matrix
No. Matrix addition (or subtraction) is defined only for matrices of the same dimensions.
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.
3x1 matrix
7 x 6A+
Yes it is possible. The resulting matrix would be of the 2x3 order.
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
-1 2 2 0 x = -1 -4 1 -6 Answer: a = 0 and b = -6
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.