I have the two equation $ac - bd = 1$ and $ad + bc = 0$. How can I solve these two equations for $c$ and $d$ in terms of $a$ and $b$?
Solving two equations for two unknown in terms of two variables
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algebra-precalculus
systems-of-equations
2 Answers
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Just use substitution. From the first, $c=\frac {1+bd}a$. Plug that into the second and solve for $d$.
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For this one, you can use Cramer's formulae (not recommended in the general case). The augmented matrix of the system (with $c, d$ as unknowns) is $$\begin{bmatrix}\!\!\!\begin{array}{rr|r}a&-b&1\\b&a&0\end{array}\!\!\!\end{bmatrix}, $$ hence the solutions, if the determinant of the system $D=a^2+b^2$ is non-zero, are $$c=\frac{\begin{bmatrix}\!\!\!\begin{array}{rr|r}1&-b\\0&a\end{array}\!\!\!\end{bmatrix}}{D}=\frac a{a^2+b^2},\qquad d= \frac{\begin{bmatrix}\!\!\!\begin{array}{rr|r}a&1\\b&0\end{array}\!\!\!\end{bmatrix}}{D}=-\frac b{a^2+b^2}.$$ No solution on $\mathbf R$ if $a^2+b^2=0$, since it implies $a=b=0$. On $\mathbf C$, if $a^2+b^2=0$, $a$, hence $b\ne 0$, the system has rank $1$ and is equivalent to the unique equation $ad+bc=0$, whence $$d=-\frac{bc}a=\pm\frac{iac}a=\pm ic.$$