حل المعادلات الثلاث بإيجاد قيم X, Y, Z وبطريقة المحددات في كل مما يأتي: 4y+z=0 , 2x+z=-8 , 5x+6y=z-3

$$\begin{gathered} \begin{array}{c}0+4\mathrm{y}+\mathrm{z}=0\\ 2\mathrm{x}+0+\mathrm{z}=-8\\ 5\mathrm{x}+6\mathrm{y}+2\mathrm{z}=4\end{array} \\ \begin{array}{rl}& \mathrm{x}=\frac{\mathrm{\Delta x}}{\mathrm{\Delta }}=\frac{(0+16-48)-(0+0+64)}{(0+20+12)-(0+0-16)}\\ & \mathrm{x}=\frac{-32-64}{32+16}=\frac{-96}{48}=-2\\ & \mathrm{x}=-2\end{array} \\ \begin{array}{rl}& \mathrm{y}=\frac{\mathrm{\Delta y}}{\mathrm{\Delta }}=\frac{(0+0+8)-(-40+0+0)}{48}\\ & \mathrm{y}=\frac{8+40}{48}=\frac{48}{48}=1\\ & \mathrm{y}=1\end{array} \\ \begin{array}{rl}& \mathrm{z}=\frac{\mathrm{\Delta z}}{\mathrm{\Delta }}=\frac{(0+(-160)+0)-(0+0+32)}{48}\\ & \mathrm{z}=\frac{-160-32}{48}=\frac{-192}{48}=-4\\ & \mathrm{z}=-4\end{array} \end{gathered}$$

تمارين (2-3)

(1)- جد قيمة كل مما يأتي وبين أي منها هو محدد لمصفوفة منفردة:

أ- $\left|\begin{array}{cc}\cos \mathrm{x}& \sin \mathrm{x}\\ \sin \mathrm{x}& -\cos \mathrm{x}\end{array}\right|$

$\begin{array}{rl}& \left[\right(\cos \mathrm{x})\times (-\cos \mathrm{x}\left)\right]-\left[\right(\sin \mathrm{x})\times (\sin \mathrm{x}\left)\right]\\ & ={\cos }^2\mathrm{x}-{\sin }^2\mathrm{x}\\ & ={\cos }^2\mathrm{x}+{\sin }^2\mathrm{x}=1\end{array}$

ب- $\left|\begin{array}{cc}\frac{1}{3}& \frac{1}{6}\\ 5& \frac{1}{2}\end{array}\right|$

$\begin{array}{rl}& [\left(\frac{1}{3}\right)\times \left(\frac{1}{2}\right)]-[\left(\frac{1}{6}\right)\times \left(\frac{5}{1}\right)]\\ & =\frac{1}{6}-\frac{5}{6}=\frac{1-5}{2}=\frac{-4}{6}=\frac{-2}{3}\end{array}$

ج- $\left|\begin{array}{cc}2\sqrt{2}& \sqrt{3}\\ \sqrt{3}& \sqrt{2}\end{array}\right|$

$\begin{array}{rl}& (2\sqrt{2}\times \sqrt{2})-(\sqrt{3}\times \sqrt{3})\\ & =\left(2\sqrt{4}\right)-\left(\sqrt{9}\right)=(2\times 2)-3=1\end{array}$

د- $\left|\begin{array}{cc}4& 3\\ -6& -8\end{array}\right|$

$\begin{array}{rl}& (4\times 8)-(3\times -6)=(-32)-(-18)\\ & =-32+18=-14\end{array}$

هـ- $\left|\begin{array}{ccc}1& 5& 2\\ -2& 4& 3\\ 3& 1& 1\end{array}\right|$

$\begin{array}{rl}& \left[\right(4\times 1)-(3\times 1\left)\right]-5\left[\right(-2\times 1)-(3\times 3\left)\right]+2\left[\right(-2\times 1)-(4\times 3\left)\right]\\ & =1[4-3]-5[-2-9]+2[-2-12]\\ & =1\left[1\right]-5[-11]+2[-14]\\ & =1+55-28=28\end{array}$

و- $\left|\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ -1& 0& 1\end{array}\right|$

$\begin{array}{rl}& \left[\right(5\times 1)-(6\times 0\left)\right]-2\left[\right(4\times 1)-(6\times -1\left)\right]+3\left[\right(4\times 0)-(5\times -1\left)\right]\\ & =1[5-0]-2[4-(-6\left)\right]+3[0+5]\\ & =1\left[5\right]-2[4+6]+3\left[5\right]\\ & =1\left[5\right]-2\left[10\right]+3\left[5\right]\\ & =5-20+15=0\end{array}$

تعتبر محددة لمصفوفة منفردة لأن الناتج = 0

(2)- حل المعادلات الآتية وجد قيمة x في كل مما يأتي:

أ- $\left|\begin{array}{cc}3\mathrm{x}& 3\\ 9& 4\mathrm{x}\end{array}\right|=0$

$\begin{array}{rl}& 12{\mathrm{x}}^2-27=0\\ & 12{\mathrm{x}}^2=27\to 12{\mathrm{x}}^2=27\\ & {\mathrm{x}}^2=\frac{27}{12}\to \mathrm{x}=\frac{\sqrt{27}}{\sqrt{12}}=\frac{3\sqrt{3}}{2\sqrt{2}}=\frac{3}{2}\\ & \mathrm{x}=\frac{3}{2}\end{array}$

ب- $\left|\begin{array}{cc}\mathrm{x}& \mathrm{x}\\ \mathrm{x}-1& \mathrm{x}-5\end{array}\right|=8$

$\begin{array}{rl}& \left[\mathrm{x}\right(\mathrm{x}-5\left)\right]-\left[\mathrm{x}\right(\mathrm{x}-1\left)\right]=8\\ & [{\mathrm{x}}^2-5\mathrm{x}]-[{\mathrm{x}}^2-\mathrm{x}]=8\\ & {\mathrm{x}}^2-5\mathrm{x}-{\mathrm{x}}^2+\mathrm{x}=8\\ & -5\mathrm{x}+\mathrm{x}=8\\ & -4\mathrm{x}=8\\ & \mathrm{x}=\frac{8}{-4}=-2\\ & \mathrm{x}=-2\end{array}$

ج- $\left|\begin{array}{ccc}\mathrm{x}& 2& -1\\ 1& 0& 6\\ 5& -1& 8\end{array}\right|=3$

$\begin{array}{rl}& \mathrm{x}[0-(-6\left)\right]-2[8-30]+-1[-1-\\ & \mathrm{x}[0+6]-2[8-30]-1[-1]=3\\ & \mathrm{x}\left[6\right]-2[-22]-1[-1]=3\\ & 6\mathrm{x}+44+1=3\to 6\mathrm{x}+44-2=0\\ & 6\mathrm{x}+42=0\to 6\mathrm{x}=-42\\ & \mathrm{x}=\frac{-42}{6}=7\\ & \mathrm{x}=7\end{array}$

(3)- حل المعادلتين الآتيتين في كل مما يأتي وبطريقة كرامر:

أ- 5x+3=4y, 3x+y=5

$$\begin{gathered} \begin{array}{rl}& \mathrm{\Delta }=\left|\begin{array}{ll}{\mathrm{a}}_1& {\mathrm{b}}_1\\ {\mathrm{a}}_2& {\mathrm{b}}_2\end{array}\right|=\left|\begin{array}{rr}5& -4\\ 3& 1\end{array}\right|\\ & \mathrm{\Delta }=(5\times 1)-(-4\times 3)=5+12=17\end{array} \\ \begin{array}{rl}& \mathrm{\Delta x}=\left|\begin{array}{ll}{\mathrm{c}}_1& {\mathrm{b}}_1\\ {\mathrm{c}}_2& {\mathrm{b}}_2\end{array}\right|=\left|\begin{array}{cr}-3& -4\\ 5& 1\end{array}\right|\\ & \mathrm{\Delta x}=(-3\times 1)-(-4\times 5)=-3+20=17\end{array} \\ \begin{array}{rl}& \mathrm{\Delta y}=\left|\begin{array}{ll}{\mathrm{a}}_1& {\mathrm{c}}_1\\ {\mathrm{a}}_2& {\mathrm{c}}_2\end{array}\right|=\left|\begin{array}{rr}5& -3\\ 3& 5\end{array}\right|\\ & \mathrm{\Delta y}=(5\times 5)-(-3\times 3)=25+9=34\\ & \mathrm{x}=\frac{\mathrm{\Delta x}}{\mathrm{\Delta }}=\frac{17}{17}=1 , \mathrm{y}=\frac{\mathrm{\Delta y}}{\mathrm{\Delta }}=\frac{34}{17}=2\\ & \mathrm{S}=\left\{\right(1,2\left)\right\}\end{array} \end{gathered}$$

ب- 2x-3=3y , x-1=2y

$\begin{array}{rl}& \mathrm{\Delta }=\left|\begin{array}{ll}{\mathrm{a}}_1& {\mathrm{b}}_1\\ {\mathrm{a}}_2& {\mathrm{b}}_2\end{array}\right|=\left|\begin{array}{rr}2& -3\\ 1& -2\end{array}\right|\\ & \mathrm{\Delta }=(-4\times 1)-(-3\times 1)=-4+3=1\\ & \mathrm{\Delta x}=\left|\begin{array}{ll}{\mathrm{c}}_1& {\mathrm{b}}_1\\ {\mathrm{c}}_2& {\mathrm{b}}_2\end{array}\right|=\left|\begin{array}{rr}3& -3\\ 1& -2\end{array}\right|\\ & \mathrm{\Delta x}=(3\times -2)-(1\times -3)=-6+3=-3\\ & \mathrm{\Delta y}=\left|\begin{array}{ll}{\mathrm{a}}_1& {\mathrm{c}}_1\\ {\mathrm{a}}_2& {\mathrm{c}}_2\end{array}\right|=\left|\begin{array}{rr}2& 3\\ 1& 1\end{array}\right|\\ & \mathrm{\Delta y}=(2\times 1)-(1\times 3)=2-3=-1\\ & \mathrm{x}=\frac{\mathrm{\Delta x}}{\mathrm{\Delta }}=\frac{-3}{-1}=3 , \mathrm{y}=\frac{\mathrm{\Delta y}}{\mathrm{\Delta }}=\frac{-1}{-1}=1\\ & \mathrm{S}=\left\{\right(3,1\left)\right\}\end{array}$

ج- 4y+2x=0 , 3x+5y=-1

$\begin{array}{rl}& \mathrm{\Delta }=\left|\begin{array}{ll}{\mathrm{a}}_1& {\mathrm{b}}_1\\ {\mathrm{a}}_2& {\mathrm{b}}_2\end{array}\right|=\left|\begin{array}{rr}2& 4\\ 3& 5\end{array}\right|\\ & \mathrm{\Delta }=(5\times 2)-(3\times 4)=10-12=-2\\ & \mathrm{\Delta x}=\left|\begin{array}{ll}{\mathrm{c}}_1& {\mathrm{b}}_1\\ {\mathrm{c}}_2& {\mathrm{b}}_2\end{array}\right|=\left|\begin{array}{rr}0& 4\\ -1& 5\end{array}\right|\\ & \mathrm{\Delta x}=(0\times 5)-(-1\times 4)=0-(-4)=0+4=4\\ & \mathrm{\Delta y}=\left|\begin{array}{ll}{\mathrm{a}}_1& {\mathrm{c}}_1\\ {\mathrm{a}}_2& {\mathrm{c}}_2\end{array}\right|=\left|\begin{array}{rr}2& 0\\ 3& -1\end{array}\right|\\ & \mathrm{\Delta y}=(2\times -1)-(3\times 0)=-2-0=-2\\ & \mathrm{x}=\frac{\mathrm{\Delta x}}{\mathrm{\Delta }}=\frac{4}{-2}=-2 , \mathrm{y}=\frac{\mathrm{\Delta y}}{\mathrm{\Delta }}=\frac{-2}{-2}=1\\ & \mathrm{S}=\left\{\right(-2,1\left)\right\}\end{array}$

د- 2x+3y=6 , x+y=1

$\begin{array}{rl}& \mathrm{\Delta }=\left|\begin{array}{ll}{\mathrm{a}}_1& {\mathrm{b}}_1\\ {\mathrm{a}}_2& {\mathrm{b}}_2\end{array}\right|=\left|\begin{array}{rr}2& 3\\ 1& 1\end{array}\right|\\ & \mathrm{\Delta }=(2\times 1)-(1\times 3)=2-3=-1\\ & \mathrm{\Delta x}=\left|\begin{array}{ll}{\mathrm{c}}_1& {\mathrm{b}}_1\\ {\mathrm{c}}_2& {\mathrm{b}}_2\end{array}\right|=\left|\begin{array}{rr}6& 3\\ 1& 1\end{array}\right|\\ & \mathrm{\Delta x}=(6\times 1)-(1\times 3)=6-3=3\\ & \mathrm{\Delta y}=\left|\begin{array}{ll}{\mathrm{a}}_1& {\mathrm{c}}_1\\ {\mathrm{a}}_2& {\mathrm{c}}_2\end{array}\right|=\left|\begin{array}{rr}2& 6\\ 1& 1\end{array}\right|\\ & \mathrm{\Delta y}=(2\times 1)-(1\times 6)=2-6=-4\\ & \mathrm{x}=\frac{\mathrm{\Delta x}}{\mathrm{\Delta }}=\frac{3}{-1}=-3 , \mathrm{y}=\frac{\mathrm{\Delta y}}{1}=\frac{-4}{-1}=4\\ & \mathrm{S}=\left\{\right(-3,4\left)\right\}\end{array}$

(4)- حل المعادلات الثلاث بإيجاد قيم X, Y, Z وبطريقة المحددات في كل مما يأتي:

أ- 3x+y-z=2, 2x+3y+z=11, x-y+3z=8

$$\begin{gathered} \begin{array}{rl}& 3\mathrm{x}+\mathrm{y}-\mathrm{z}=2\\ & 2\mathrm{x}+3\mathrm{y}+\mathrm{z}=11\\ & \mathrm{x}-\mathrm{y}+3\mathrm{z}=8\end{array} \\ \mathrm{x}=\frac{\mathrm{\Delta x}}{\mathrm{\Delta }}=\frac{(18+8+11)-(-24-2+33)}{(27+1+2)-(-3-3+6)} \\ \begin{array}{rl}& \mathrm{x}=\frac{37-7}{30}=\frac{30}{30}=1\\ & \mathrm{x}=1\end{array} \\ \begin{array}{rl}& \mathrm{y}=\frac{\mathrm{\Delta y}}{\mathrm{\Delta }}=\frac{(99+2-16)-(-11+24+12)}{30}\\ & \mathrm{y}=\frac{85-25}{30}=\frac{60}{30}=2\\ & \mathrm{y}=2\end{array} \\ \begin{array}{rl}& \mathrm{z}=\frac{\mathrm{\Delta z}}{\mathrm{\Delta }}=\frac{(72+11-4)-(6-33+16)}{30}\\ & \mathrm{z}=\frac{79+11}{30}=\frac{90}{30}=3\\ & \mathrm{z}=3\end{array} \end{gathered}$$

ب- 4y+z=0 , 2x+z=-8 , 5x+6y=z-3

$$\begin{gathered} \begin{array}{c}0+4\mathrm{y}+\mathrm{z}=0\\ 2\mathrm{x}+0+\mathrm{z}=-8\\ 5\mathrm{x}+6\mathrm{y}+2\mathrm{z}=4\end{array} \\ \begin{array}{rl}& \mathrm{x}=\frac{\mathrm{\Delta x}}{\mathrm{\Delta }}=\frac{(0+16-48)-(0+0+64)}{(0+20+12)-(0+0-16)}\\ & \mathrm{x}=\frac{-32-64}{32+16}=\frac{-96}{48}=-2\\ & \mathrm{x}=-2\end{array} \\ \begin{array}{rl}& \mathrm{y}=\frac{\mathrm{\Delta y}}{\mathrm{\Delta }}=\frac{(0+0+8)-(-40+0+0)}{48}\\ & \mathrm{y}=\frac{8+40}{48}=\frac{48}{48}=1\\ & \mathrm{y}=1\end{array} \\ \begin{array}{rl}& \mathrm{z}=\frac{\mathrm{\Delta z}}{\mathrm{\Delta }}=\frac{(0+(-160)+0)-(0+0+32)}{48}\\ & \mathrm{z}=\frac{-160-32}{48}=\frac{-192}{48}=-4\\ & \mathrm{z}=-4\end{array} \end{gathered}$$

ج- x+3y=2z-2 , 4x+2y=z-3 , 2x-y+z=0

$$\begin{gathered} \begin{array}{rl}& \mathrm{x}+3\mathrm{y}-2\mathrm{z}=-2\\ & 4\mathrm{x}+2\mathrm{y}-\mathrm{z}=-3\\ & 2\mathrm{x}+\mathrm{y}-\mathrm{z}=0\end{array} \\ \begin{array}{rl}& \mathrm{x}=\frac{\mathrm{\Delta x}}{\mathrm{\Delta }}=\frac{(-4+0-6)-(0-2-9)}{(2-6+8)-(-8+1+12)}\\ & \mathrm{x}=\frac{-10+11}{4-5}=\frac{1}{-1}=-1\\ & \mathrm{x}=-1\end{array} \\ \begin{array}{rl}& \mathrm{y}=\frac{\mathrm{\Delta y}}{\mathrm{\Delta }}=\frac{(-3+4-8)-(12+0-8)}{-1}\\ & \mathrm{y}=\frac{-7-4}{-1}=\frac{-11}{-1}=11\\ & \mathrm{y}=11\end{array} \\ \begin{array}{rl}& \mathrm{z}=\frac{\mathrm{\Delta z}}{\mathrm{\Delta }}=\frac{(0-18+8)-(-8+3+0)}{-1}\\ & \mathrm{z}=\frac{-10+5}{-1}=\frac{-5}{-1}=5\\ & \mathrm{z}=5\end{array} \end{gathered}$$

د- 3x+y-z=-1 , 5x+2y+z=8 , x-3y-4z=-5