Instructor's Solutions Manual, Section 1A
Exercise 10
**10** Explain why there does not exist $\lambda \in C$ such that
1(2 - 31,5 + 4i, -6 + 7i) = (12 - 5i,7 + 22i, —32 - 9i).
SOLUTION. The equation above is equivalent to the equation
(1(2 - 31), 1(5 + 4i), 1(-6 + 7i)) = (12 - 5i, 7 + 221, -32 - 9i),
which is equivalent to the three equations
1(2 - 3і) = 12 - 5i, 1(5 + 4) = 7 + 221, 1(-6 + 71) = -32 - 91.
The first equation above implies that
12 - 5i
1 =
2 - 3і
12 - 5i
＝
1 2 + 3і
2- 3і
2 + 31
= (24 + 15) + (36 - 10)
22 + 32
= 3 + 2i.
The choice of 1 forced by the equation above indeed satisfies the second
required equation 1(5 + 4i) = 7 + 22i.
However, with this choice of 1 we have
1(-6 + 7i) = -32 + 9i, which shows that the third required equation is not
satisfied.
Thus no choice of 1 € C satisfies all three required equations.
Edward Frenkel
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Commentary:
$\textbf{Exercise 10.}$ Explain why there does not exist $u\in \mathbb{C}$ such that $(u - 2i, u + 3i, -u + 7i) = (4 - 2i, 5 + 3i, -1 - 7i).$
$\textbf{Solution 10.}$ The equation above is equivalent to the equation $((u - 2i), (u + 3i), -(u + 7i)) = (4 - 2i, 5 + 3i, -1 - 7i),$ which is equivalent to the three equations $(u - 2i) = 4 - 2i, \quad (u + 3i) = 5 + 3i, \quad -(u + 7i) = -1 - 7i.$ The first equation above implies that $u = \frac{4 - 2i}{2 - 2i} = \frac{4 - 2i}{2 - 2i} \cdot \frac{2 + 2i}{2 + 2i} = \frac{4 + 4i}{2^2 + 2^2} = 1 + i.$ The choice of $u$ forced by the equation above indeed satisfies the second required equation $(u + 3i) = 5 + 3i$. However, with this choice of $u$ we have $-(u + 7i) = -8 + i$, which shows that the third required equation is not satisfied.
Thus no choice of $u\in \mathbb{C}$ satisfies all three required equations.
$\textit{Commentary:}$ This exercise demonstrates that a system of equations over the complex numbers may have no solution.
It presents a system of three equations in one complex variable $u$, and shows that no value of $u$ can simultaneously satisfy all three equations.
The proof proceeds by solving for $u$ using the first equation, and then checking whether this value of $u$ satisfies the other two equations. The second equation is satisfied, but the third is not, proving that the system has no solution.
This example illustrates the importance of checking the consistency of a system of equations before attempting to solve it. In the complex numbers (as in any field), a linear system of equations can have zero, one, or infinitely many solutions, depending on the coefficients of the equations.
$\textit{Example:}$ Consider the following system of equations over $\mathbb{C}$:
$\begin{align*} z + (1+i)w &= 2 \ (2-i)z - w &= 1+3i \end{align*}$ This system has a unique solution, which can be found by substitution or elimination: $z = \frac{7-i}{5}, \quad w = \frac{3+4i}{5}$
In $\mathbb{R}$, a system of linear equations can also have zero, one, or infinitely many solutions.
The same is true in $\mathbb{F}_p$ for prime $p$.
The consistency and number of solutions of a linear system can be determined by studying the rank of the coefficient matrix and the augmented matrix of the system.
In the field of complex numbers, systems of polynomial equations are of particular interest.
The Fundamental Theorem of Algebra guarantees that a single nonconstant polynomial equation always has a solution in $\mathbb{C}$, but a system of polynomial equations may have no solution, a finite number of solutions, or infinitely many solutions. The study of such systems is a major topic in algebraic geometry.
The solvability of systems of equations over various fields is also a central topic in linear algebra and has numerous applications in science and engineering.
Techniques such as Gaussian elimination, Cramer's rule, and matrix inversion are used to solve linear systems, while more advanced methods from commutative algebra and algebraic geometry are employed for systems of polynomial equations.
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