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 $x^2+y^2-(w+z)^2$
Answer & explanation
Correct answer: option 3
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 $x^2+y^2-(w+z)^2=2^2+4^2-(1+3)^2=4$