-1

\begin{vmatrix} 2 & 4 & 4 & \dots & 4 & 4 \\ 4 & 4 & 4 & \dots & 4 & 4 \\ 4 & 4 & 6 & \dots & 4 & 4 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 4 & 4 & 4 & \dots & 2n-2 & 4 \\ 4 & 4 & 4 & \dots & 4 & 2n \end{vmatrix}

I tried to get a triangular matrix, subtracting the lines, but it did not work out

Efim Grob
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  • see this: https://math.stackexchange.com/questions/2110766/calculating-determinant-with-different-numbers-on-diagonal-and-x-everywhere-else – Marco Sep 26 '18 at 21:34

2 Answers2

3

\begin{eqnarray*} \begin{vmatrix} 2 & 4 & 4 & \dots & 4 & 4 \\ 4 & 4 & 4 & \dots & 4 & 4 \\ 4 & 4 & 6 & \dots & 4 & 4 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 4 & 4 & 4 & \dots & 2n-2 & 4 \\ 4 & 4 & 4 & \dots & 4 & 2n \end{vmatrix} \end{eqnarray*} Subtract the second row off all the other rows to get \begin{eqnarray*} \begin{vmatrix} -2 & 0 & 0 & \dots & 0 & 0 \\ 4 & 4 & 4 & \dots & 4 & 4 \\ 0 & 0 & 2 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 2n-6 & 0 \\ 0 & 0 & 0 & \dots & 4 & 2n-4 \end{vmatrix} \end{eqnarray*} The product on the leading diagonal is $\color{red}{-2^{n+1} (n-2)!}$.

Donald Splutterwit
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$$ Q^T D Q = H $$ $$\left( \begin{array}{rr} 1 & 0 \\ 2 & 1 \\ \end{array} \right) \left( \begin{array}{rr} 2 & 0 \\ 0 & - 4 \\ \end{array} \right) \left( \begin{array}{rr} 1 & 2 \\ 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rr} 2 & 4 \\ 4 & 4 \\ \end{array} \right) $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 2 & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & - 4 & 0 \\ 0 & 0 & 2 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & 4 & 4 \\ 4 & 4 & 4 \\ 4 & 4 & 6 \\ \end{array} \right) $$

$$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 \\ 2 & 1 & 1 & 0 \\ 2 & 1 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & - 4 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 4 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 2 & 2 & 2 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 2 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 \\ 4 & 4 & 6 & 4 \\ 4 & 4 & 4 & 8 \\ \end{array} \right) $$

$$\left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 & 0 \\ 2 & 1 & 1 & 0 & 0 \\ 2 & 1 & 0 & 1 & 0 \\ 2 & 1 & 0 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrrrr} 2 & 0 & 0 & 0 & 0 \\ 0 & - 4 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 0 & 6 \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrr} 2 & 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 & 4 \\ 4 & 4 & 6 & 4 & 4 \\ 4 & 4 & 4 & 8 & 4 \\ 4 & 4 & 4 & 4 & 10 \\ \end{array} \right) $$

$$\left( \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 & 0 & 0 \\ 2 & 1 & 1 & 0 & 0 & 0 \\ 2 & 1 & 0 & 1 & 0 & 0 \\ 2 & 1 & 0 & 0 & 1 & 0 \\ 2 & 1 & 0 & 0 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrrrrr} 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & - 4 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 6 & 0 \\ 0 & 0 & 0 & 0 & 0 & 8 \\ \end{array} \right) \left( \begin{array}{rrrrrr} 1 & 2 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrrr} 2 & 4 & 4 & 4 & 4 & 4 \\ 4 & 4 & 4 & 4 & 4 & 4 \\ 4 & 4 & 6 & 4 & 4 & 4 \\ 4 & 4 & 4 & 8 & 4 & 4 \\ 4 & 4 & 4 & 4 & 10 & 4 \\ 4 & 4 & 4 & 4 & 4 & 12 \\ \end{array} \right) $$

Will Jagy
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