Variable System of Equation

Question

Suppose we have three systems of three linear equations $E_1$, $E_2$ and $E_3$ such that $$aE_1+bE_2=E_3$$ and $$cE_2+dE_3=E_1$$ where $a$ and $b$ are non-zero constants. Express $c$ and $d$ in terms of $a$ and $b$.

My solution -

$$\begin{matrix} aE_{ 1 }+bE_{ 2 }-E_{ 3 } & = & 0 \\ cE_{ 2 }+dE_{ 3 }-E_{ 1 } & = & 0 \end{matrix}$$

Hence,

$$aE_{ 1 }+bE_{ 2 }-E_{ 3 }=cE_{ 2 }+dE_{ 3 }-E_{ 1 }$$

On comparing the coefficients, $b = c$ and $d = -1$. Is this result correct?

The book states that $d = \frac{1}{a}$ and $c=-\frac{b}{a}$.

Answer 1

Your result is incorrect because $$aE_1 + bE_2 - E_3 = cE_2 + dE_3 - E_1$$ does not imply the coefficients of the equations are equal. Two distinct linear combinations of three equations may be equal without implying this condition; however, for linear combinations of two equations, this is true.

To solve the problem, from the first equation, we have $$ aE_1 + bE_2 = E_3 \\ \implies E_1 = \frac{E_3 - bE_2}{a} = \frac{1}{a}E_3 + \frac{-b}{a}E_2 $$ But from the second equation, $$E_1 = cE_2 + dE_3$$ and hence $cE_2 + dE_3 = \frac{1}{a}E_3 + \frac{-b}{a}E_2$. Since these expressions are equal to each other and have only two 'variables' (read: equations), you can say $c = \frac{-b}{a}$ and $d = \frac{1}{a}$.

Answer 2

The way I went about the problem achieves the result of the book. I did the following:

$$aE_1+bE_2=E_3 (1)$$ and $$cE_2+dE_3=E_1 (2)$$

Rearranging equation $1$ to where $E_1$ is the subject I get:

$$ E_1 = \frac{E_3 - bE_2}{a} (3)$$

From there I can say equation $2$ and $3$ are equal, to get:

$$\frac{E_3 - bE_2}{a} = cE_2+dE_3$$

$$\frac {1}{a} E_3 - E_2\frac{b}{a} = cE_2+dE_3$$

$$\therefore c = \frac{1}{a}, d = \frac {-b}{a}$$