I'm currently revising for a maths module that I am taking as part of my physics degree.
The final part of the matrices section of a paper I was doing included this question:
Solve the this set of simultaneous equations: $$ x + log_2y +z(z+1) = 0 \\ 3x + log_2y^2 + \frac{3}{2}z(z+1) = 0 \\2x + log_2\frac{1}{y}- \frac{1}{2}z(z+1) = 4 $$
I can solve simple systems of linear simultaneous equations by using matrices however this threw me. I tried this matrix: $$ \left[ \begin{array}{ccc|c} 1 & 2^y & z+1 & 0 \\ 3 & 2^{y^2} & \frac{3}{2}(z+1) & 0 \\ 2 & 2^\frac{1}{y} & -\frac{1}{2}(z+1) & 4 \\ \end{array} \right] $$ However I am pretty sure that not only is this attempt wrong but also that I have no idea how to solve this.
Thanks! Sean.
Making the substitution $\tilde{y}=\log_2 y$ and $\tilde{z}=z(z+1)$ leads to the equivalent linear system $$ \begin{eqnarray} x &+& &\tilde{y} &+& &\tilde{z} &= 0,\\ 3x &+& 2&\tilde{y} &+& \frac{3}{2}&\tilde{z} &= 0,\\ 2x&-&&\tilde{y}&-&\frac{1}{2}&\tilde{z} &= 4 \end{eqnarray} $$ solved by $x=1$, $\tilde{y}=-3$, $\tilde{z}=2$. Hence $y=2^{\tilde{y}}=2^{-3}=1/8$ and $z$ is obtained by solving $z(z+1)=2$, that is, $z=1$ or $z=-2$.