The Leslie Matrix is ：L = [ 2.2 4.2, 0.1 0]. Use the L to calculate L3
Question
The Leslie Matrix is ：L = [ 2.2 4.2, 0.1 0]. Use the L to calculate L3
Solution
To calculate L^3, we need to multiply the matrix L by itself three times. Here are the steps:
Step 1: Calculate L^2 L^2 = L * L = [2.2 4.2; 0.1 0] * [2.2 4.2; 0.1 0] = [2.22.2 + 4.20.1, 2.24.2 + 4.20; 0.12.2 + 00.1, 0.14.2 + 00] = [4.84, 9.24; 0.22, 0.42]
Step 2: Calculate L^3 L^3 = L^2 * L = [4.84, 9.24; 0.22, 0.42] * [2.2 4.2; 0.1 0] = [4.842.2 + 9.240.1, 4.844.2 + 9.240; 0.222.2 + 0.420.1, 0.224.2 + 0.420] = [10.648, 20.328; 0.484, 0.924]
So, L^3 = [10.648, 20.328; 0.484, 0.924]
Similar Questions
Find LU-decomposition of the following matrix.𝐴 = [2 01 2].
#<GRADED> def l2distance(X,Z=None): """ function D=l2distance(X,Z) Computes the Euclidean distance matrix. Syntax: D=l2distance(X,Z) Input: X: nxd data matrix with n vectors (rows) of dimensionality d Z: mxd data matrix with m vectors (rows) of dimensionality d Output: Matrix D of size nxm D(i,j) is the Euclidean distance of X(i,:) and Z(j,:) call with only one input: l2distance(X)=l2distance(X,X) """ if Z is None: Z=X; n,d1=X.shape m,d2=Z.shape assert (d1==d2), "Dimensions of input vectors must match!" # Your code goes here .. raise NotImplementedError('Your code goes here!') return D # ... until here #</GRADED> without using for-loop
What will be the output of the following statement?l = [1,2,3,4,5]ans = 3 in lprint(ans)
matrix A use Gauss-Jordan eli
Let L1 = L (a∗baa∗) and L2 = L (aba∗). Find L1/L2
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.