Finding the value of an expression given three equations.

Question

$$x+y+2z=22$$ $$3x-2y+z=6$$ $$7x+3y-5z=1$$

above are three equations and we have to find the value of $$x^2+y^2+z^2=?$$

Answer 1

$$ \begin{cases} x+y+2z=22\\ 3x-2y+z=6\\ 7x+3y-5z=1 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} y=22+(-x-2z)\\ 3x-2y+z=6\\ 7x+3y-5z=1 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} y=22+(-x-2z)\\ 3x-2(22+(-x-2z))+z=6\\ 7x+3(22+(-x-2z))-5z=1 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} y=22+(-x-2z)\\ -44+5x+5z=6\\ 66+4x-11z=1 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} y=22+(-x-2z)\\ x=10-z\\ 66+4x-11z=1 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} y=22+(-x-2z)\\ x=10-z\\ 66+4(10-z)-11z=1 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} y=22+(-x-2z)\\ x=10-z\\ 106-15z=1 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} y=22+(-x-2z)\\ x=10-z\\ z=7 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} y=22+(-x-2z)\\ x=10-7\\ z=7 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} y=22+(-x-2z)\\ x=3\\ z=7 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} y=22+(-3-2\cdot7)\\ x=3\\ z=7 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} y=5\\ x=3\\ z=7 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} x=3\\ y=5\\ z=7 \end{cases} $$

---

So:

$$x^2+y^2+z^2=3^2+5^2+7^2=9+25+49=83$$

Answer 2

Hint: $$\det \begin{vmatrix}1&1&2\\3&-2&1\\7&3&-5\end{vmatrix}\neq 0$$ so the system has a unique solution. Find it and substitute in $x^2+y^2+z^2$. To check you should find $(x,y,z)=(3,5,7)$. Can you solve the system of three equations with the three unknowns (by performing elementary row operations)?