Worksheet18: Dickson 101
Current Work (editable)
CHAPTER VIII Determinants; Systems of Linear Equations 80. Solution of Two Linear Equations by Determinants of Order 2. Assume that there is a pair of numbers x and y for which (1) { a_{1} x + b_{1} y = k_{1}, { a_{2} x + b_{2} y = k_{2}. Multiply the members of the first equation by b_{2} and those of the second equation by -b_{1}, and add the resulting equations. We get (a_{1} b_{2} - a_{2} b_{1})x = k_{1} b_{2} - k_{2} b_{1}. Employing the respective multipliers -a_{2} and a_{1}, we get (a_{1} b_{2} - a_{2} b_{1})y = a_{1} k_{2} - a_{2} k_{1}. The common multiplier of x and y is (2) a_{1} b_{2} - a_{2} b_{1}, and is denoted by the symbol (2') |a_{1} b_{1}| |a_{2} b_{2}|, which is called a determinant of the second order, and also called the determinant of the coefficients of x and y in equations (1). The results above may now be written in the form (3) |a_{1} b_{1}| |a_{2} b_{2}| x = |k_{1} b_{1}| |k_{2} b_{2}|, |a_{1} b_{1}| |a_{2} b_{2}| y = |a_{1} k_{1}| |a_{2} k_{2}|. We shall call k_{1} and k_{2} the known terms of our equations (1). Hence, if D is the determinant of the coefficients of the unknowns, the product of D by any one of the unknowns is equal to the determinant obtained from D by substituting the known terms in place of the coefficients of that unknown.
Answer Key (non-editable)
\Chapter{VIII}{Determinants; Systems of Linear Equations} \Par{80. Solution of Two Linear Equations by Determinants of Order $2$.} Assume that there is a pair of numbers $x$ and $y$ for which \[ \Tag{(1)} \left\{ \begin{aligned} a_{1} x + b_{1} y &= k_{1}, \\ a_{2} x + b_{2} y &= k_{2}. \end{aligned} \right. \] Multiply the members of the first equation by $b_{2}$ and those of the second equation by $-b_{1}$, and add the resulting equations. We get \[ (a_{1} b_{2} - a_{2} b_{1})x = k_{1} b_{2} - k_{2} b_{1}. \] Employing the respective multipliers $-a_{2}$ and $a_{1}$, we get \[ (a_{1} b_{2} - a_{2} b_{1})y = a_{1} k_{2} - a_{2} k_{1}. \] The common multiplier of $x$ and $y$ is \[ \Tag{(2)} a_{1} b_{2} - a_{2} b_{1}, \] and is denoted by the symbol \[ \Tag{(2')} \left|\begin{array}{cc} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|, \] which is called a \emph{determinant of the second order}, and also called the determinant of the coefficients of $x$ and $y$ in equations \Eq{(1)}. The results above may now be written in the form \[ \Tag{(3)} \left|\begin{array}{cc} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right| x = \left|\begin{array}{cc} k_{1} & b_{1} \\ k_{2} & b_{2} \end{array}\right|,\qquad \left|\begin{array}{cc} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right| y = \left|\begin{array}{cc} a_{1} & k_{1} \\ a_{2} & k_{2} \end{array}\right|. \] We shall call $k_{1}$ and $k_{2}$ the known terms of our equations \Eq{(1)}. Hence, \begin{Thm}if $D$ is the determinant of the coefficients of the unknowns, the product of $D$ by any one of the unknowns is equal to the determinant obtained from $D$ by substituting the known terms in place of the coefficients of that unknown\end{Thm}.
Comparison of Current Work and Answer Key
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