Class 12: Determinants – Exercise 6.4

$\displaystyle \textbf{Question 1: }\text{Solve the following system of linear equations by Cramer's rule:}$
$\displaystyle x-2y=4$
$\displaystyle -3x+5y=-7$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle x-2y=4$
$\displaystyle -3x+5y=-7$
$\displaystyle \therefore D=\begin{vmatrix}1&-2\\-3&5\end{vmatrix}=5-6=-1\neq0$
$\displaystyle D_1=\begin{vmatrix}4&-2\\-7&5\end{vmatrix}=20-14=6$
$\displaystyle D_2=\begin{vmatrix}1&4\\-3&-7\end{vmatrix}=-7+12=5$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{6}{-1}=-6$
$\displaystyle y=\frac{D_2}{D}=\frac{5}{-1}=-5$
$\displaystyle \therefore x=-6\text{ and }y=-5$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Solve the following system of linear equations by Cramer's rule:}$
$\displaystyle 2x-y=1$
$\displaystyle 7x-2y=-7$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 2x-y=1$
$\displaystyle 7x-2y=-7$
$\displaystyle \therefore D=\begin{vmatrix}2&-1\\7&-2\end{vmatrix}=-4+7=3$
$\displaystyle D_1=\begin{vmatrix}1&-1\\-7&-2\end{vmatrix}=-2-7=-9$
$\displaystyle D_2=\begin{vmatrix}2&1\\7&-7\end{vmatrix}=-14-7=-21$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{-9}{3}=-3$
$\displaystyle y=\frac{D_2}{D}=\frac{-21}{3}=-7$
$\displaystyle \therefore x=-3\text{ and }y=-7$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Solve the following system of linear equations by Cramer's rule:}$
$\displaystyle 2x-y=17$
$\displaystyle 3x+5y=6$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 2x-y=17$
$\displaystyle 3x+5y=6$
$\displaystyle \therefore D=\begin{vmatrix}2&-1\\3&5\end{vmatrix}=10+3=13\neq0$
$\displaystyle D_1=\begin{vmatrix}17&-1\\6&5\end{vmatrix}=85+6=91$
$\displaystyle D_2=\begin{vmatrix}2&17\\3&6\end{vmatrix}=12-51=-39$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{91}{13}=7$
$\displaystyle y=\frac{D_2}{D}=\frac{-39}{13}=-3$
$\displaystyle \therefore x=7\text{ and }y=-3$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Solve the following system of linear equations by Cramer's rule:}$
$\displaystyle 3x+y=19$
$\displaystyle 3x-y=23$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 3x+y=19$
$\displaystyle 3x-y=23$
$\displaystyle \therefore D=\begin{vmatrix}3&1\\3&-1\end{vmatrix}=-3-3=-6\neq0$
$\displaystyle D_1=\begin{vmatrix}19&1\\23&-1\end{vmatrix}=-19-23=-42$
$\displaystyle D_2=\begin{vmatrix}3&19\\3&23\end{vmatrix}=69-57=12$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{-42}{-6}=7$
$\displaystyle y=\frac{D_2}{D}=\frac{12}{-6}=-2$
$\displaystyle \therefore x=7\text{ and }y=-2$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Solve the following system of linear equations by Cramer's rule:}$
$\displaystyle 2x-y=-2$
$\displaystyle 3x+4y=3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 2x-y=-2$
$\displaystyle 3x+4y=3$
$\displaystyle \therefore D=\begin{vmatrix}2&-1\\3&4\end{vmatrix}=8+3=11\neq0$
$\displaystyle D_1=\begin{vmatrix}-2&-1\\3&4\end{vmatrix}=-8+3=-5$
$\displaystyle D_2=\begin{vmatrix}2&-2\\3&3\end{vmatrix}=6+6=12$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{-5}{11}=-\frac{5}{11}$
$\displaystyle y=\frac{D_2}{D}=\frac{12}{11}$
$\displaystyle \therefore x=-\frac{5}{11}\text{ and }y=\frac{12}{11}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Solve the following system of linear equations by Cramer's rule:}$
$\displaystyle 3x+ay=4$
$\displaystyle 2x+ay=2,\quad a\neq0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 3x+ay=4$
$\displaystyle 2x+ay=2$
$\displaystyle \therefore D=\begin{vmatrix}3&a\\2&a\end{vmatrix}=3a-2a=a\neq0$
$\displaystyle D_1=\begin{vmatrix}4&a\\2&a\end{vmatrix}=4a-2a=2a$
$\displaystyle D_2=\begin{vmatrix}3&4\\2&2\end{vmatrix}=6-8=-2$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{2a}{a}=2$
$\displaystyle y=\frac{D_2}{D}=\frac{-2}{a}=-\frac{2}{a}$
$\displaystyle \therefore x=2\text{ and }y=-\frac{2}{a}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Solve the following system of linear equations by Cramer's rule:}$
$\displaystyle 2x+3y=10$
$\displaystyle x+6y=4$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 2x+3y=10$
$\displaystyle x+6y=4$
$\displaystyle \therefore D=\begin{vmatrix}2&3\\1&6\end{vmatrix}=12-3=9\neq0$
$\displaystyle D_1=\begin{vmatrix}10&3\\4&6\end{vmatrix}=60-12=48$
$\displaystyle D_2=\begin{vmatrix}2&10\\1&4\end{vmatrix}=8-10=-2$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{48}{9}=\frac{16}{3}$
$\displaystyle y=\frac{D_2}{D}=\frac{-2}{9}=-\frac{2}{9}$
$\displaystyle \therefore x=\frac{16}{3}\text{ and }y=-\frac{2}{9}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Solve the following system of linear equations by Cramer's rule:}$
$\displaystyle 5x+7y=-2$
$\displaystyle 4x+6y=-3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 5x+7y=-2$
$\displaystyle 4x+6y=-3$
$\displaystyle \therefore D=\begin{vmatrix}5&7\\4&6\end{vmatrix}=30-28=2\neq0$
$\displaystyle D_1=\begin{vmatrix}-2&7\\-3&6\end{vmatrix}=-12+21=9$
$\displaystyle D_2=\begin{vmatrix}5&-2\\4&-3\end{vmatrix}=-15+8=-7$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{9}{2}$
$\displaystyle y=\frac{D_2}{D}=\frac{-7}{2}=-\frac{7}{2}$
$\displaystyle \therefore x=\frac{9}{2}\text{ and }y=-\frac{7}{2}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Solve the following system of linear equations by Cramer's rule:}$
$\displaystyle 9x+5y=10$
$\displaystyle 3y-2x=8$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations can be written as}$
$\displaystyle 9x+5y=10$
$\displaystyle -2x+3y=8$
$\displaystyle \therefore D=\begin{vmatrix}9&5\\-2&3\end{vmatrix}=27+10=37\neq0$
$\displaystyle D_1=\begin{vmatrix}10&5\\8&3\end{vmatrix}=30-40=-10$
$\displaystyle D_2=\begin{vmatrix}9&10\\-2&8\end{vmatrix}=72+20=92$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{-10}{37}=-\frac{10}{37}$
$\displaystyle y=\frac{D_2}{D}=\frac{92}{37}$
$\displaystyle \therefore x=-\frac{10}{37}\text{ and }y=\frac{92}{37}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Solve the following system of linear equations by Cramer's rule:}$
$\displaystyle x+2y=1$
$\displaystyle 3x+y=4$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle x+2y=1$
$\displaystyle 3x+y=4$
$\displaystyle \therefore D=\begin{vmatrix}1&2\\3&1\end{vmatrix}=1-6=-5\neq0$
$\displaystyle D_1=\begin{vmatrix}1&2\\4&1\end{vmatrix}=1-8=-7$
$\displaystyle D_2=\begin{vmatrix}1&1\\3&4\end{vmatrix}=4-3=1$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{-7}{-5}=\frac{7}{5}$
$\displaystyle y=\frac{D_2}{D}=\frac{1}{-5}=-\frac{1}{5}$
$\displaystyle \therefore x=\frac{7}{5}\text{ and }y=-\frac{1}{5}$
$\displaystyle \\$

$\displaystyle \text{Solve the following system of linear equations by Cramer's rule.}$

$\displaystyle \textbf{Question 11: }3x+y+z=2,\;2x-4y+3z=-1,\;4x+y-3z=-11$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 3x+y+z=2$
$\displaystyle 2x-4y+3z=-1$
$\displaystyle 4x+y-3z=-11$
$\displaystyle \therefore D=\begin{vmatrix}3&1&1\\2&-4&3\\4&1&-3\end{vmatrix}=3(12-3)-2(-3-1)+4(3+4)=27+8+28=63\neq0$
$\displaystyle D_1=\begin{vmatrix}2&1&1\\-1&-4&3\\-11&1&-3\end{vmatrix}=2(12-3)+1(-3-1)-11(3+4)=18-4-77=-63$
$\displaystyle D_2=\begin{vmatrix}3&2&1\\2&-1&3\\4&-11&-3\end{vmatrix}=3(3+33)-2(-6+11)+4(6+1)=108-10+28=126$
$\displaystyle D_3=\begin{vmatrix}3&1&2\\2&-4&-1\\4&1&-11\end{vmatrix}=3(44+1)-2(-11-2)+4(-1+8)=135+26+28=189$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{-63}{63}=-1$
$\displaystyle y=\frac{D_2}{D}=\frac{126}{63}=2$
$\displaystyle z=\frac{D_3}{D}=\frac{189}{63}=3$
$\displaystyle \therefore x=-1,\;y=2,\;z=3$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }x-4y-z=11,\;2x-5y+2z=39,\;-3x+2y+z=1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle x-4y-z=11$
$\displaystyle 2x-5y+2z=39$
$\displaystyle -3x+2y+z=1$
$\displaystyle \therefore D=\begin{vmatrix}1&-4&-1\\2&-5&2\\-3&2&1\end{vmatrix}=1(-5-4)-(-4)(2+6)+(-1)(4-15)=-9+32+11=34\neq0$
$\displaystyle D_1=\begin{vmatrix}11&-4&-1\\39&-5&2\\1&2&1\end{vmatrix}=11(-5-4)-(-4)(39-2)+(-1)(78+5)=-99+148-83=-34$
$\displaystyle D_2=\begin{vmatrix}1&11&-1\\2&39&2\\-3&1&1\end{vmatrix}=1(39-2)-11(2+6)+(-1)(2+117)=37-88-119=-170$
$\displaystyle D_3=\begin{vmatrix}1&-4&11\\2&-5&39\\-3&2&1\end{vmatrix}=1(-5-78)-(-4)(2+117)+11(4-15)=-83+476-121=272$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{-34}{34}=-1$
$\displaystyle y=\frac{D_2}{D}=\frac{-170}{34}=-5$
$\displaystyle z=\frac{D_3}{D}=\frac{272}{34}=8$
$\displaystyle \therefore x=-1,\;y=-5,\;z=8$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }6x+y-3z=5,\;x+3y-2z=5,\;2x+y+4z=8$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 6x+y-3z=5$
$\displaystyle x+3y-2z=5$
$\displaystyle 2x+y+4z=8$
$\displaystyle \therefore D=\begin{vmatrix}6&1&-3\\1&3&-2\\2&1&4\end{vmatrix}=6(12+2)-1(4+4)-3(1-6)=84-8+15=91\neq0$
$\displaystyle D_1=\begin{vmatrix}5&1&-3\\5&3&-2\\8&1&4\end{vmatrix}=5(12+2)-1(20+16)-3(5-24)=70-36+57=91$
$\displaystyle D_2=\begin{vmatrix}6&5&-3\\1&5&-2\\2&8&4\end{vmatrix}=6(20+16)-5(4+4)-3(8-10)=216-40+6=182$
$\displaystyle D_3=\begin{vmatrix}6&1&5\\1&3&5\\2&1&8\end{vmatrix}=6(24-5)-1(8-10)+5(1-6)=114+2-25=91$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{91}{91}=1$
$\displaystyle y=\frac{D_2}{D}=\frac{182}{91}=2$
$\displaystyle z=\frac{D_3}{D}=\frac{91}{91}=1$
$\displaystyle \therefore x=1,\;y=2,\;\text{and }z=1$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }x+y=5,\;y+z=3,\;x+z=4$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations can be written as}$
$\displaystyle x+y+0z=5$
$\displaystyle 0x+y+z=3$
$\displaystyle x+0y+z=4$
$\displaystyle \therefore D=\begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}=1(1-0)-1(0-1)+0(0-1)=1+1=2\neq0$
$\displaystyle D_1=\begin{vmatrix}5&1&0\\3&1&1\\4&0&1\end{vmatrix}=5(1-0)-1(3-4)+0(0-4)=5+1=6$
$\displaystyle D_2=\begin{vmatrix}1&5&0\\0&3&1\\1&4&1\end{vmatrix}=1(3-4)-5(0-1)+0(0-3)=-1+5=4$
$\displaystyle D_3=\begin{vmatrix}1&1&5\\0&1&3\\1&0&4\end{vmatrix}=1(4-0)-1(0-3)+5(0-1)=4+3-5=2$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{6}{2}=3$
$\displaystyle y=\frac{D_2}{D}=\frac{4}{2}=2$
$\displaystyle z=\frac{D_3}{D}=\frac{2}{2}=1$
$\displaystyle \therefore x=3,\;y=2,\;\text{and }z=1$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }2y-3z=0,\;x+3y=-4,\;3x+4y=3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations can be written as}$
$\displaystyle 0x+2y-3z=0$
$\displaystyle x+3y+0z=-4$
$\displaystyle 3x+4y+0z=3$
$\displaystyle \therefore D=\begin{vmatrix}0&2&-3\\1&3&0\\3&4&0\end{vmatrix}=0-2(0-0)-3(4-9)=15\neq0$
$\displaystyle D_1=\begin{vmatrix}0&2&-3\\-4&3&0\\3&4&0\end{vmatrix}=0-2(0-0)-3(-16-9)=75$
$\displaystyle D_2=\begin{vmatrix}0&0&-3\\1&-4&0\\3&3&0\end{vmatrix}=0-0-3(3+12)=-45$
$\displaystyle D_3=\begin{vmatrix}0&2&0\\1&3&-4\\3&4&3\end{vmatrix}=0-2(3+12)+0=-30$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{75}{15}=5$
$\displaystyle y=\frac{D_2}{D}=\frac{-45}{15}=-3$
$\displaystyle z=\frac{D_3}{D}=\frac{-30}{15}=-2$
$\displaystyle \therefore x=5,\;y=-3,\;\text{and }z=-2$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }5x-7y+z=11,\;6x-8y-z=15,\;3x+2y-6z=7$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 5x-7y+z=11$
$\displaystyle 6x-8y-z=15$
$\displaystyle 3x+2y-6z=7$
$\displaystyle \therefore D=\begin{vmatrix}5&-7&1\\6&-8&-1\\3&2&-6\end{vmatrix}$
$\displaystyle =5(48+2)+7(-36+3)+(12+24)$
$\displaystyle =250-231+36=55\neq0$
$\displaystyle D_1=\begin{vmatrix}11&-7&1\\15&-8&-1\\7&2&-6\end{vmatrix}$
$\displaystyle =11(48+2)+7(-90+7)+(30+56)$
$\displaystyle =550-581+86=55$
$\displaystyle D_2=\begin{vmatrix}5&11&1\\6&15&-1\\3&7&-6\end{vmatrix}$
$\displaystyle =5(-90+7)-11(-36+3)+(42-45)$
$\displaystyle =-415+363-3=-55$
$\displaystyle D_3=\begin{vmatrix}5&-7&11\\6&-8&15\\3&2&7\end{vmatrix}$
$\displaystyle =5(-56-30)+7(42-45)+11(12+24)$
$\displaystyle =-430-21+396=-55$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{55}{55}=1$
$\displaystyle y=\frac{D_2}{D}=\frac{-55}{55}=-1$
$\displaystyle z=\frac{D_3}{D}=\frac{-55}{55}=-1$
$\displaystyle \therefore x=1,\;y=-1,\;\text{and }z=-1$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }2x-3y-4z=29,\;-2x+5y-z=-15,\;3x-y+5z=-11$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 2x-3y-4z=29$
$\displaystyle -2x+5y-z=-15$
$\displaystyle 3x-y+5z=-11$
$\displaystyle \therefore D=\begin{vmatrix}2&-3&-4\\-2&5&-1\\3&-1&5\end{vmatrix}$
$\displaystyle =2(25-1)+3(-10+3)-4(2-15)$
$\displaystyle =48-21+52=79\neq0$
$\displaystyle D_1=\begin{vmatrix}29&-3&-4\\-15&5&-1\\-11&-1&5\end{vmatrix}$
$\displaystyle =29(25-1)+3(-75-11)-4(15+55)$
$\displaystyle =696-258-280=158$
$\displaystyle D_2=\begin{vmatrix}2&29&-4\\-2&-15&-1\\3&-11&5\end{vmatrix}$
$\displaystyle =2(-75-11)-29(-10+3)-4(22+45)$
$\displaystyle =-172+203-268=-237$
$\displaystyle D_3=\begin{vmatrix}2&-3&29\\-2&5&-15\\3&-1&-11\end{vmatrix}$
$\displaystyle =2(-55-15)+3(22+45)+29(2-15)$
$\displaystyle =-140+201-377=-316$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{158}{79}=2$
$\displaystyle y=\frac{D_2}{D}=\frac{-237}{79}=-3$
$\displaystyle z=\frac{D_3}{D}=\frac{-316}{79}=-4$
$\displaystyle \therefore x=2,\;y=-3,\;\text{and }z=-4$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }x+y=1,\;x+z=-6,\;x-y-2z=3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations can be written as}$
$\displaystyle x+y+0z=1$
$\displaystyle x+0y+z=-6$
$\displaystyle x-y-2z=3$
$\displaystyle \therefore D=\begin{vmatrix}1&1&0\\1&0&1\\1&-1&-2\end{vmatrix}=1(0+1)-1(-2-1)+0(-1-0)=1+3=4\neq0$
$\displaystyle D_1=\begin{vmatrix}1&1&0\\-6&0&1\\3&-1&-2\end{vmatrix}=1(0+1)-1(12-3)+0(6-0)=1-9=-8$
$\displaystyle D_2=\begin{vmatrix}1&1&0\\1&-6&1\\1&3&-2\end{vmatrix}=1(12-3)-1(-2-1)+0(3+6)=9+3=12$
$\displaystyle D_3=\begin{vmatrix}1&1&1\\1&0&-6\\1&-1&3\end{vmatrix}=1(0-6)-1(3+6)+1(-1-0)=-6-9-1=-16$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{-8}{4}=-2$
$\displaystyle y=\frac{D_2}{D}=\frac{12}{4}=3$
$\displaystyle z=\frac{D_3}{D}=\frac{-16}{4}=-4$
$\displaystyle \therefore x=-2,\;y=3,\;\text{and }z=-4$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }x+y+z+1=0,\;ax+by+cz+d=0,\;a^2x+b^2y+c^2z+d^2=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations can be written as}$
$\displaystyle x+y+z=-1$
$\displaystyle ax+by+cz=-d$
$\displaystyle a^2x+b^2y+c^2z=-d^2$
$\displaystyle D=\begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix}$
$\displaystyle =\begin{vmatrix}1&0&0\\a&a-b&b-c\\a^2&a^2-b^2&b^2-c^2\end{vmatrix}$
$\displaystyle \left[\text{Applying }C_2\to C_1-C_2,\;C_3\to C_2-C_3\text{ simultaneously using the original columns}\right]$
$\displaystyle =(a-b)(b-c)\begin{vmatrix}1&0&0\\a&1&1\\a^2&a+b&b+c\end{vmatrix}$
$\displaystyle =(a-b)(b-c)(c-a)\neq0\qquad\ldots(1)$
$\displaystyle \text{where }a,\;b,\;c\text{ are distinct.}$
$\displaystyle D_1=\begin{vmatrix}-1&1&1\\-d&b&c\\-d^2&b^2&c^2\end{vmatrix}=-\begin{vmatrix}1&1&1\\d&b&c\\d^2&b^2&c^2\end{vmatrix}$
$\displaystyle D_1=-(d-b)(b-c)(c-d)$
$\displaystyle \left[\text{Replacing }a\text{ by }d\text{ in }(1)\right]$
$\displaystyle D_2=\begin{vmatrix}1&-1&1\\a&-d&c\\a^2&-d^2&c^2\end{vmatrix}=-\begin{vmatrix}1&1&1\\a&d&c\\a^2&d^2&c^2\end{vmatrix}$
$\displaystyle D_2=-(a-d)(d-c)(c-a)$
$\displaystyle \left[\text{Replacing }b\text{ by }d\text{ in }(1)\right]$
$\displaystyle D_3=\begin{vmatrix}1&1&-1\\a&b&-d\\a^2&b^2&-d^2\end{vmatrix}=-\begin{vmatrix}1&1&1\\a&b&d\\a^2&b^2&d^2\end{vmatrix}$
$\displaystyle D_3=-(a-b)(b-d)(d-a)$
$\displaystyle \left[\text{Replacing }c\text{ by }d\text{ in }(1)\right]$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=-\frac{(d-b)(b-c)(c-d)}{(a-b)(b-c)(c-a)}$
$\displaystyle y=\frac{D_2}{D}=-\frac{(a-d)(d-c)(c-a)}{(a-b)(b-c)(c-a)}$
$\displaystyle z=\frac{D_3}{D}=-\frac{(a-b)(b-d)(d-a)}{(a-b)(b-c)(c-a)}$
$\displaystyle \therefore x=-\frac{(d-b)(c-d)}{(a-b)(c-a)},\quad y=-\frac{(a-d)(d-c)}{(a-b)(b-c)},\quad z=-\frac{(b-d)(d-a)}{(b-c)(c-a)}$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }x+y+z+w=2,\;x-2y+2z+2w=-6,\;2x+y-2z+2w=-5,\;3x-y+3z-3w=-3$
$\displaystyle \text{Answer:}$
$\displaystyle D=\begin{vmatrix}1&1&1&1\\1&-2&2&2\\2&1&-2&2\\3&-1&3&-3\end{vmatrix}$
$\displaystyle =1\begin{vmatrix}-2&2&2\\1&-2&2\\-1&3&-3\end{vmatrix}-1\begin{vmatrix}1&2&2\\2&-2&2\\3&3&-3\end{vmatrix}+1\begin{vmatrix}1&-2&2\\2&1&2\\3&-1&-3\end{vmatrix}-1\begin{vmatrix}1&-2&2\\2&1&-2\\3&-1&3\end{vmatrix}$
$\displaystyle =4-48-35-15=-94\neq0$
$\displaystyle D_1=\begin{vmatrix}2&1&1&1\\-6&-2&2&2\\-5&1&-2&2\\-3&-1&3&-3\end{vmatrix}$
$\displaystyle =2(4)-1(-84)+1(64)-1(-32)$
$\displaystyle =8+84+64+32=188$
$\displaystyle D_2=\begin{vmatrix}1&2&1&1\\1&-6&2&2\\2&-5&-2&2\\3&-3&3&-3\end{vmatrix}$
$\displaystyle =1(-84)-2(48)+1(-33)-1(69)$
$\displaystyle =-84-96-33-69=-282$
$\displaystyle D_3=\begin{vmatrix}1&1&2&1\\1&-2&-6&2\\2&1&-5&2\\3&-1&-3&-3\end{vmatrix}$
$\displaystyle =1(-64)-1(-33)+2(-35)-1(40)$
$\displaystyle =-64+33-70-40=-141$
$\displaystyle D_4=\begin{vmatrix}1&1&1&2\\1&-2&2&-6\\2&1&-2&-5\\3&-1&3&-3\end{vmatrix}$
$\displaystyle =1(-32)-1(-69)+1(40)-2(15)$
$\displaystyle =-32+69+40-30=47$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{188}{-94}=-2$
$\displaystyle y=\frac{D_2}{D}=\frac{-282}{-94}=3$
$\displaystyle z=\frac{D_3}{D}=\frac{-141}{-94}=\frac{3}{2}$
$\displaystyle w=\frac{D_4}{D}=\frac{47}{-94}=-\frac{1}{2}$
$\displaystyle \therefore x=-2,\;y=3,\;z=\frac{3}{2},\;\text{and }w=-\frac{1}{2}$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }2x-3z+w=1,\;x-y+2w=1,\;-3y+z+w=1,\;x+y+z=1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations can be written as}$
$\displaystyle 2x+0y-3z+w=1$
$\displaystyle x-y+0z+2w=1$
$\displaystyle 0x-3y+z+w=1$
$\displaystyle x+y+z+0w=1$
$\displaystyle D=\begin{vmatrix}2&0&-3&1\\1&-1&0&2\\0&-3&1&1\\1&1&1&0\end{vmatrix}$
$\displaystyle =2\begin{vmatrix}-1&0&2\\-3&1&1\\1&1&0\end{vmatrix}-3\begin{vmatrix}1&-1&2\\0&-3&1\\1&1&0\end{vmatrix}-\begin{vmatrix}1&-1&0\\0&-3&1\\1&1&1\end{vmatrix}$
$\displaystyle =2(-7)-3(4)-(-5)=-21\neq0$
$\displaystyle D_1=\begin{vmatrix}1&0&-3&1\\1&-1&0&2\\1&-3&1&1\\1&1&1&0\end{vmatrix}$
$\displaystyle =\begin{vmatrix}-1&0&2\\-3&1&1\\1&1&0\end{vmatrix}-3\begin{vmatrix}1&-1&2\\1&-3&1\\1&1&0\end{vmatrix}-\begin{vmatrix}1&-1&0\\1&-3&1\\1&1&1\end{vmatrix}$
$\displaystyle =-7-3(6)+4=-21$
$\displaystyle D_2=\begin{vmatrix}2&1&-3&1\\1&1&0&2\\0&1&1&1\\1&1&1&0\end{vmatrix}$
$\displaystyle =2\begin{vmatrix}1&0&2\\1&1&1\\1&1&0\end{vmatrix}-\begin{vmatrix}1&0&2\\0&1&1\\1&1&0\end{vmatrix}-3\begin{vmatrix}1&1&2\\0&1&1\\1&1&0\end{vmatrix}-\begin{vmatrix}1&1&0\\0&1&1\\1&1&1\end{vmatrix}$
$\displaystyle =2(-1)-(-3)-3(-2)-1=6$
$\displaystyle D_3=\begin{vmatrix}2&0&1&1\\1&-1&1&2\\0&-3&1&1\\1&1&1&0\end{vmatrix}$
$\displaystyle =2\begin{vmatrix}-1&1&2\\-3&1&1\\1&1&0\end{vmatrix}+\begin{vmatrix}1&-1&2\\0&-3&1\\1&1&0\end{vmatrix}-\begin{vmatrix}1&-1&1\\0&-3&1\\1&1&1\end{vmatrix}$
$\displaystyle =2(-6)+4-(-2)=-6$
$\displaystyle D_4=\begin{vmatrix}2&0&-3&1\\1&-1&0&1\\0&-3&1&1\\1&1&1&1\end{vmatrix}$
$\displaystyle =2\begin{vmatrix}-1&0&1\\-3&1&1\\1&1&1\end{vmatrix}-3\begin{vmatrix}1&-1&1\\0&-3&1\\1&1&1\end{vmatrix}-\begin{vmatrix}1&-1&0\\0&-3&1\\1&1&1\end{vmatrix}$
$\displaystyle =2(-4)-3(-2)-(-5)=3$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{-21}{-21}=1$
$\displaystyle y=\frac{D_2}{D}=\frac{6}{-21}=-\frac{2}{7}$
$\displaystyle z=\frac{D_3}{D}=\frac{-6}{-21}=\frac{2}{7}$
$\displaystyle w=\frac{D_4}{D}=\frac{3}{-21}=-\frac{1}{7}$
$\displaystyle \therefore x=1,\;y=-\frac{2}{7},\;z=\frac{2}{7},\;\text{and }w=-\frac{1}{7}$
$\displaystyle \\$

$\displaystyle \text{Solve the following system of linear equations is inconsistent.}$

$\displaystyle \textbf{Question 22: }2x-y=5,\;4x-2y=7$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 2x-y=5$
$\displaystyle 4x-2y=7$
$\displaystyle D=\begin{vmatrix}2&-1\\4&-2\end{vmatrix}=-4+4=0$
$\displaystyle D_1=\begin{vmatrix}5&-1\\7&-2\end{vmatrix}=-10+7=-3$
$\displaystyle D_2=\begin{vmatrix}2&5\\4&7\end{vmatrix}=14-20=-6$
$\displaystyle \text{Since }D=0,\;D_1\neq0\text{ and }D_2\neq0,$
$\displaystyle \text{the given system of linear equations is inconsistent.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }3x+y=5,\;-6x-2y=9$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 3x+y=5$
$\displaystyle -6x-2y=9$
$\displaystyle D=\begin{vmatrix}3&1\\-6&-2\end{vmatrix}=-6+6=0$
$\displaystyle D_1=\begin{vmatrix}5&1\\9&-2\end{vmatrix}=-10-9=-19$
$\displaystyle D_2=\begin{vmatrix}3&5\\-6&9\end{vmatrix}=27+30=57$
$\displaystyle \text{Since }D=0,\;D_1\neq0\text{ and }D_2\neq0,$
$\displaystyle \text{the given system of linear equations is inconsistent.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }3x-y+2z=3,\;2x+y+3z=5,\;x-2y-z=1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 3x-y+2z=3$
$\displaystyle 2x+y+3z=5$
$\displaystyle x-2y-z=1$
$\displaystyle D=\begin{vmatrix}3&-1&2\\2&1&3\\1&-2&-1\end{vmatrix}=3(-1+6)+1(-2-3)+2(-4-1)=0$
$\displaystyle D_1=\begin{vmatrix}3&-1&2\\5&1&3\\1&-2&-1\end{vmatrix}=3(-1+6)+1(-5-3)+2(-10-1)=-15$
$\displaystyle D_2=\begin{vmatrix}3&3&2\\2&5&3\\1&1&-1\end{vmatrix}=3(-5-3)-3(-2-3)+2(2-5)=-15$
$\displaystyle D_3=\begin{vmatrix}3&-1&3\\2&1&5\\1&-2&1\end{vmatrix}=3(1+10)+1(2-5)+3(-4-1)=15$
$\displaystyle \text{Since }D=0,\;D_1\neq0,\;D_2\neq0\text{ and }D_3\neq0,$
$\displaystyle \text{the given system of linear equations is inconsistent.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }3x-y+2z=6,\;2x-y+z=2,\;3x+6y+5z=20$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations is}$
$\displaystyle 3x-y+2z=6$
$\displaystyle 2x-y+z=2$
$\displaystyle 3x+6y+5z=20$
$\displaystyle D=\begin{vmatrix}3&-1&2\\2&-1&1\\3&6&5\end{vmatrix}$
$\displaystyle =3(-5-6)+1(10-3)+2(12+3)$
$\displaystyle =-33+7+30=4\neq0$
$\displaystyle \therefore\text{ the system is consistent and has a unique solution.}$
$\displaystyle D_1=\begin{vmatrix}6&-1&2\\2&-1&1\\20&6&5\end{vmatrix}$
$\displaystyle =6(-5-6)+1(10-20)+2(12+20)$
$\displaystyle =-66-10+64=-12$
$\displaystyle D_2=\begin{vmatrix}3&6&2\\2&2&1\\3&20&5\end{vmatrix}$
$\displaystyle =3(10-20)-6(10-3)+2(40-6)$
$\displaystyle =-30-42+68=-4$
$\displaystyle D_3=\begin{vmatrix}3&-1&6\\2&-1&2\\3&6&20\end{vmatrix}$
$\displaystyle =3(-20-12)+1(40-6)+6(12+3)$
$\displaystyle =-96+34+90=28$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{-12}{4}=-3$
$\displaystyle y=\frac{D_2}{D}=\frac{-4}{4}=-1$
$\displaystyle z=\frac{D_3}{D}=\frac{28}{4}=7$
$\displaystyle \therefore x=-3,\;y=-1,\;\text{and }z=7$
$\displaystyle \\$

$\displaystyle \text{Show that each of the following systems of linear equations has infinitely many} \\ \text{solutions and solve them }(26\text{-}30).$

$\displaystyle \textbf{Question 26: }x-y+z=3,\;2x+y-z=2,\;-x-2y+2z=1$
$\displaystyle \text{Answer:}$
$\displaystyle D=\begin{vmatrix}1&-1&1\\2&1&-1\\-1&-2&2\end{vmatrix}$
$\displaystyle =1(2-2)+1(4-1)+1(-4+1)=0$
$\displaystyle D_1=\begin{vmatrix}3&-1&1\\2&1&-1\\1&-2&2\end{vmatrix}$
$\displaystyle =3(2-2)+1(4+1)+1(-4-1)=0$
$\displaystyle D_2=\begin{vmatrix}1&3&1\\2&2&-1\\-1&1&2\end{vmatrix}$
$\displaystyle =1(4+1)-3(4-1)+1(2+2)=0$
$\displaystyle D_3=\begin{vmatrix}1&-1&3\\2&1&2\\-1&-2&1\end{vmatrix}$
$\displaystyle =1(1+4)+1(2+2)+3(-4+1)=0$
$\displaystyle \therefore D=D_1=D_2=D_3=0$
$\displaystyle \text{To examine the solutions, add the first and second equations.}$
$\displaystyle (x-y+z)+(2x+y-z)=3+2$
$\displaystyle 3x=5$
$\displaystyle \Rightarrow x=\frac{5}{3}$
$\displaystyle \text{Substituting }x=\frac{5}{3}\text{ in }x-y+z=3,\text{ we get}$
$\displaystyle \frac{5}{3}-y+z=3$
$\displaystyle \Rightarrow z-y=\frac{4}{3}$
$\displaystyle \Rightarrow z=y+\frac{4}{3}$
$\displaystyle \text{Let }y=\lambda,\text{ where }\lambda\in\mathbb{R}.$
$\displaystyle \therefore x=\frac{5}{3},\quad y=\lambda,\quad z=\lambda+\frac{4}{3}$
$\displaystyle \text{These values also satisfy the third equation.}$
$\displaystyle \therefore\text{ the system has infinitely many solutions.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }x+2y=5,\;3x+6y=15$
$\displaystyle \text{Answer:}$
$\displaystyle D=\begin{vmatrix}1&2\\3&6\end{vmatrix}=6-6=0$
$\displaystyle D_1=\begin{vmatrix}5&2\\15&6\end{vmatrix}=30-30=0$
$\displaystyle D_2=\begin{vmatrix}1&5\\3&15\end{vmatrix}=15-15=0$
$\displaystyle \therefore D=D_1=D_2=0$
$\displaystyle \text{To examine the solutions, consider the first equation.}$
$\displaystyle x+2y=5$
$\displaystyle \text{Let }y=\lambda,\text{ where }\lambda\in\mathbb{R}.$
$\displaystyle \therefore x=5-2\lambda$
$\displaystyle \text{The second equation is three times the first equation.}$
$\displaystyle \therefore x=5-2\lambda,\quad y=\lambda,\quad\lambda\in\mathbb{R}$
$\displaystyle \text{Hence, the system has infinitely many solutions.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }x+y-z=0,\;x-2y+z=0,\;3x+6y-5z=0$
$\displaystyle \text{Answer:}$
$\displaystyle D=\begin{vmatrix}1&1&-1\\1&-2&1\\3&6&-5\end{vmatrix}$
$\displaystyle =1(10-6)-1(-5-3)-1(6+6)=0$
$\displaystyle D_1=\begin{vmatrix}0&1&-1\\0&-2&1\\0&6&-5\end{vmatrix}=0$
$\displaystyle D_2=\begin{vmatrix}1&0&-1\\1&0&1\\3&0&-5\end{vmatrix}=0$
$\displaystyle D_3=\begin{vmatrix}1&1&0\\1&-2&0\\3&6&0\end{vmatrix}=0$
$\displaystyle \therefore D=D_1=D_2=D_3=0$
$\displaystyle \text{To examine the solutions, add the first and second equations.}$
$\displaystyle (x+y-z)+(x-2y+z)=0$
$\displaystyle 2x-y=0$
$\displaystyle \Rightarrow y=2x$
$\displaystyle \text{Substituting }y=2x\text{ in }x+y-z=0,\text{ we get}$
$\displaystyle x+2x-z=0$
$\displaystyle \Rightarrow z=3x$
$\displaystyle \text{Let }x=\lambda,\text{ where }\lambda\in\mathbb{R}.$
$\displaystyle \therefore x=\lambda,\quad y=2\lambda,\quad z=3\lambda$
$\displaystyle \text{These values also satisfy the third equation.}$
$\displaystyle \text{Hence, the system has infinitely many solutions.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }2x+y-2z=4,\;x-2y+z=-2,\;5x-5y+z=-2$
$\displaystyle \text{Answer:}$
$\displaystyle D=\begin{vmatrix}2&1&-2\\1&-2&1\\5&-5&1\end{vmatrix}$
$\displaystyle =2(-2+5)-1(1-5)-2(-5+10)=0$
$\displaystyle D_1=\begin{vmatrix}4&1&-2\\-2&-2&1\\-2&-5&1\end{vmatrix}$
$\displaystyle =4(-2+5)-1(-2+2)-2(10-4)=0$
$\displaystyle D_2=\begin{vmatrix}2&4&-2\\1&-2&1\\5&-2&1\end{vmatrix}$
$\displaystyle =2(-2+2)-4(1-5)-2(-2+10)=0$
$\displaystyle D_3=\begin{vmatrix}2&1&4\\1&-2&-2\\5&-5&-2\end{vmatrix}$
$\displaystyle =2(4-10)-1(-2+10)+4(-5+10)=0$
$\displaystyle \therefore D=D_1=D_2=D_3=0$
$\displaystyle \text{To examine the solutions, multiply the second equation by }2\text{ and subtract it from the first equation.}$
$\displaystyle (2x+y-2z)-2(x-2y+z)=4-2(-2)$
$\displaystyle 5y-4z=8$
$\displaystyle \Rightarrow y=\frac{8+4z}{5}$
$\displaystyle \text{Substituting this in }x-2y+z=-2,\text{ we get}$
$\displaystyle x=\frac{6+3z}{5}$
$\displaystyle \text{Let }z=5\lambda-2,\text{ where }\lambda\in\mathbb{R}.$
$\displaystyle \therefore x=3\lambda,\quad y=4\lambda,\quad z=5\lambda-2$
$\displaystyle \text{These values also satisfy the third equation.}$
$\displaystyle \text{Hence, the system has infinitely many solutions.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }x-y+3z=6,\;x+3y-3z=-4,\;5x+3y+3z=10$
$\displaystyle \text{Answer:}$
$\displaystyle D=\begin{vmatrix}1&-1&3\\1&3&-3\\5&3&3\end{vmatrix}$
$\displaystyle =1(9+9)+1(3+15)+3(3-15)=0$
$\displaystyle D_1=\begin{vmatrix}6&-1&3\\-4&3&-3\\10&3&3\end{vmatrix}$
$\displaystyle =6(9+9)+1(-12+30)+3(-12-30)=0$
$\displaystyle D_2=\begin{vmatrix}1&6&3\\1&-4&-3\\5&10&3\end{vmatrix}$
$\displaystyle =1(-12+30)-6(3+15)+3(10+20)=0$
$\displaystyle D_3=\begin{vmatrix}1&-1&6\\1&3&-4\\5&3&10\end{vmatrix}$
$\displaystyle =1(30+12)+1(10+20)+6(3-15)=0$
$\displaystyle \therefore D=D_1=D_2=D_3=0$
$\displaystyle \text{To examine the solutions, subtract the first equation from the second equation.}$
$\displaystyle (x+3y-3z)-(x-y+3z)=-4-6$
$\displaystyle 4y-6z=-10$
$\displaystyle \Rightarrow y=\frac{3z-5}{2}$
$\displaystyle \text{Substituting this in }x-y+3z=6,\text{ we get}$
$\displaystyle x=\frac{7-3z}{2}$
$\displaystyle \text{Let }z=2\lambda+1,\text{ where }\lambda\in\mathbb{R}.$
$\displaystyle \therefore x=2-3\lambda,\quad y=3\lambda-1,\quad z=2\lambda+1$
$\displaystyle \text{These values also satisfy the third equation.}$
$\displaystyle \text{Hence, the system has infinitely many solutions.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: }$
$\displaystyle \text{A salesman has the following record of sales during three months for three items }A,\;B\text{ and }C$
$\displaystyle \text{which have different rates of commission.}$
$\displaystyle \begin{array}{|c|c|c|c|c|}\hline \text{Month}&A&B&C&\text{Total commission drawn (in Rs.)}\\ \hline \text{January}&90&100&20&800\\ \text{February}&130&50&40&900\\ \text{March}&60&100&30&850\\ \hline \end{array}$
$\displaystyle \text{Find the rates of commission on items }A,\;B\text{ and }C\text{ using the determinant method.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let Rs. }x,\;\text{Rs. }y\text{ and Rs. }z\text{ be the rates of commission per item on }A,\;B\text{ and }C,\text{ respectively.}$
$\displaystyle \text{From the given data, we obtain}$
$\displaystyle 90x+100y+20z=800$
$\displaystyle 130x+50y+40z=900$
$\displaystyle 60x+100y+30z=850$
$\displaystyle \text{Dividing each equation by }10,\text{ we get}$
$\displaystyle 9x+10y+2z=80$
$\displaystyle 13x+5y+4z=90$
$\displaystyle 6x+10y+3z=85$
$\displaystyle D=\begin{vmatrix}9&10&2\\13&5&4\\6&10&3\end{vmatrix}$
$\displaystyle =9(15-40)-10(39-24)+2(130-30)$
$\displaystyle =-225-150+200=-175\neq0$
$\displaystyle D_1=\begin{vmatrix}80&10&2\\90&5&4\\85&10&3\end{vmatrix}$
$\displaystyle =80(15-40)-10(270-340)+2(900-425)$
$\displaystyle =-2000+700+950=-350$
$\displaystyle D_2=\begin{vmatrix}9&80&2\\13&90&4\\6&85&3\end{vmatrix}$
$\displaystyle =9(270-340)-80(39-24)+2(1105-540)$
$\displaystyle =-630-1200+1130=-700$
$\displaystyle D_3=\begin{vmatrix}9&10&80\\13&5&90\\6&10&85\end{vmatrix}$
$\displaystyle =9(425-900)-10(1105-540)+80(130-30)$
$\displaystyle =-4275-5650+8000=-1925$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{-350}{-175}=2$
$\displaystyle y=\frac{D_2}{D}=\frac{-700}{-175}=4$
$\displaystyle z=\frac{D_3}{D}=\frac{-1925}{-175}=11$
$\displaystyle \therefore\text{ the commission rates on items }A,\;B\text{ and }C\text{ are Rs. }2,\;\text{Rs. }4\text{ and Rs. }11\text{ per item, respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 32: }$
$\displaystyle \text{An automobile company uses three types of steel }S_1,\;S_2\text{ and }S_3$
$\displaystyle \text{for producing three types of cars }C_1,\;C_2\text{ and }C_3.$
$\displaystyle \text{The steel requirements, in tons, for each type of car are given below:}$
$\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Steel/Cars}&C_1&C_2&C_3\\ \hline S_1&2&3&4\\ S_2&1&1&2\\ S_3&3&2&1\\ \hline \end{array}$
$\displaystyle \text{Using Cramer's rule, find the number of cars of each type that can be produced using}$
$\displaystyle 29,\;13\text{ and }16\text{ tons of steel of types }S_1,\;S_2\text{ and }S_3,\text{ respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }x,\;y\text{ and }z\text{ be the numbers of cars of types }C_1,\;C_2\text{ and }C_3,\text{ respectively.}$
$\displaystyle \text{From the given data, we obtain}$
$\displaystyle 2x+3y+4z=29$
$\displaystyle x+y+2z=13$
$\displaystyle 3x+2y+z=16$
$\displaystyle D=\begin{vmatrix}2&3&4\\1&1&2\\3&2&1\end{vmatrix}$
$\displaystyle =2(1-4)-3(1-6)+4(2-3)$
$\displaystyle =-6+15-4=5\neq0$
$\displaystyle D_1=\begin{vmatrix}29&3&4\\13&1&2\\16&2&1\end{vmatrix}$
$\displaystyle =29(1-4)-3(13-32)+4(26-16)$
$\displaystyle =-87+57+40=10$
$\displaystyle D_2=\begin{vmatrix}2&29&4\\1&13&2\\3&16&1\end{vmatrix}$
$\displaystyle =2(13-32)-29(1-6)+4(16-39)$
$\displaystyle =-38+145-92=15$
$\displaystyle D_3=\begin{vmatrix}2&3&29\\1&1&13\\3&2&16\end{vmatrix}$
$\displaystyle =2(16-26)-3(16-39)+29(2-3)$
$\displaystyle =-20+69-29=20$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{D_1}{D}=\frac{10}{5}=2$
$\displaystyle y=\frac{D_2}{D}=\frac{15}{5}=3$
$\displaystyle z=\frac{D_3}{D}=\frac{20}{5}=4$
$\displaystyle \therefore 2\text{ cars of type }C_1,\;3\text{ cars of type }C_2\text{ and }4\text{ cars of type }C_3\text{ can be produced.}$
$\displaystyle \\$