Practicing Success
If $\begin{bmatrix} x+y+z+w\\x+y\\x+z\\x+w\end{bmatrix} =\begin{bmatrix}10 \\6\\5\\3\end{bmatrix},$ find the value of $x^2+y^2-(w+z)^2$ |
16 20 4 8 |
4 |
The correct answer is Option (3) → 4 $\begin{bmatrix} x+y+z+w\\x+y\\x+z\\x+w\end{bmatrix} =\begin{bmatrix}10 \\6\\5\\3\end{bmatrix}$ from given matrix $(x+y)+(x+z)+(z+w)-(x+y+z+w)=6+5+3-10$ so $2x=4⇒x=2$ so $x+y=6⇒y=4,x+w=3⇒w=1$ $x+z=5⇒z=3$ so $x^2+y^2-(w+z)^2=2^2+4^2-(1+3)^2=4$ |