539: Determinant of Square Matrix Whose Last Row Is All 1 and Whose Each Other Row Is All 0 Except Row Number + 1 Column 1 Is -1 to Power of Dimension + 1

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description/proof of that determinant of square matrix whose last row is all 1 and whose each other row is all 0 except row number + 1 column 1 is -1 to power of dimension + 1

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Topics

About: matrix

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of determinant of square matrix.
• The reader admits the Laplace expansion of determinant.

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Target Context

• The reader will have a description and a proof of the proposition that the determinant of any square matrix whose last row is all $1$ and whose each other row is all $0$ except the row number $+1$ column $1$ is $-1$ to the power of the dimension $+1$.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$M$: $\in \{\text{ the n x n matrices }\}$, the last row is all $1$ and each other $j$-th row is $0$ except the $j+1$-th column $1$
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Statements:
$detM=(-1{)}^{n+1}$.
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2: Natural Language Description

For any $\text{ n x n }$ matrix, $M$, whose last row is all $1$ and whose each other $j$-th row is $0$ except the $j+1$-th column $1$, the determinant is $detM=(-1{)}^{n+1}$.

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3: Proof

$M=\left(\begin{array}{cccccc}0& 1& 0& 0& ...& 0\\ 0& 0& 1& 0& ...& 0\\ ...\\ 1& 1& 1& 1& ...& 1\end{array}\right)$.

Let us prove it inductively.

Let the determinant of the $n$-dimensional matrix be $f(n)$.

When $n=1$, $M=\left(\begin{array}{c}1\end{array}\right)$, and $f(1)=detM=1=(-1{)}^{1+1}$.

When $n=2$, $M=\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right)$, and $f(2)=detM=-1=(-1{)}^{2+1}$.

Let us suppose that for $n=m-1$, $detM=(-1{)}^{m-1+1}$. For $n=m$, let us expand $detM$ by the Laplace expansion of determinant by the 1st row. The 1st row has only 1 nonzero column, the 2nd column, 1. The cofactor of the component is $(-1{)}^{1+2}f(m-1)$, because $M$ with the 1st row and the 2nd column removed is nothing but the matrix for the case, $m-1$. So, $f(m)=-1f(m-1)$.

So, $f(n)=(-1{)}^{n+1}$.

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References

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