Parker Glynn-Adey

Q8(b) from 2014 Final Winter

Posted in Math by pgadey on 2016/04/07

 

Exercise 8b Assume {A} is an {n \times n} matrix with {n \geq 3} and has the property that each entry in the first row equals the sum of the {n-1} entries in the column below it. Prove that {\det(A) = 0}.


You want to understand the wording of the question. What does it mean to say: “ each entry in the first row equals the sum of the {n-1} entries in the column below it”. Here are some examples to get you started. This example has the property you want:

\displaystyle \left[ \begin{matrix} 2 & 3 & 100\\ 1 & 1 & 25\\ 1 & 2 & 75\\ \end{matrix} \right]

This example does not have the property you want:

\displaystyle \left[ \begin{matrix} 1 & 3 & 0\\ 0 & 1 & 25\\ 0 & 2 & 75\\ \end{matrix} \right]

because {1 \neq 0 + 0} and {0 \neq 25 + 75}.

To be a bit more formal, consider the following matrix:

\displaystyle \left[ \begin{matrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\\ \end{matrix} \right]

We say that “the each entry in the first row is the sum of the {n-1} entries in the column below it” if: {a_{11} = a_{21} + a_{31}} and {a_{12} = a_{22} + a_{32}} and {a_{13} = a_{23} + a_{33}}. The left hand side of each equation is “an entry in the first row” and the right hand side is “the sum of the {n-1} entries in the column below it”.

 

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