Can you think of any 2x2 matrices d for which cd = dc for all 2x2 matrices c? give 2 different examples of such matricesd.

1 answer:

6 0

For any arbitrary 2x2 matrices $\mathbf C$ and $\mathbf D$ , only one choice of $\mathbf D$ exists to satisfy $\mathbf{CD}=\mathbf{DC}$ , which is the identity matrix.

There is no other matrix that would work unless we place some more restrictions on $\mathbf C$ . One such restriction would be to ensure that $\mathbf C$ is not singular, or its determinant is non-zero. Then this matrix has an inverse, and taking $\mathbf D=\mathbf C^{-1}$ we'd get equality.

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et $R = \ZZ[\sqrt{-n}]$ where $n$ is a squarefree integer $> 3$. Prove that $2$, $\sqrt{-n}$, and $1 + \sqrt{-n}$ are all irr

melisa1 [442]

What the hell is that?

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3 years ago

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Andre45 [30]

For this case, we must indicate which of the given functions is not defined for $x = 0$

By definition, we know that:

$f (x) = \sqrt {x}$ has a domain from 0 to infinity.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .

While $y = \sqrt {x + 2}$ if it is defined for $x = 0.$

$f(x)=\sqrt[3]{x}$ , your domain is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root.

So, we have:

$y = \sqrt [3] {x-2}$ with x = 0: $y = \sqrt [3] {- 2}$ is defined.