De determinant
\(\begin{vmatrix} x+y& x-y \\ x-y& x+y \end{vmatrix}\)
is gelijk aan
|
A. 0 |
| B. 2y2 |
| C. − 2y2 |
| D. 4xy |
| E. x2 + y2 |
[ 5-7400 - op net sinds 12.8.13-(E)-20.10.2026 ]
Translation in E N G L I S H
Evaluate
\(\begin{vmatrix} x+y& x-y \\ x-y& x+y \end{vmatrix}\)
|
A. 0 |
| B. 2y2 |
| C. − 2y2 |
| D. 4xy |
| E. x2 + y2 |
Oplossing - Solution
1ste manier :
\(\begin{vmatrix} x+y& x-y \\ x-y& x+y \end{vmatrix}=(x+y)^2-(x-y)^2=x^2+2xy+y^2-x^2+2xy-y^2=4xy\)
2de manier :
\(\begin{vmatrix} x+y& x-y \\ x-y& x+y \end{vmatrix}=(x+y)^2-(x-y)^2=(x+y+x-y)(x+y-x+y)\\=2x.2y=4xy\)
3de manier :
\(\begin{vmatrix} x+y& x-y \\ x-y& x+y \end{vmatrix}\overset{R_1\to R_1+R_2}{=}\begin{vmatrix}2x&2x\\x-y&x+y\end{vmatrix}=2\cdot \begin{vmatrix}x&x\\x-y&x+y\end{vmatrix}\\\overset{R_2\to R_2-R_1}{=}2\cdot \begin{vmatrix}x&x\\-y&+y\end{vmatrix}=2xy\begin{vmatrix}1&1\\-1&1\end{vmatrix}=2xy.2=4xy\)
4de manier :
\(\small\begin{vmatrix}x\!+\!y& x\!-\!y \\ x-y& x+y \end{vmatrix}\!=\!\begin{vmatrix}x&x\!-\!y\\x&x\!+\!y\end{vmatrix}\!+\!\begin{vmatrix}y&x\!-\!y\\-y&x\!+\!y\end{vmatrix}=\begin{vmatrix}x&x\\x&x\end{vmatrix}\!+\!\begin{vmatrix}x&-y\\x&y\end{vmatrix} \!+\!\begin{vmatrix}y&x\\-y&x\end{vmatrix} \!+\!\begin{vmatrix}y&-y\\-y&y\end{vmatrix} \\\\=0+2xy+2xy+0=4xy\)