Linear Algebra Determinant proof

TanWu
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Homework Statement
(a) Show that a matrix ##\left(\begin{array}{ll}e & g \\ 0 & f\end{array}\right)## has determinant equal to the product of the elements on the leading diagonal. Can you generalise this idea to any ##n \times n## matrix?
Relevant Equations
##\left(\begin{array}{ll}e & g \\ 0 & f\end{array}\right)##
I have a doubt about this problem.

(a) Show that a matrix ##\left(\begin{array}{ll}e & g \\ 0 & f\end{array}\right)## has determinant equal to the product of the elements on the leading diagonal. Can you generalize this idea to any ##n \times n## matrix? The first part is simple, it is just ef.

I have a doubt about what ##n \times n## matrix they want generalized too, for example do they want a upper triangular ##n \times n## matrix like the one the author as written or a lower triangular, or general matrix, etc.

I express gratitude to those who help.
 
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The determinant of an upper or lower triangular matrix is equal to the product of the elements on the leading diagonal.

An upper triangular matrix is a square matrix whose entries below the leading diagonal are zero.

The claim follows quickly provided you are familiar with the Laplace expansion of the determinant.

Also, good job on using much more helpful titles. :cool:
 
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Thank you Sir. Apologize, I am not familiar with that.
 
In the "Relevant Equations" section, you should state how you have defined the determinant or any already-proven fact(s) that you use in your proof.
In general, you should work on stating your proofs in a more formal way. Where they are used, your proof should state what definitions, theorems, or lemas you are using.
 
note that geometrically this is the fact that the area of a parallelogram equals that of the rectangle with same height and base.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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