Worksheet19: Dickson 102
Current Work (editable)
If D $$ 0, relations (3) uniquely determine values of x and y: x = k_{1} b_{2} - k_{2} b_{1} / D, y = a_{1} k_{2} - a_{2} k_{1} / D, and these values satisfy equations (1); for example, a_{1} x + b_{1} y = (a_{1} b_{2} - a_{2} b_{1}) k_{1} / D = k_{1}. Hence our equations (1) have been solved by determinants when D $$ 0. We shall treat in §96 the more troublesome case in which D = 0. Example. For 2x - 3y = -4, 6x - 2y = 2, we have |2 -3| |6 -2| x = |-4 -3| | 2 -2|, 14x = 14, x = 1, 14y = | 2 -4| | 6 2| = 28, y = 2. EXERCISES Solve by determinants the following systems of equations: 1. 8x - y = 34, x + 8y = 53. 2. 3x + 4y = 10, 4x + y = 9. 3. ax + by = a^{2}, bx - ay = ab. 81. Solution of Three Linear Equations by Determinants of Order 3. Consider a system of three linear equations a_{1} x + b_{1} y + c_{1} z = k_{1}, (4) a_{2} x + b_{2} y + c_{2} z = k_{2}, a_{3} x + b_{3} y + c_{3} z = k_{3}. Multiply the members of the first, second and third equations by (5) b_{2} c_{3} - b_{3} c_{2}, b_{3} c_{1} - b_{1} c_{3}, b_{1} c_{2} - b_{2} c_{1}, respectively, and add the resulting equations. We obtain an equation in which the coefficients of y and z are found to be zero, while the coefficient of x is (6) a_{1} b_{2} c_{3} - a_{1} b_{3} c_{2} + a_{2} b_{3} c_{1} - a_{2} b_{1} c_{3} + a_{3} b_{1} c_{2} - a_{3} b_{2} c_{1}.
Answer Key (non-editable)
If $D \neq 0$, relations \Eq{(3)} uniquely determine values of $x$ and $y$: \[ x = \frac{k_{1} b_{2} - k_{2} b_{1}}{D},\qquad y = \frac{a_{1} k_{2} - a_{2} k_{1}}{D}, \] and these values satisfy equations \Eq{(1)}; for example, \[ a_{1} x + b_{1} y = \frac{(a_{1} b_{2} - a_{2} b_{1}) k_{1}}{D} = k_{1}. \] Hence our equations \Eq{(1)} have been solved by determinants when $D \neq 0$. We shall treat in §96 the more troublesome case in which $D = 0$. \begin{Example} For $2x - 3y = -4$, $6x - 2y = 2$, we have \begin{gather*} \left|\begin{array}{rr} 2 & -3 \\ 6 & -2 \end{array}\right| x = \left|\begin{array}{rr} -4 & -3 \\ 2 & -2 \end{array}\right|,\qquad 14x = 14, \qquad x = 1, \\ % 14y = \left|\begin{array}{rr} 2 & -4 \\ 6 & 2 \end{array}\right| = 28, \qquad y = 2. \end{gather*} \end{Example} \begin{Exercises} Solve by determinants the following systems of equations: \begin{itemize} \item[1.] $\begin{aligned}[t] 8x - y &= 34, \\ x + 8y &= 53. \end{aligned}$ \item[2.] $\begin{aligned}[t] 3x + 4y &= 10, \\ 4x + y &= 9. \end{aligned}$ \item[3.] $\begin{aligned}[t] ax + by &= a^{2}, \\ bx - ay &= ab. \end{aligned}$ \end{itemize} \end{Exercises} \Par{81. Solution of Three Linear Equations by Determinants of Order $3$.} Consider a system of three linear equations \[ \Tag{(4)} \begin{aligned} a_{1} x + b_{1} y + c_{1} z &= k_{1}, \\ a_{2} x + b_{2} y + c_{2} z &= k_{2}, \\ a_{3} x + b_{3} y + c_{3} z &= k_{3}. \end{aligned} \] Multiply the members of the first, second and third equations by \[ \Tag{(5)} b_{2} c_{3} - b_{3} c_{2},\qquad b_{3} c_{1} - b_{1} c_{3},\qquad b_{1} c_{2} - b_{2} c_{1}, \] respectively, and add the resulting equations. We obtain an equation in which the coefficients of $y$ and $z$ are found to be zero, while the coefficient of $x$ is \[ \Tag{(6)} a_{1} b_{2} c_{3} - a_{1} b_{3} c_{2} + a_{2} b_{3} c_{1} - a_{2} b_{1} c_{3} + a_{3} b_{1} c_{2} - a_{3} b_{2} c_{1}. \]
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