Why the nᵗʰ root of the determinant of a matrix equals to the norm?

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I was doing some algebra exercise and I noticed that to normalize an orthogonal matrix $M_{n \times n}$, I can just divide the matrix using the nᵗʰ root of the absolute value of the determinant.

I know that the determinant can be defined as how much the size of the area, volume,[...] created with the column vectors increases compared to the one created with canonical vectors, and I also know that to normalize the vector I can divide it by the norm of vectors of the matrix, so I am asking myself why it works when I use the the nᵗʰ root of the determinant in the place of the norm to get the normalized form of the matrix?

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Which number you should divide by depends on what you want, i.e. what exactly it means for a matrix to be "normalized" in your context.

If you want a matrix whose norm is $1$, divide by the matrix by its norm. If you want a matrix whose determinant has magnitude $1$, divide the matrix by the $n$th root of its determinant.