Verify Property 1 for Delta=left|begin{array} | vedclass

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(N/A) Solution: Expanding the determinant along the first row,we have:$\Delta = 2\begin{vmatrix} 0 & 4 \ 5 & -7 \end{vmatrix} - (-3)\begin{vmatrix} 6

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(N/A) Solution: Expanding the determinant along the first row,we have:
$\Delta = 2\begin{vmatrix} 0 & 4 \\ 5 & -7 \end{vmatrix} - (-3)\begin{vmatrix} 6 & 4 \\ 1 & -7 \end{vmatrix} + 5\begin{vmatrix} 6 & 0 \\ 1 & 5 \end{vmatrix}$
$= 2(0 - 20) + 3(-42 - 4) + 5(30 - 0)$
$= -40 - 138 + 150 = -28$
By interchanging rows and columns,we get the transpose determinant $\Delta_{1}$:
$\Delta_{1} = \begin{vmatrix} 2 & 6 & 1 \\ -3 & 0 & 5 \\ 5 & 4 & -7 \end{vmatrix}$
Expanding along the first column:
$\Delta_{1} = 2\begin{vmatrix} 0 & 5 \\ 4 & -7 \end{vmatrix} - (-3)\begin{vmatrix} 6 & 1 \\ 4 & -7 \end{vmatrix} + 5\begin{vmatrix} 6 & 1 \\ 0 & 5 \end{vmatrix}$
$= 2(0 - 20) + 3(-42 - 4) + 5(30 - 0)$
$= -40 - 138 + 150 = -28$
Clearly,$\Delta = \Delta_{1}$.
Hence,Property $1$ (the value of the determinant remains unchanged if its rows and columns are interchanged) is verified.

Verify Property $1$ for $\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$

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