Question

Chapter Name :-

Answer 1

Step-by-step explanation:

This is a inverse. matrix because we have alternating

zeroes and the w symbol it looks similar to the matrix,

[ 1 0]

[ 0 1]

Except, we just replace 1 with q. So we also know that if we multiply a inverse like matrix by itself, a infinite number times, it going to stay the same.

However, since we have variables in place for 1, we would raise the matrix to an nth power.

Because remeber that a inverse matrix has 1, an that

$1 {}^{n} = 1$

So since we have a variable, we just raised it to the nth number.

We know that Zeroes raised to any power is 0, so w raised to the

^20226397596 is infact, w^20226397596, so our matrix is just

[w^20026397596 0]

[ 0 w^20026397596]

Answer 2

The value for $A^{20226397596}$ is        $\left[\begin{array}{cc}\omega^{20226397596}&0\\0&\omega^{20226397596}\end{array}\right]$.

Procedure - Determination of the power of a matrix

Let be $A$ a diagonal matrix. By linear algebra we know that the n-th power of diagonal matrix of the form $\left[\begin{array}{cc}\omega&0\\0&\omega\end{array}\right]$ is equal to:

$A^{n} = \left[\begin{array}{cc}\omega^{n}&0\\0&\omega^{n}\end{array}\right]$, $\forall\, n \ge 1$, $n\in \mathbb{N}$ $\forall \,\omega \in \mathbb{R}$ (1)

Hence, we have the following result for $A^{20226397596}$:

$A^{20226397596} = \left[\begin{array}{cc}\omega^{20226397596}&0\\0&\omega^{20226397596}\end{array}\right]$

The value for $A^{20226397596}$ is        $\left[\begin{array}{cc}\omega^{20226397596}&0\\0&\omega^{20226397596}\end{array}\right]$. $\blacksquare$

To learn more on matrices, we kindly invite to check this verified question: brainly.com/question/4470545