Then you can perform GPU-accelerated matrix multiplication as follows: import cupy as cp A_gpu = cp.random.rand() B_gpu = cp.random.rand(1000, 1000) # Matrix multiplication on the GPU result_gpu = A_gpu @ B_gpu. By simply replacing NumPy with CuPy, you can execute the same code on a GPU. Multiplythis coefficient by the constant term of the divisor with the opposite sign. Write this product under the next coefficient and add them. Continue multiplying the constant term of the divisor with the opposite sign by the obtained sum and add the result to the next coefficient in the line. Continue Till the End: Tryto write it a little bit neater. X to the fifth DX. Pause the video and try to figure it out. So, here the realization is well, if you just rewrite all this as one exponent, so this is equal to the indefinite integral of X to the fifth to the one third, I just rewrote the cube root as the one third power DX, which is the same thing as the Linearmeans we can multiply by a constant, but no powers or other functions. The constant can be any real number (0, 1, any whole number, fraction, negatives, etc.). For a square matrix the determinant can help: a non-zero determinant tells us that all rows (or columns) are linearly independent, so it is "full rank" and its rank equals the Step2: Go ahead and define the function multiply_matrix (A,B). This function takes in two matrices A and B as inputs and returns the product matrix C if matrix multiplication is valid. def multiply_matrix( A, B): global C if A. shape [1] == B. shape [0]: C = np. zeros (( A. shape [0], B. shape [1]), dtype = int) for row in range ( rows): for AdvancedMath questions and answers. 1. Let A be a 2x3 matrix, B a 3x2 matrix, and C a 2x2 matrix. Clearly indicate which of the expressions below are defined and which are not. For those which are defined, indicate the number of rows and columns of the matrix being expressed. No justification is needed. i) CT (AB)*C ii) C2AB-1 (A + B)Ğ (where B0fthc.

can you multiply a 2x3 and 2x3 matrix