MHB Finding the Matrix Aij: Cubic Root of 2, -4 Power of j & Sqrt of 3)

[leprofece]

Hello friends(a(
Find the matriz A ij) if we know that a) aij = (cubic root (2) )^{i} + {-4}^{j})\{\sqrt{3}} (View attachment 3438

 

[Sudharaka]

Hello friends(a(
Find the matriz A ij) if we know that a) aij = (cubic root (2) )^{i} + {-4}^{j})\{\sqrt{3}} (View attachment 3438

Hello leprofece, :)

So you are given the general form of the elements of the matrix.

\[a_{ij} = (\sqrt[3]{2})^{i}+\frac{(-4)^{j}}{\sqrt{3}}\]

I cannot read the attachment clearly but I assume that the size of the matrix is \(2\times 3\). That means it has 2 rows and 3 columns. Therefore $i$ runs from 1 to 2 and $j$ runs from 1 to 3. Hence you can find all the elements in the matrix by substituting values for $i$ and $j$.

Example: $a_{12}= (\sqrt[3]{2})^{1}+\frac{(-4)^{2}}{\sqrt{3}}=\sqrt[3]{2}+\frac{16}{\sqrt{3}}\approx 10.498$