Question 15 Marks
A salesman has the following record of sales during three months for three items A, B and C which have different rates of commission.
|
Month
|
Sale of units
|
Total commission drawn (in Rs.)
|
|
|
A
|
B
|
C
|
|
|
Jan
|
90
|
100
|
20
|
800
|
|
Feb
|
130
|
50
|
40
|
900
|
|
March
|
60
|
100
|
30
|
850
|
Find out the rates of commission on items A, B and C by using determinant method.
AnswerLet the rates of commissions on iteams A, B and C be x, y and z respectively.
Then we can express the given modal as system of linear equations
90x + 100y + 20z = 800
130x + 50y + 40z = 900
60x + 100y + 30z = 850
We will solve this using the Cramer's rule
Here,
$\text{D}=\begin{vmatrix}90&100&20\\130&50&40\\60&100&30\end{vmatrix}=\begin{vmatrix}-170&0&-60\\130&50&40\\-200&0&-50\end{vmatrix}$
$=50(8500-12000)=-175000$
$\text{D}_1=\begin{vmatrix}800&100&20\\900&50&40\\60&100&30\end{vmatrix}=\begin{vmatrix}-170&0&-60\\130&50&40\\-950&0&-50\end{vmatrix}$
$=50(50000-57000)=-350000$
$\text{D}_2=\begin{vmatrix}90&800&20\\130&900&40\\60&850&30\end{vmatrix}=\begin{vmatrix}90&800&20\\-50&-700&0\\-75&-350&0\end{vmatrix}$
$=20(17500-52500)=-700000$
$\text{D}_3=\begin{vmatrix}90&100&800\\130&50&900\\60&100&850\end{vmatrix}=\begin{vmatrix}-170&0&-1000\\130&50&900\\-200&0&-950\end{vmatrix}$
$=50(161500-200000)=-1925000$
$\therefore\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-350000}{-175000}=2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-700000}{-175000}=4$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-1925000}{-175000}=11$
$\therefore$ The rates of commission of iteam A, B and C are 2%, 4% and 11% respectively.
View full question & answer→Question 25 Marks
Solve the following systems of linear equations by cramer's rule:
$x + y + z + 1 = 0,$
$ax + by + cz + d = 0,$
$a^2x + b^2y + x^2z + d^2 = 0$
AnswerThese equations can be written as
$x + y + z + 1 = 0$
$ax + by + cz + d = 0$
$a^2x + b^2y + x^2z + d^2 = 0$
$\text{D}=\begin{vmatrix}1&1&1\\\text{a}&\text{b}&\text{c}\\\text{a}^2&\text{b}^2&\text{c}^2 \end{vmatrix}$
$=\begin{vmatrix}1&0&0\\\text{a}&\text{a}-\text{b}&\text{b}-\text{c}\\\text{a}^2&\text{a}^2-\text{b}^2&\text{b}^2-\text{c}^2 \end{vmatrix} [$Applying $C_2 → C_1 - C_2, C_3 → C_2 - C_3]$
Taking $(b - a)$ and $(c - a)$ common from $C_1$ and $C_2,$ respectively, we get
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}1&0&0\\\text{a}&1&1\\\text{a}^2&\text{a}+\text{b}&\text{b}+\text{c}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})\ ....(\text{i})$
$\text{D}_1=\begin{vmatrix}-1&1&1\\-\text{d}&\text{b}&\text{c}\\-\text{d}^2&\text{b}^2&\text{c}^2\end{vmatrix}=-\begin{vmatrix}1&1&1\\\text{d}&\text{b}&\text{c}\\\text{d}^2&\text{b}^2&\text{c}^2\end{vmatrix}$
$\text{D}_1=-(\text{d}-\text{b})(\text{b}-\text{c})(\text{c}-\text{d})$ [Replacing a by d in eq. (i)]
$\text{D}_2=\begin{vmatrix}1&-1&1\\\text{a}&-\text{d}&\text{c}\\\text{a}^2&-\text{d}^2&\text{c}^2\end{vmatrix}=-\begin{vmatrix}1&1&1\\\text{a}&\text{d}&\text{c}\\\text{a}^2&\text{d}^2&\text{c}^2\end{vmatrix}$
$\text{D}_2=-(\text{a}-\text{d})(\text{d}-\text{c})(\text{c}-\text{a})$
$\text{D}_3=\begin{vmatrix}1&1&-1\\\text{a}&\text{b}&-\text{d}\\\text{a}^2&\text{b}^2&-\text{d}^2\end{vmatrix}=-\begin{vmatrix}1&1&1\\\text{a}&\text{d}&\text{d}\\\text{a}^2&\text{b}^2&\text{d}^2\end{vmatrix}$
$\text{D}_3=-(\text{a}-\text{b})(\text{b}-\text{d})(\text{d}-\text{a})$
Thus,
$\text{x}=\frac{\text{D}_1}{\text{D}}=-\frac{(\text{d}-\text{b})(\text{b}-\text{c})(\text{c}-\text{d})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=-\frac{(\text{a}-\text{d})(\text{d}-\text{c})(\text{c}-\text{a})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{(\text{a}-\text{b})(\text{b}-\text{d})(\text{d}-\text{a})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
View full question & answer→Question 35 Marks
Solve the following systems of linear equations by cramer's rule:
$6x + y - 3z = 5,$
$x + 3y - 2z = 5,$
$2x + y + 4z = 8$
AnswerLet $\text{D}=\begin{vmatrix}6&1&-3\\1&3&-2\\2&1&4\end{vmatrix}$
Expanding along $R_1$
$=6(14)-1(8)-3(-5)$
$=84-8+15=91$
Also, $\text{D}_1=\begin{vmatrix}5&1&-3\\5&-3&-2\\8&1&4\end{vmatrix}$
Expanding along $R_1$
$=5(14)-1(36)-3(-19)$
$=70-36+57=91$
Again $\text{D}_2=\begin{vmatrix}6&5&-3\\1&5&-2\\2&8&4\end{vmatrix}$
Expanding along $R_1$
$=6(36)-5(8)-3(-2)$
$=216-40+6=182$
Also $\text{D}_3=\begin{vmatrix}6&1&5\\1&3&5\\2&1&8\end{vmatrix}$
Expanding along $R_1$
$=6(19)-1(-2)+5(-5)$
$=114+2-25=91$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{91}{91}=1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{182}{91}=2$
Also $\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{91}{91}=1$
Hence, $x = 1, y = 2, z = 1$
View full question & answer→Question 45 Marks
Solve the following systems of linear equations by cramer's rule:
5x - 7y + z = 11,
6x - 8y - z = 15,
3x + 2y - 6z = 7
Answer$\text{D}=\begin{vmatrix}5&-7&1\\6&-8&-1\\3&2&-6 \end{vmatrix}$
$=5(48+2)+7(-36+3)+1(12+24)$
$=5(50)+7(-33)+1(36)=55$
$\text{D}_1=\begin{vmatrix}11&-7&1\\15&-8&-1\\7&2&-6 \end{vmatrix}$
$=11(48+2)+7(-90+7)+1(30+36)$
$=11(50)+7(-83)+1(86)=55$
$\text{D}_2=\begin{vmatrix}5&11&1\\6&15&-1\\3&7&-6 \end{vmatrix}$
$=5(-90+7)-11(-36+3)+1(42-45)$
$=5(-83)-11(-33)+1(-3)=-55$
$\text{D}_3=\begin{vmatrix}5&-7&11\\6&-8&15\\3&2&7 \end{vmatrix}$
$=5(-56-30)+7(42-45)+11(12+24)$
$=5(-86)+7(-3)+11(36)=-55$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{55}{55}=1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-55}{55}=-1$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-55}{55}=-1$
$\therefore\text{x}=1,\text{ y}=-1$ and $\text{z}=-1$
View full question & answer→Question 55 Marks
If $\sin(\text{xy})+\frac{\text{y}}{\text{x}}=\text{x}^2-\text{y}^2,$ find $\frac{\text{dy}}{\text{dx}}$
AnswerWe have, $\sin(\text{xy})+\frac{\text{y}}{\text{x}}=\text{x}^2-\text{y}^2$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sin\text{ xy})+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{\text{d}}{\text{dx}}(\text{x}^2)-\frac{\text{d}}{\text{dx}}(\text{y}^2)$
$\Rightarrow \cos(\text{xy})\frac{\text{d}}{\text{dx}}(\text{xy})+\Bigg\{\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\frac{\text{d}}{\text{dx}}(\text{x})}{\text{x}^2}\Bigg\}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \cos(\text{xy})\Big\{\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big\}+\Bigg\{\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}(1)}{\text{x}^2}\Bigg\}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \cos(\text{xy})\Big\{\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)\Big\}+\frac{1}{\text{x}^2}\Big(\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\Big)=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \text{x}\cos(\text{xy})\frac{\text{dy}}{\text{dx}}+\text{y}\cos(\text{xy})+\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}^2}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\Big\{\text{x}\cos(\text{xy})+\frac{1}{\text{x}}+2\text{y}\Big\}=\frac{\text{y}}{\text{x}^2}-\text{y}\cos(\text{xy})+2\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}^2\cos(\text{xy})+1+2\text{xy}}{\text{x}}\Big\}=\frac{1}{\text{x}^2}\big(\text{y}-\text{x}^2\text{y}\cos(\text{xy})+2\text{x}^2\big)$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{2\text{x}^3+\text{y}-\text{x}^2\text{y}\cos(\text{xy})}{\text{x}\big(\text{x}^2\cos(\text{xy})+1+2\text{xy}\big)}$
View full question & answer→Question 65 Marks
Solve the following systems of linear equations by cramer's rule:
$x + y = 5,$
$y + z = 3,$
$x + z = 4$
AnswerLet $\text{D}=\begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}$
Expanding along $R_1$
$=1(1)-1(-1)+0(-1)$
$=1+1+0=2$
Also $\text{D}_1=\begin{vmatrix}5&1&0\\3&1&1\\4&0&1\end{vmatrix}$
Expanding along $R_1$
$=5(1)-1(-1)+0(-4)$
$=5+1+0=6$
Again $\text{D}_2=\begin{vmatrix}1&5&0\\0&3&1\\1&4&1\end{vmatrix}$
Expanding along $R_1$
$=1(-1)-5(-1)+0(-3)$
$=-1+5+0=4$
Also $\text{D}_3=\begin{vmatrix}1&1&5\\0&1&3\\1&0&4\end{vmatrix}$
$=1(4)-1(-3)+5(-1)$
$=4+3-5=2$
Now $\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{6}{2}=3$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{4}{2}=2$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{2}{2}=1$
Hence, $x = 3, y = 2, z = 1$
View full question & answer→Question 75 Marks
Solve the following systems of linear equations by cramer's rule:
x + y = 1,
x + z = -6,
x - y - 2z = 3
AnswerThese equations can be written as
x + y + 0z = 1
x + 0y + z = -6
x - y - 2z = 3
$\text{D}=\begin{vmatrix}1&1&0\\1&0&1\\1&-1&-2 \end{vmatrix}$
$=1(0+1)-1(-2-1)+0(-1-0)=4$
$\text{D}_1=\begin{vmatrix}1&1&0\\-6&0&1\\3&-1&-2 \end{vmatrix}$
$=1(0+1)-1(12-3)+0(6-0)=-8$
$\text{D}_2=\begin{vmatrix}1&1&0\\1&-6&1\\1&3&-2 \end{vmatrix}$
$=1(12-3)-1(-2-1)+0(3+6)=12$
$\text{D}_3=\begin{vmatrix}1&1&0\\1&0&-6\\1&-1&3 \end{vmatrix}$
$=1(0-6)-1(3+6)+1(-1-0)=-16$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-8}{4}=-2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{12}{4}=3$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-16}{4}=-4$
$\therefore\text{x}=-2,\text{y}=3$ and $\text{z}=-4$
View full question & answer→Question 85 Marks
Solve the following systems of linear equations by cramer's rule:
$x + y + z + w = 2,$
$x - 2y + 2z + 2w = -6,$
$2x + y - 2z + 2w = -5,$
$3x - y + 3z - 3w = -3$
AnswerHere, $\text{D}=\begin{vmatrix}1&1&1&1\\1&-2&2&2\\2&1&-2&2\\3&-1&3&-3\end{vmatrix}$
$\therefore\text{D}=\begin{vmatrix}1&0&0&0\\1&-3&1&1\\2&-1&-4&0\\3&-4&0&-6\end{vmatrix}=1\begin{vmatrix}-3&1&1\\-1&-4&0\\-4&0&-6\end{vmatrix}$
$[C_2 → C_2 - C_1, C_3 → C_3 - C_1, C_4 → C_4 - C_1]$
$\therefore\text{D}=\begin{vmatrix}0&0&1\\-1&-4&0\\-22&6&-6\end{vmatrix}$$[C_1 → C_1 + 3C_3, C_2 → C_2 - C_3]$
$=1(-6-88)=-94$
$\text{D}_1=\begin{vmatrix}2&1&1&1\\-6&-2&2&2\\-5&1&-2&2\\-3&-1&3&-3\end{vmatrix}=188$
$\text{D}_2=\begin{vmatrix}1&2&1&1\\1&-6&2&2\\2&-5&-2&2\\3&-3&3&-3\end{vmatrix}=-282$
$\text{D}_3=\begin{vmatrix}1&1&2&1\\1&-2&-6&2\\2&-1&-5&-2\\3&-1&-3&-3\end{vmatrix}=-141$
$\text{D}_4=\begin{vmatrix}1&1&1&2\\1&-2&2&-6\\2&1&-2&-5\\3&-1&3&-3\end{vmatrix}=47$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{188}{-94}=-2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-282}{-94}=3$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-141}{94}=\frac{3}{2}$
$\text{w}=\frac{\text{D}_4}{\text{D}}=\frac{47}{-94}=-\frac{1}{2}$
Hence, $\text{x}=-2,\text{y}=3,\text{z}=\frac{3}{2},\text{w}=-\frac{1}{2}$
View full question & answer→Question 95 Marks
Show that the following systems of linear equations has infinite number of solutions and solve:
x - y + z = 3,
2x + y - z = 2,
-x - 2y + 2z = 1
AnswerWe have,
$\text{D}=\begin{vmatrix}1&-1&1\\2&1&-1\\-1&-2&2\end{vmatrix}=\begin{vmatrix}0&-1&0\\3&1&0\\-3&-2&0\end{vmatrix}=0$
$\text{D}_1=\begin{vmatrix}3&-1&1\\2&1&-1\\1&-2&2\end{vmatrix}=\begin{vmatrix}3&-1&0\\2&1&0\\1&-2&0\end{vmatrix}=0$
$\text{D}_2=\begin{vmatrix}1&3&1\\2&2&-1\\-1&1&2\end{vmatrix}=\begin{vmatrix}1&0&0\\2&-4&-3\\-1&4&3\end{vmatrix}=1(-12+12)=0$
$\text{D}_3=\begin{vmatrix}1&-1&3\\2&1&2\\-1&-2&1\end{vmatrix}=\begin{vmatrix}1&0&0\\2&3&-4\\-1&-3&4\end{vmatrix}=1(12-12)=0$
$\therefore\text{D}=\text{D}_1=\text{D}_2=\text{D}_3=0$
So, either the system is consistent with infinitely many solutions or it is inconsistent.
Consider the first two equations, written as
x - y = 3 - z
2x + y = 2 + z
to solve these equation, written as
Here,
$\text{D}=\begin{vmatrix}1&-1\\2&1\end{vmatrix}=1+2=3$
$\text{D}_1=\begin{vmatrix}3-\text{z}&-1\\2+\text{z}&1\end{vmatrix}=(3-\text{z})+(2+\text{z})=5$
$\text{D}_2=\begin{vmatrix}1&3-\text{z}\\1&2+\text{z}\end{vmatrix}=(2+\text{z})-(6-2\text{z})=-4+3\text{z}$
$\therefore\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{5}{3}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-4+3\text{z}}{3}$
Let z = k, then the equations have the solution.
$\text{x}=\frac{5}{3},\text{ y}=\frac{-4+3\text{k}}{3},\text{ z}=\text{ k}$
View full question & answer→Question 105 Marks
Solve the following systems of linear equations by cramer's rule:
2x - 3y - 4z = 29,
-2x + 5y - z = -15,
3x - y + 5z = -11
AnswerGiven, 2x - 3y - 4z = 29
-2x + 5y - z = -15
3x - y + 5z = -11
$\text{D}=\begin{vmatrix}2&-3&-4\\-2&5&-1\\3&-1&5 \end{vmatrix}$
$=2(25-1)+3(-10+3)-4(2-15)$
$=2(24)+3(-7)-4(-13)=79$
$\text{D}_1=\begin{vmatrix}29&-3&-4\\-15&5&-1\\-11&-1&5 \end{vmatrix}$
$=29(25-1)+3(-72-11)-4(15+55)$
$=29(24)+3(-86)-4(70)=158$
$\text{D}_2=\begin{vmatrix}2&29&-4\\-2&-15&-1\\3&-11&5 \end{vmatrix}$
$=2(-75-11)-29(-10+3)-4(22+45)$
$=2(-86)-29(-7)-4(67)=-237$
$\text{D}_3=\begin{vmatrix}2&-3&29\\-2&5&-15\\3&-1&-11 \end{vmatrix}$
$=2(-55-15)+3(22+45)+29(2-15)$
$=2(-70)+3(67)+29(-13)=-316$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{158}{79}=2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-237}{79}=-3$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-316}{79}=-4$
$\therefore\text{x}=2,\text{y}=-3$ and $\text{z}=-4$
View full question & answer→Question 115 Marks
Solve the following systems of linear equations by cramer's rule:
x - 4y - z = 11,
2x - 5y + 2z = 39,
-3x + 2y + z = 1
AnswerGiven, x - 4y - z = 11
2x - 5y + 2z = 39
-3x + 2y + z = 1
$\text{D}=\begin{vmatrix}1&-4&-1\\2&-5&2\\-3&2&1\end{vmatrix}$
$=1(-5-4)-(-4)(2+6)+(-1)(4-15)$
$=1(-9)-(-4)(8)+(-1)(-11)=34$
$\text{D}_1=\begin{vmatrix}11&-4&-1\\39&-5&2\\1&2&1\end{vmatrix}$
$=11(-5-4)-(-4)(39-2)+(-1)(78+5)$
$=11(-9)-(-4)(37)+(-1)(83)=-34$
$\text{D}_2=\begin{vmatrix}1&11&-1\\2&39&2\\-3&1&1\end{vmatrix}$
$=1(39-2)-11(2+6)+(-1)(2+117)$
$=1(37)-11(8)+(-1)(119)=-170$
$\text{D}_3=\begin{vmatrix}1&-4&11\\2&-5&39\\-3&2&1\end{vmatrix}$
$=1(-5-78)-(-4)(2+117)+11(4-15)$
$=1(-83)-(-4)(119)+11(-11)=272$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-34}{34}=-1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-170}{34}=-5$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{272}{34}=8$
$\therefore\text{x}=-1,\text{ y}=-5$ and $\text{z}=8$
View full question & answer→Question 125 Marks
Solve the following systems of linear equations by cramer's rule:
$3x + y + z = 2,$
$2x - 4y + 3z = -1,$
$4x + y - 3z = -11$
AnswerLet $\text{D}=\begin{vmatrix}3&1&1\\2&-4&3\\4&1&-3\end{vmatrix}$
Expanding along $R_1$
$=3(9)+(-1)(-18)+1(18)$
$=27+18+18=63$
Again $\text{D}_2=\begin{vmatrix}3&2&1\\2&-1&3\\4&-11&-3\end{vmatrix}$
Expanding along $R_1$
$=3(3+33)-2(-18)+1(-22+4)$
$=108+36-18=126$
Also $\text{D}_3=\begin{vmatrix}3&1&2\\2&-4&-1\\4&1&-11\end{vmatrix}$
Expanding along $R_1$
$=3(45)-1(-18)+2(18)$
$=135+18+36=189$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-63}{63}=-1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{126}{63}=2$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{189}{63}=3$
View full question & answer→Question 135 Marks
An automobile company uses three types of steel $S_1, S_2$ and $S_3$ for producing three types of cars $C_1, C_2$ and $C_3.$ Steel requirements (in tons) for each type of cars are given below:
|
Steel
|
Cars
|
|
|
$C_1$
|
$C_2$
|
$C_3$
|
|
$S_1$
|
$2$
|
$3$
|
$4$
|
|
$S_2$
|
$1$
|
$1$
|
$2$
|
|
$S_3$
|
$3$
|
$2$
|
$1$
|
Using Cramer's rule, find the number of cars of each type which can be produced using $29, 13$ and $16$ tons of steel of three types respectively. AnswerExpressing the given information as a system of linear equations we get,
$2x +3y + 4z = 29$
$x + y + 2z = 13$
$3x + 2y + z = 16$
Where $x, y, z$ is the number of cars $C_1, C_2$ and $C_3$ produced.
We use Cramer's rule to solve this system.
Here,
$\text{D}=\begin{vmatrix}2&3&4\\1&1&2\\3&2&1\end{vmatrix}=\begin{vmatrix}-10&-5&0\\-5&-3&0\\3&2&1\end{vmatrix}=1(30-25)=5$
$\text{D}_1=\begin{vmatrix}29&3&4\\13&1&2\\16&2&1\end{vmatrix}=\begin{vmatrix}-35&-5&0\\-19&-3&0\\16&2&1\end{vmatrix}=1(105-95)=10$
$\text{D}_2=\begin{vmatrix}0&29&4\\1&13&2\\3&16&1\end{vmatrix}=\begin{vmatrix}-10&-35&0\\-5&-19&0\\3&16&1\end{vmatrix}=(190-175)=15$
$\text{D}_3=\text{D}=\begin{vmatrix}2&3&29\\1&1&13\\3&2&16\end{vmatrix}=\begin{vmatrix}-2&0&0\\1&1&13\\3&2&16\end{vmatrix}=-2(16-26)=20$
$\therefore\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{10}{5}=2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{15}{5}=3$
and $\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{20}{5}=4$
Hence, the number of cars produced of the type $C_1, C_2$ and $C_3$ are $2, 3$ and $4$ respectively.
View full question & answer→Question 145 Marks
Solve the following systems of linear equations by cramer's rule:
2x - 3z + w = 1,
x - y + 2w = 1,
-3y + z + w = 1,
x + y + z = 1
Answer$\text{D}=\begin{vmatrix}2&0&-3&1\\1&-1&0&2\\0&-3&1&1\\1&1&1&0\end{vmatrix}$
$=2\begin{vmatrix}-1&0&2\\-3&1&1\\1&1&0\end{vmatrix}-0-3\begin{vmatrix}1&-1&2\\0&-3&1\\1&1&0\end{vmatrix}-1\begin{vmatrix}1&-1&0\\0&-3&1\\1&1&1\end{vmatrix}$
= 2[-1(0 - 1) - 0(0 - 1) + 2(-3 - 1)] - 3[1(0 - 1) + 1(0 - 1) + 2(0 + 3)] - 1[1(-3 - 1) + 1(0 - 1) + 0(0 + 3)]
= -21
$\text{D}_1=\begin{vmatrix}1&0&-3&1\\1&-1&0&2\\1&-3&1&1\\1&1&1&0\end{vmatrix}$
$=\begin{vmatrix}-1&0&2\\-3&1&1\\1&1&0\end{vmatrix}-0-3\begin{vmatrix}1&-1&2\\0&-3&1\\1&1&0\end{vmatrix}-1\begin{vmatrix}1&-1&0\\0&-3&1\\1&1&1\end{vmatrix}$
= 2[-1(0 - 1) - 0(0 - 1) + 2(-3 - 1)] - 3[1(0 - 1) + 1(0 - 1) + 2(0 + 3)] - 1[1(-3 - 1) + 1(0 - 1) + 0(0 + 3)]
= -21
$\text{D}_2=\begin{vmatrix}2&1&3&1\\1&1&0&2\\0&1&1&1\\1&1&1&0\end{vmatrix}$
$=2\begin{vmatrix}1&0&2\\1&1&1\\1&1&0\end{vmatrix}-1\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}-1\begin{vmatrix}1&1&0\\0&1&1\\1&1&1\end{vmatrix}$
= 2[1(0 - 1) + 2(1 - 1)] - 1[1(0 - 1) + 2(0 - 1)] - 3[1(0 - 1) - 1(0 - 1) + 2(0 - 1)] - 1[1(1 - 1) - 1(0 - 1)]
= 6
$\text{D}_3=2\begin{vmatrix}-1&1&2\\-3&1&1\\1&1&0\end{vmatrix}-0+1\begin{vmatrix}1&-1&2\\0&-3&1\\1&1&0\end{vmatrix}-1\begin{vmatrix}1&-1&1\\0&-3&1\\1&1&1\end{vmatrix}$
= 2[-1(0 - 1) - 1(0 - 1) + 2(-3 - 1)] + 1[1(0 - 1) + 1(0 - 1) + 2(0 + 3)] - 1[1(-3 - 1) + 1(0 - 1) + 1(0 + 3)]
= -6
$\text{D}_4=\begin{vmatrix}2&0&-3&1\\1&-1&0&1\\0&-3&1&1\\1&1&1&1\end{vmatrix}$
$=2\begin{vmatrix}-1&0&1\\-3&1&1\\1&1&1\end{vmatrix}-0-3\begin{vmatrix}1&-1&1\\0&-3&1\\1&1&1\end{vmatrix}-1\begin{vmatrix}1&-1&0\\0&-3&1\\1&1&1\end{vmatrix}$
= 2[-1(1 - 1) + 1(-3 - 1)] - 3[1(-3 - 1) + 1(0 - 1) + 1(0 + 3)] - 1[1(-3 - 1) + 1(0 - 1)]
= 3
So, by Cramer's rule, we obtain
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{21}{21}=1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{6}{-21}=-\frac{2}{7}$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{3}{-21}=-\frac{1}{7}$
$\text{w}=\frac{\text{D}_4}{\text{D}}=\frac{3}{-21}=-\frac{1}{7}$
Hence, $\text{x}=1,\text{ y}=-\frac{2}{7},\text{ z}=\frac{2}{7},\text{ w}=-\frac{1}{7}$
View full question & answer→Question 155 Marks
Show that the following systems of linear equations has infinite number of solutions and solve:
$2x + y - 2z = 0,$
$x - 2y + z = -2,$
$5x - 5y + z = -2$
Answer$=\begin{vmatrix}2&1&-2\\1&-2&1\\5&5&1\end{vmatrix}=\begin{vmatrix}12&9&-12\\-4&-3&1\\0&0&1\end{vmatrix}=1(-36+36)=0$
$\text{D}_1=\begin{vmatrix}4&1&-2\\-2&-2&1\\-2&-5&1\end{vmatrix}=\begin{vmatrix}0&1&-2\\0&-2&1\\0&-5&1\end{vmatrix}=0$
$\text{D}_2=\begin{vmatrix}2&4&-2\\1&-2&1\\5&-2&1\end{vmatrix}=\begin{vmatrix}2&0&-2\\1&0&1\\5&0&1\end{vmatrix}=0$
$\text{D}_3=\begin{vmatrix}2&1&4\\1&-2&-2\\5&-5&-2\end{vmatrix}=\begin{vmatrix}4&-3&0\\1&-2&-2\\4&-3&0\end{vmatrix}=2(-12+12)=0$
So, $\text{D}=\text{D}_1=\text{D}_2=\text{D}_3=0$
So, the given system is either inconsistent or has infinite solutions.
Consider the $2^{nd}$ and $3^{rd}$ equations, written as
$x - 2y = -2 - z$
$5x - 5y = -2 - z$
Then,
$\text{D}=\begin{vmatrix}1&-2\\5&-5\end{vmatrix}=-5-(-10)=5$
$\text{D}_2=\begin{vmatrix}-2-\text{z}&-2\\-2-\text{z}&-5\end{vmatrix}$
$=(2+\text{z})+(5)-2(2+\text{z})=3(2+\text{z})=6+3\text{z}$
$\text{D}_2=\begin{vmatrix}1&-(2+\text{z})\\5&-(2+\text{z})\end{vmatrix}$
$=-(2+\text{z})+5(2+\text{z})=4(2+\text{z})=8+4\text{z}$
$\therefore\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{6+3\text{z}}{5}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{8+4\text{z}}{5}$
Let $z = k,$ then
$\text{x}=\frac{6+3\text{k}}{5},\text{ y}=\frac{8+4\text{k}}{5},\text{ z}=\text{k}$ are the infinite solution of the given system of equations.
View full question & answer→Question 165 Marks
Solve the following systems of linear equations by cramer's rule:
2y - 3z = 0,
x + 3y = -4,
3x + 4y = 3
AnswerThese equations can be written as
0x + 2y - 3z = 0
x + 3y + 0z = -4
3x + 4y + 0z= 3
$\text{D}=\begin{vmatrix}0&2&-3\\1&3&0\\3&4&0 \end{vmatrix}$
$=0(0-0)-2(0-0)-3(4-9)$
$=15$
$\text{D}_1=\begin{vmatrix}0&2&-3\\-4&3&0\\3&4&0 \end{vmatrix}$
$=0(0-0)-0(0-0)-3(3+12)$
$=-45$
$\text{D}_3=\begin{vmatrix}0&2&0\\1&3&-4\\3&4&3 \end{vmatrix}$
$=0(9+16)-2(3+12)-0(4-9)$
$=-30$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{75}{15}=5$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-45}{15}=-3$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-30}{12}=-2$
$\therefore\text{x}=5,\text{y}=-3$ and $\text{z}=-2$
View full question & answer→Question 175 Marks
Solve the following determinant equations:
$\begin{vmatrix}\text{x}+1&3&5\\2&\text{x}+2&5\\2&3&\text{x}+4\end{vmatrix}=0$
AnswerLet $\begin{vmatrix}\text{x}+1&3&5\\2&\text{x}+2&5\\2&3&\text{x}+4\end{vmatrix}$
$=\begin{vmatrix}\text{x}+9&3&5\\\text{x}+9&\text{x}+2&5\\\text{x}+9&3&\text{x}+4\end{vmatrix} [$Applying $C_1 = C_1 + C_2 + C_3]$
$=(\text{x}+9)\begin{vmatrix}1&3&5\\1&\text{x}+2&5\\1&3&\text{x}+4\end{vmatrix}$
$=(\text{x}+9)\begin{vmatrix}1&3&5\\0&\text{x}-1&0\\1&3&\text{x}+4\end{vmatrix}=0[ $Applying $R_2 → R_2 - R_1]$
$=(\text{x}+9)\begin{vmatrix}1&3&5\\0&\text{x}-1&0\\1&0&\text{x}-1\end{vmatrix} [$Applying $R_3 → R_3 - R_1]$
$=(\text{x}+9)(\text{x}-1)^2=0$
$\text{x}=-9,1,1$
View full question & answer→Question 185 Marks
Prove that:
$\begin{vmatrix}\text{a}+\text{b}+2\text{c}&\text{a}&\text{b}\\\text{c}&\text{b}+\text{c}+2\text{a}&\text{b}\\\text{c}&\text{a}&\text{c}+\text{a}+2\text{b} \end{vmatrix}=2(\text{a}+\text{b}+\text{c})^3$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}+\text{b}+2\text{c}&\text{a}&\text{b}\\\text{c}&\text{b}+\text{c}+2\text{a}&\text{b}\\\text{c}&\text{a}&\text{c}+\text{a}+2\text{b} \end{vmatrix}$
$=2(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{a}&\text{b}\\1&\text{b}+\text{c}+2\text{a}&\text{b}\\1&\text{a}&\text{c}+\text{a}+2\text{b} \end{vmatrix} [$Taking out $2(a + b + c)$ common from $C_1]$
$=2(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{a}&\text{b}\\1&\text{b}+\text{c}+\text{a}&0\\0&-\text{b}-\text{c}-\text{a}&\text{c}+\text{a}+\text{b} \end{vmatrix} [$Applying $R_2 → R_2 - R_1$ and $R_2 → R_2 - R_3]$
$=2(\text{a}+\text{b}+\text{c})(\text{a}+\text{b}+\text{c})(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{a}&\text{b}\\0&1&0\\0&-1&0\end{vmatrix} [$Taking out $(a + b + c)$ common from $R_2$ and $R_3]$
$=2(\text{a}+\text{b}+\text{c})^3\{1(1-0)\} [$Expanding along $C_1]$
$=2(\text{a}+\text{b}+\text{c})^3$
$=\text{R.H.S}$
View full question & answer→Question 195 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sin^223^{\circ}&\sin^267^{\circ}&\cos180^{\circ}\\-\sin^267^{\circ}&-\sin^223^{\circ}&\cos^2180^{\circ}\\\cos180^{\circ}&\sin^223^{\circ}&\sin^267^{\circ}\end{vmatrix}$
Answer$\begin{vmatrix}\sin^223^{\circ}&\sin^267^{\circ}&\cos180^{\circ}\\-\sin^267^{\circ}&-\sin^223^{\circ}&\cos^2180^{\circ}\\\cos180^{\circ}&\sin^223^{\circ}&\sin^267^{\circ}\end{vmatrix}$
$=\begin{vmatrix}\sin^223^{\circ}&\sin^2(90-23)^{\circ}&-1\\-\sin^2(90-23)^{\circ}&-\sin^223^{\circ}&1\\1&\sin^223^{\circ}&\sin^2(90-23)^{\circ}\end{vmatrix}$
$=\begin{vmatrix}\sin^223^{\circ}&\cos^223^{\circ}&-1\\-\cos^223^{\circ}&-\sin^223^{\circ}&1\\-1&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$
$=\begin{vmatrix}\sin^223^{\circ}+\cos^223^{\circ}&\cos^223^{\circ}&-1\\-\cos^223^{\circ}-\sin^223^{\circ}&-\sin^223^{\circ}&1\\-1+\sin^223^{\circ}&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix} [$Applying $C_1 → C_1 + C_2]$
$=\begin{vmatrix}1&1&-1\\-1&-\sin^223^{\circ}&1\\-\cos^223^{\circ}&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$
$=(-1)\begin{vmatrix}1&1&-1\\-1&-\sin^223^{\circ}&1\\-\cos^223^{\circ}&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$
$=0$
View full question & answer→Question 205 Marks
Prove that:
$\begin{vmatrix}\text{a}^2&\text{a}^2-(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&\text{b}^2-(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&\text{c}^2-(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^2&\text{a}^2-(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&\text{b}^2-(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&\text{c}^2-(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$
$=\begin{vmatrix}\text{a}^2&-(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&-(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&-(\text{a}-\text{b})^2&\text{ab}\end{vmatrix} [$Applying $C_2 → C_2 - C1]$
$=(-1)\begin{vmatrix}\text{a}^2&(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$
$=-\begin{vmatrix}\text{a}^2&\text{b}^2+\text{c}^2&\text{bc}\\\text{b}^2&\text{c}^2+\text{a}^2&\text{ca}\\\text{c}^2&\text{a}^2+\text{b}^2&\text{ab}\end{vmatrix} [$Applying $C_2 → C_2 - 2C_1]$
$=-\begin{vmatrix}\text{a}^2+\text{b}^2+\text{c}^2&\text{b}^2+\text{c}^2&\text{bc}\\\text{b}^2+\text{c}^2+\text{a}^2&\text{c}^2+\text{a}^2&\text{ca}\\\text{c}^2+\text{a}^2+\text{b}&\text{a}^2+\text{b}^2&\text{ab}\end{vmatrix} [$Applying $C_1 → C_1 + C_2]$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)\begin{vmatrix}1&\text{b}^2+\text{c}^2&\text{bc}\\1&\text{c}^2+\text{a}^2&\text{ca}\\1&\text{a}^2+\text{b}^2&\text{ab}\end{vmatrix}$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)\begin{vmatrix}1&\text{b}^2+\text{c}^2&\text{bc}\\0&\text{a}^2-\text{b}^2&\text{c}(\text{a}-\text{b})\\0&\text{a}^2-\text{c}^2&\text{b}(\text{a}-\text{c})\end{vmatrix}$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{a}-\text{c})\begin{vmatrix}1&\text{b}^2+\text{c}^2&\text{bc}\\0&\text{a}+\text{b}&\text{c}\\0&\text{a}+\text{c}&\text{b}\end{vmatrix} $
$[$Taking $(a - b)$ common from $R_2$ and $(a - c)$ common from $R_3]$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{c}-\text{a})\times\left\{1\times\begin{vmatrix}\text{a}+\text{b}&\text{c}\\\text{a}+\text{c}&\text{b}\end{vmatrix}\right\}$
$[\because(\text{c}-\text{a})=-(\text{a}-\text{c})] [$Expanding along $C_1]$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{c}-\text{a})(\text{ab}+\text{b}^2-\text{ac}-\text{c}^2)$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{c}-\text{a})\{\text{a}(\text{b}-\text{c})+(\text{b}+\text{c})(\text{b}-\text{c})\}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
$=\text{R.H.S}$
Hence prove.
View full question & answer→Question 215 Marks
Evaluate:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{c}&\text{a}&\text{b}\\\text{b}&\text{c}&\text{a}\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{c}&\text{a}&\text{b}\\\text{b}&\text{c}&\text{a}\end{vmatrix}$
Applying $C_1 → C_1 + C_2 + C_3$ we get,
$\triangle=\begin{vmatrix}\text{a}+\text{b}+\text{c}&\text{b}&\text{c}\\\text{a}+\text{b}+\text{c}&\text{a}&\text{b}\\\text{a}+\text{b}+\text{c}&\text{c}&\text{a}\end{vmatrix}$
Taking $(a + b + c)$ common, we have
$\triangle=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{b}&\text{c}\\1&\text{a}&\text{b}\\1&\text{c}&\text{a}\end{vmatrix}$
Applying $R_2 → R_2 - R_1, R_3 - R_1,$ we get
$\triangle=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{b}&\text{c}\\0&\text{a}-\text{b}&\text{b}-\text{c}\\0&\text{c}-\text{b}&\text{a}-\text{c}\end{vmatrix}$
$\Rightarrow\triangle=(\text{a}+\text{b}+\text{c})[(\text{a}-\text{b})(\text{a}-\text{c})-(\text{b}-\text{c})(\text{c}-\text{b})]$
$\Rightarrow\triangle=(\text{a}+\text{b}+\text{c})\big[\text{a}^2-\text{ac}-\text{ab}+\text{bc}+\text{b}^2+\text{c}^2-2\text{ab}\big]$
$\Rightarrow\triangle=(\text{a}+\text{b}+\text{c})\big[\text{a}^2+\text{b}^2+\text{c}^2-\text{ac}-\text{ab}-\text{bc}\big]$
View full question & answer→Question 225 Marks
Prove that: $\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\(\text{a}+2)(\text{a}+3)&\text{a}+3&1\\(\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}=-2$
Answer$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\(\text{a}+2)(\text{a}+3)&\text{a}+3&1\\(\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}=-2$
$\text{L.H.S}=\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\(\text{a}+2)(\text{a}+3)&\text{a}+3&1\\(\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}$ Apply $R_3 → R_3 - R_2$
$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\(\text{a}+2)(\text{a}+3)&\text{a}+3&1\\(\text{a}+3)2&1&0\end{vmatrix}$ Apply $R_2 → R_2 - R_1$
$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\(\text{a}+2)2&1&0\\(\text{a}+3)2&1&0\end{vmatrix}$ $=[(2\text{a}+4)(1)-(1)(2\text{a}+6)]$ $=-2$ $=\text{R.H.S}$
View full question & answer→Question 235 Marks
Evaluate:
$\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
When a = b, the first two rows become identical. Hence, a - b is a factor. Similarly, when b = c and c = a, the second and third and third and first rows become indetical. Hence, b - c and c - a are also factors. The degree of product of the diagonal elements is 3. Hence, there are no other factors.
$\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
$=\lambda(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$ [Where $\lambda$ is a constant]
$\begin{vmatrix}1&0&2\\1&1&0\\1&2&0\end{vmatrix}=2\lambda$ $[$Putting a = 0, b = 1 and c = 2 to find $\lambda]$
$\Rightarrow2=2\lambda$
$\Rightarrow\lambda=1$
Hence,
$\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$
View full question & answer→Question 245 Marks
Prove that:
$\begin{vmatrix}1&1+\text{p}&1+\text{p}+\text{q}\\2&3+2\text{p}&4+3\text{p}+2\text{p}\\3&6+3\text{p}&10+6\text{p}+3\text{q}\end{vmatrix}=1$
Answer$=\begin{vmatrix}1&1&1\\2&3&4\\3&2&10\end{vmatrix}+\begin{vmatrix}1&1&\text{p}\\2&3&3\text{p}\\3&6&6\text{p}\end{vmatrix}+(\text{pq})\begin{vmatrix}1&1&1\\2&2&2\\3&3&3\end{vmatrix}$
$=\begin{vmatrix}1&1&1\\2&3&4\\3&2&10\end{vmatrix}+(\text{p})\begin{vmatrix}1&1&\text{p}\\2&3&3\\3&6&6\end{vmatrix}+0$
$=\begin{vmatrix}1&1&1\\2&3&4\\3&2&10\end{vmatrix}+0$
$[\because$ Value of determinant with two identical columns is zero$]$
$=\begin{vmatrix}1&0&0\\2&1&2\\3&3&7\end{vmatrix} [$Applying $C_2 → C_2 - C_1$ and $C_3 → C_3 - C_1]$
$=\left\{1\times\begin{vmatrix}1&2\\3&7\end{vmatrix}\right\}[$ Expanding along $R_1]$
$=7-6$
$=1$
$=\text{R.H.S}$
View full question & answer→Question 255 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}$
Answer$\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}$
$=\begin{vmatrix}\sin\alpha\sin\delta&\cos\alpha\cos\delta&\cos(\alpha+\delta)\\\sin\beta\sin\delta&\cos\beta\cos\delta&\cos(\beta+\delta)\\\sin\gamma\sin\delta&\cos\gamma\cos\delta&\cos(\gamma+\delta)\end{vmatrix}$$[\text{Applying} \text{ C}_1\rightarrow\sin\delta\text{ C}_1\text{ and}\text{ C}_2\rightarrow\cos\delta\text{ C}_2]$
$=\begin{vmatrix}\sin\alpha\sin\delta&\cos(\alpha+\delta)&\cos(\alpha+\delta)\\\sin\beta\sin\delta&\cos(\beta+\delta)&\cos(\beta+\delta)\\\sin\gamma\sin\delta&\cos(\gamma+\delta)&\cos(\gamma+\delta)\end{vmatrix} [$Applying $C_2 → C_2 - C_1]$
$=0$
View full question & answer→Question 265 Marks
Prove the following identities:
$\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=(5\text{x}+\lambda)(\lambda-\text{x})^2$
Answer$\text{L.H.S}=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}-\text{x}-\lambda&\text{x}+\lambda-2\text{x}&0\\2\text{x}-\text{x}-\lambda&0&\text{x}+\lambda-2\text{x}\end{vmatrix} [$Applying $R_2 → R_2 - R_1$ and $R_3 → R_3 - R_1]$
$=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\-(\lambda-\text{x})&\lambda-\text{x}&0\\-(\lambda-\text{x})&0&\lambda-\text{x}\end{vmatrix}$
$=(\lambda-\text{x})^2\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\-1&1&0\\-1&0&1\end{vmatrix}$ [Taking $(\lambda-\text{x})$ common from $R_2$ and $(\lambda-\text{x})$ common from $R_3]$
$=(\lambda-\text{x})^2[-1(-2\text{x})+1(\text{x}+\lambda+2\text{x})]$ [Expanding along last row]
$=(\lambda-\text{x})^2(\lambda+5\text{x})$
$=\text{R.H.S}$
$\because\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=(\lambda-\text{x})^2(\lambda+5\text{x})$
View full question & answer→Question 275 Marks
Prove the following identities:
$\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}=4\text{xyz}$
Answer$\text{L.H.S}=\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
Applying $R_1 → R_1 - R_2$
$=\begin{vmatrix}\text{y}&-\text{x}&\text{y}-\text{x}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
Applying $R_1 → R_1 - R_3$
$=\begin{vmatrix}0&-2\text{x}&-2\text{x}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
$=2\text{x}[\text{z}(\text{x}+\text{y})-\text{xy}]-2\text{x}[\text{zx}-\text{y}(\text{z}+\text{x})]$
$=2\text{x}[\text{zx}+\text{zy}-\text{xy}-\text{zx}+\text{yz}+\text{yx}]$
$=4\text{xyz}$
View full question & answer→Question 285 Marks
Prove that:
$\begin{vmatrix}\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\end{vmatrix}=2\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\end{vmatrix}$
Using the property of determinants that if each element of a row or column is expressed as the sum of two or more quantities, the determinant is expressed as the sum of two or more determinants, we get
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}+(-1)\begin{vmatrix}\text{a}&\text{c}&\text{b}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{b}&\text{a} \end{vmatrix}$
$[$Applying $\text{C}_1\leftrightarrow\text{C}_3$ in second determinant to get negative value of the deteminant$]$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}+(-1)(-1)\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}$ $[$Applying $\text{C}_2\leftrightarrow\text{C}_3]$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}=\text{R.H.S}$
View full question & answer→Question 295 Marks
Solve the following systems of linear equations by cramer's rule:
3x + ay = 4,
2x + ay = 2, $\text{a}\neq0$
AnswerGiven, 3x + ay = 4
2x + ay = 2
Using Cramer's rule, we get
$\text{D}=\begin{vmatrix}3&\text{a}\\2&\text{a}\end{vmatrix}=3\text{a}-2\text{a}=\text{a}$
$\text{D}_1=\begin{vmatrix}4&\text{a}\\2&\text{a}\end{vmatrix}=4\text{a}-2\text{a}=2\text{a}$
$\text{D}_2=\begin{vmatrix}3&4\\2&2\end{vmatrix}=6-8=-2$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{2\text{a}}{\text{a}}=2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-2}{\text{a}}=-\frac{2}{\text{a}}$
$\therefore\text{x}=2$ and $\text{y}=-\frac{2}{\text{a}}$
View full question & answer→Question 305 Marks
Using properties of determinants prove that:
$\begin{vmatrix}\text{x}+4&2\text{x}&2\text{x}\\2\text{x}&\text{x}+4&2\text{x}\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$
$=(5\text{x}+4)(4-\text{x})^2$
Answer$\text{L.H.S}=\begin{vmatrix}\text{x}+4&2\text{x}&2\text{x}\\2\text{x}&\text{x}+4&2\text{x}\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$
$=\begin{vmatrix}5\text{x}+4&5\text{x}+4&5\text{x}+4\\2\text{x}&\text{x}+4&2\text{x}\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$ $[$Applying $R_1 → R_1 + R_2 + R_3]$
$=5\text{x}+4\begin{vmatrix}1&1&1\\2\text{x}&\text{x}+4&2\text{x}\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix} [$Take out $5x + 4$ common from $R_1]$
$=5\text{x}+4\begin{vmatrix}1&0&0\\2\text{x}&4-\text{x}&0\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}[ $Applying $C_2 → C_2 - C_1$ and $C_3 → C_3 - C_1]$
$=5\text{x}+4(4-\text{x})^2$
View full question & answer→Question 315 Marks
If the points $(x, -2), (5, 2), (8, 8)$ are collinear, find $x$ using determinants.
AnswerThe points $(k, -2), (5, 2), (8, 8)$ are collinear.
$\begin{vmatrix}\text{x}&-2&1\\5&2&1\\8&8&1\end{vmatrix}=0$
$\triangle=\begin{vmatrix}\text{x}&-2&1\\5&2&1\\8&8&1\end{vmatrix}$
$=\begin{vmatrix}\text{x}&-2&1\\5-\text{x}&4&0\\8&8&1\end{vmatrix} [$Applying $R_2 → R_2 - R_1]$
$=\begin{vmatrix}\text{x}&-2&1\\5-\text{x}&4&0\\8-\text{x}&10&0\end{vmatrix} [$Applying $R_3 → R_3 - R_1]$
$=\begin{vmatrix}5-\text{x}&4\\8-\text{x}&10\end{vmatrix}$
$=50-10\text{x}-32+4\text{x}$
$=18-6\text{x}=0$
$\Rightarrow\text{x}=3$
View full question & answer→Question 325 Marks
Evaluate:
$\begin{vmatrix}\text{x}+\lambda&\text{x}&\text{x}\\\text{x}&\text{x}+\lambda&\text{x}\\\text{x}&\text{x}&\text{x}+\lambda\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}\text{x}+\lambda&\text{x}&\text{x}\\\text{x}&\text{x}+\lambda&\text{x}\\\text{x}&\text{x}&\text{x}+\lambda\end{vmatrix}$
$=\begin{vmatrix}\lambda&0&\text{x}\\-\lambda&\lambda&\text{x}\\0&-\lambda&\text{x}+\lambda\end{vmatrix}$ $[$Applying $\left.C_1 \rightarrow C_1-C_2, C_2 \rightarrow C_2-C_3\right]$
$=\begin{vmatrix}\lambda&0&\text{x}\\-\lambda&0&2\text{x}+\lambda\\0&-\lambda&\text{x}+\lambda\end{vmatrix} [$Applying $R_1→ R_2+ R_3]$
$=\lambda\begin{vmatrix}0&2\text{x}+\lambda\\-\lambda&\text{x}+\lambda\end{vmatrix}+\text{x}\begin{vmatrix}-\lambda&0\\0&-\lambda\end{vmatrix}$
$=\lambda\big[\lambda(2\text{x}+\lambda)\big]+\text{x}\lambda^2$
$=\lambda^2(2\text{x}+\lambda+\lambda^2\text{x})$
$=3\lambda^2\text{x}+\lambda^3$
$=\lambda^2(3\text{x}+\lambda)$
View full question & answer→Question 335 Marks
$\triangle=\begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha \end{vmatrix}$
AnswerGiven,
$\triangle=\begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha \end{vmatrix}$
$\Rightarrow\triangle=(-1)^{1+1}\cos\alpha\cos\beta(\cos\alpha\cos\beta-0)\\+(-1)^{1+2}\cos\alpha\sin\beta(-\sin\beta\cos\alpha-0)\\+(-1)^{1+3}(-\sin\alpha)(-\sin^2\beta\sin\alpha-\sin\alpha\cos^2\beta)$ [Expanding along $R_1$]
$=\cos\alpha\cos\beta(\cos\alpha\cos\beta-0)-\cos\alpha\sin\beta(-\sin\beta\cos\alpha-0)\\-\sin\alpha(\sin^2\beta\sin\alpha-\sin\alpha-\sin\alpha\cos^2\beta)$
$=\cos^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta+\sin^2\alpha\sin^2\beta+\sin^2\alpha\cos^2\beta$
$=\cos^2\alpha(\cos^2\beta+\sin^\beta)+\sin^2\alpha(\sin^2\beta+\cos^2\beta)$
$\Rightarrow\triangle=\cos^2\alpha+\sin^2\alpha$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$\Rightarrow\triangle=1$
View full question & answer→Question 345 Marks
$\begin{vmatrix}\text{a}+\text{b}+\text{c}&-\text{c}&-\text{b}\\-\text{c}&\text{a}+\text{b}+\text{c}&-\text{a}\\-\text{b}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$
$=2(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$
Answer$\text{L.H.S}=\begin{vmatrix}\text{a}+\text{b}+\text{c}&-\text{c}&-\text{b}\\-\text{c}&\text{a}+\text{b}+\text{c}&-\text{a}\\-\text{b}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$
$=\begin{vmatrix}\text{a}&-\text{c}&-\text{b}\\\text{b}&\text{a}+\text{b}+\text{c}&-\text{a}\\\text{c}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix} [$Applying $C_1 → C_1 + C_2 + C_3]$
$=\begin{vmatrix}\text{a}+\text{b}&\text{a}+\text{b}&-(\text{a}+\text{b})\\\text{b}+\text{c}&\text{b}+\text{c}&\text{b}+\text{c}\\\text{c}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$ $[$Applying $R_1 → R_1 + R_2$ and $R_2 → R_2 + R_3]$
$=(\text{a}+\text{b})(\text{b}+\text{c})\begin{vmatrix}1&1&-1\\1&1&1\\\text{c}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$
$[$Taking out common factor from $R_1$ and $R_2]$
$=(\text{a}+\text{b})(\text{b}+\text{c})\begin{vmatrix}0&0&-2\\1&1&1\\\text{c}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$ $[$Applying $R_1 → R_1- R_2]$
$=(\text{a}+\text{b})(\text{b}+\text{c})\{(-2)(-\text{a}-\text{c})\} [$Expanding along $R_1]$
$=2(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$
$=\text{R.H.S}$
Hence proved.
View full question & answer→Question 355 Marks
Without expanding, prove that:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\\\text{a}&\text{b}&\text{c}\end{vmatrix}=\begin{vmatrix}\text{y}&\text{b}&\text{q}\\\text{x}&\text{a}&\text{p}\\\text{z}&\text{c}&\text{r}\end{vmatrix}$
Answer$=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\\\text{a}&\text{b}&\text{c}\end{vmatrix}$ $\text{R}_2\leftrightarrow\text{R}_3$
$=-\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{a}&\text{b}&\text{c}\\\text{p}&\text{q}&\text{r}\end{vmatrix}$ $\text{R}_1\leftrightarrow\text{R}_2$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\end{vmatrix}$
$\begin{vmatrix}\text{y}&\text{b}&\text{q}\\\text{x}&\text{a}&\text{p}\\\text{z}&\text{c}&\text{r}\end{vmatrix}$
$=\begin{vmatrix}\text{y}&\text{x}&\text{z}\\\text{b}&\text{a}&\text{c}\\\text{q}&\text{p}&\text{r}\end{vmatrix}$ $\text{C}_1\leftrightarrow\text{C}_2$
$=-\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{a}&\text{b}&\text{c}\\\text{p}&\text{q}&\text{r}\end{vmatrix}$ $\text{R}_1\leftrightarrow\text{R}_2$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\end{vmatrix}$
View full question & answer→Question 365 Marks
Show that the following systems of linear equations is inconsistent:
3x - y + 2z = 3,
2x + y + 3z = 5,
x - 2y - z = 1
AnswerGiven,
3x - y + 2z = 3,
2x + y + 3z = 5,
x - 2y - z = 1
$\text{D}=\begin{vmatrix}3&-1&2\\2&1&3\\1&-2&-1\end{vmatrix}$
$=3(-1+6)+1(-2-3)+2(-4-1)=0$
$\text{D}_1=\begin{vmatrix}3&-1&2\\5&1&3\\1&-2&-1\end{vmatrix}$
$=3(-1+6)+1(-5-3)+2(-10-1)=-15$
$\text{D}_2=\begin{vmatrix}3&3&2\\2&5&3\\1&1&-1\end{vmatrix}$
$=3(-5-3)-3(-2-3)+2(2-5)=-15$
$\text{D}_3=\begin{vmatrix}3&-1&3\\2&1&5\\1&-2&1\end{vmatrix}$
$=3(1+10)+1(2-5)+3(-4-1)=-15$
Here, d is zero, but $D_1, D_2$ and $D_3$ are non-zero. Thus, the system of linear equations is inconsistent.
View full question & answer→Question 375 Marks
Solve the following determinant equations:
$\begin{vmatrix}3&-2&\sin(3\theta)\\-7&8&\cos(2\theta)\\-11&14&2\end{vmatrix}=0$
Answer$\begin{vmatrix}3&-2&\sin3\theta\\-7&8&\cos2\theta\\-11&14&2\end{vmatrix}=0$
$\Rightarrow3(16-14\cos2\theta)+2(-14+11\cos2\theta)\\+\sin3\theta(-98+88)=0$
$\Rightarrow20(1-\cos2\theta)+10\sin3\theta=0$
$\Rightarrow20(2\sin^2\theta)+10(3\sin\theta-4\sin^3\theta)=0$
$\Rightarrow4\sin^2\theta+3\sin\theta-4\sin^3\theta=0$
$\Rightarrow4\sin^2\theta+3-4\sin^2\theta=0$
$\Rightarrow4\sin^2\theta-4\sin\theta-3=0$
$\Rightarrow(2\sin\theta+1)(2\sin\theta-3)=0$
$\Rightarrow\sin\theta=-\frac{1}{2}$ or $\sin\theta=\frac{3}{2}=1.5$
As $\sin\theta\in[-1,1]$
$\therefore\sin\theta=-\frac{1}{2}$
$\Rightarrow\theta=\text{n}\pi+(-1)^{\text{n}}\frac{\pi}{6},\text{n }\in\text{ z}$
View full question & answer→Question 385 Marks
If a, b and c are all non-zero and $\begin{vmatrix}1+\text{a}&1&1\\1&1+\text{b}&1\\1&1&1+\text{c} \end{vmatrix}=0,$ then prove that $\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}+1=0.$
Answer$\begin{vmatrix}1+\text{a}&1&1\\1&1+\text{b}&1\\1&1&1+\text{c} \end{vmatrix}=0$
$C_1 → C_1 - C_2$
$\begin{vmatrix}\text{a}&1&1\\-\text{b}&1+\text{b}&1\\1&1&1+\text{c} \end{vmatrix}=0$
$C_2 → C_2 - C_3$
$\begin{vmatrix}\text{a}&0&1\\-\text{b}&\text{b}&1\\0&-\text{c}&1+\text{c} \end{vmatrix}=0$
Expanding along $R_1$, we get
$\text{a}(\text{b}+\text{bc}+\text{c})+1(\text{bc})=0$
$\Rightarrow\text{ab}+\text{abc}+\text{ac}+\text{bc}=0$
Dividing by abc, we get
$\frac{1}{\text{c}}+1+\frac{1}{\text{b}}+\frac{1}{\text{a}}=0$
$\therefore\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}+1=0$
View full question & answer→Question 395 Marks
Solve the following systems of linear equations by cramer's rule:
x + 2y = 1,
3x + y = 4
Answer$\text{D}=\begin{vmatrix}1&2\\3&1\end{vmatrix}=-5$
$\text{D}_1=\begin{vmatrix}1&2\\4&1\end{vmatrix}=-7$
$\text{D}_2=\begin{vmatrix}1&1\\3&4\end{vmatrix}=1$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{7}{5}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=-\frac{1}{5}$
$\therefore\text{x}=\frac{7}{5}$ and $\text{y}=-\frac{1}{5}$
View full question & answer→Question 405 Marks
Prove that: $\begin{vmatrix}(b+c)^2&a^2&{bc}\\(c+a)^2&{b}^2&{ca}\\(a+b)^2&{c}^2&{ab}\end{vmatrix}$ $=(a-b)(b-c)(c-b)(a+b+c)({a}^2+{b}^2+{c}^2)$
Answer${L.H.S}= $$\begin{vmatrix}(b+c)^2&a^2&{bc}\\(c+a)^2&b^2&{ca}\\(a+b)^2&c^2&{ab}\end{vmatrix}$
$=\begin{vmatrix}(b+c)^2-(c+a)^2&a^2-b^2&{bc}-{ca}\\(c+a)^2-(a+b)^2&b^2-c^2&{ca}-{ab}\\(a+b)^2&c^2&{ab}\end{vmatrix}$ [Applying R1 → R1 - R2 and R2 → R2 - R1]
$=\begin{vmatrix}(b+c)(b+2c+a)&(a+b)(a-b)&c(b-a)\\(c-a)(b+2a+c)&(b-c)(b+c)&a(c-b)\\(a+b)^2&c^2&{ab}\end{vmatrix}$
$=(a-b)(b-c)\begin{vmatrix}-(b+2c+a)&a+b&-c\\-(b+2a+c)&b+c&-a\\(a+b)^2&c^2&{ab}\end{vmatrix}$ [Applying x2 - y2 = (x + y)(x - y) and taking out (a - b) common from R1 and (b - c) from R2]
$=(a-b)(b-c)\begin{vmatrix}-2(b+c+a)&a+b&-c\\-2(b+a+c)&b+c&-a\\(a+b)^2-c^2&c^2&{ab}\end{vmatrix}$ [Applying C1 → C1 - C2]
$=(a-b)(b-c)\begin{vmatrix}-2(b+c+a)&a+b&-c\\-2(b+a+c)&b+c&-a\\(a+b+c)(a+b-c)&c^2&{ab}\end{vmatrix}$ [Applying x2 - y2 = (x + y)(x - y) in C1]
$=(a-b)(b-c)(a+b+c)\begin{vmatrix}-2&a+b&-c\\-2&b+c&-a\\(a+b-c)&c^2&{ab}\end{vmatrix}$ [Taking out (a + b + c) common from C1]
$=(a-b)(b-c)(a+b+c)\begin{vmatrix}-2&a+b&-c\\0&c-a&c-a\\(a+b-c)&c^2&{ab}\end{vmatrix}$ [Applying R2 → R2 - R1]
$=(a-b)(b-c)(a+b+c)(c-a)\begin{vmatrix}-2&a+b&-c\\0&1&1\\(a+b-c)&c^2&{ab}\end{vmatrix}$ [Taking out (c - a) common from R2]
$=(a-b)(b-c)(a+b+c)(c-a)\begin{vmatrix}-2&a+b+c&-c\\0&0&1\\(a+b-c)&c^2-{ab}&{ab}\end{vmatrix}$ [Applying C2 → C2 - C3]
$=(a-b)(b-c)(a+b+c)(c-a) \left\{(-1)\begin{vmatrix}-2&a+b+c&\\(a+b-c)&c^2-{ab}\end{vmatrix}\right\}$ [Expanding along R2]
$=-(a-b)(b-c)(a+b+c)(c-a)\{-2c^2+2{ab}-a^2-b^2-2{ab}+c^2\}$
$=-(a-b)(b-c)(a+b+c)(c-a)(-a^2-b^2-c^2)$ $=(a-b)(b-c)(c-b)(a+b+c)(a^2+b^2+c^2)$ $={R.H.S}$
View full question & answer→Question 415 Marks
Without expanding, show that the values of the following determinant are zero: $\begin{vmatrix}(2^{\text{x}}+2^{-\text{x}})^2&(2^{\text{x}}-2^{-\text{x}})^2&1\\(3^{\text{x}}+3^{-\text{x}})^2&(3^{\text{x}}-3^{-\text{x}})^2&1\\(4^{\text{x}}+4^{-\text{x}})^2&(4^{\text{x}}-4^{-\text{x}})^2&1\end{vmatrix}$
Answer$\begin{vmatrix}(2^{\text{x}}+2^{-\text{x}})^2&(2^{\text{x}}-2^{-\text{x}})^2&1\\(3^{\text{x}}+3^{-\text{x}})^2&(3^{\text{x}}-3^{-\text{x}})^2&1\\(4^{\text{x}}+4^{-\text{x}})^2&(4^{\text{x}}-4^{-\text{x}})^2&1\end{vmatrix}$
$=\begin{vmatrix}(2^{\text{x}}+2^{-\text{x}}+2)&(2^{\text{x}}-2^{-\text{x}}-2)&1\\(3^{\text{x}}+3^{-\text{x}}+2)&(3^{\text{x}}-3^{-\text{x}}-2)&1\\(4^{\text{x}}+4^{-\text{x}}+2)&(4^{\text{x}}-4^{-\text{x}}-2)&1\end{vmatrix}$
$=\begin{vmatrix}4&(2^{\text{x}}+2^{-\text{x}}-2)&1\\4&(3^{\text{x}}+3^{-\text{x}}-2)&1\\4&(4^{\text{x}}+4^{-\text{x}}-2)&1\end{vmatrix}$ [Applying $C_1 → C_1 - C_2$]
$=4\begin{vmatrix}1&(2^{\text{x}}+2^{-\text{x}}-2)&1\\1&(3^{\text{x}}+3^{-\text{x}}-2)&1\\1&(4^{\text{x}}+4^{-\text{x}}-2)&1\end{vmatrix}$ $=0$
View full question & answer→Question 425 Marks
Find values of k, if area of triangle is 4 square units whose vertices are:
(k, 0), (4, 0), (0, 2)
AnswerIf the area of a triangle with vertices (k, 0), (4, 0) and (0, 2) is 4 square units, then
$\triangle=\frac{1}{2}\begin{vmatrix}\text{k}&0&1\\4&0&1\\0&2&1\end{vmatrix}$
$=\frac{1}{2}\left\{(2)\times\begin{vmatrix}\text{k}&1\\4&1\end{vmatrix}\right\}$ [Expanding along $C_2$]
$=(\text{k}-4)$
Since area is always +ve, we take its absolute value, which is given as 4 square units.
$\Rightarrow(\text{k}-4)=\pm4$
$⇒ (k - 4) = 4$ or $(k - 4) = -4$
$⇒ k - 4 = 4$ or $k - 4 = -4$
$⇒ k = 8$ or $k = 0$
$⇒ k = 8, 0$
View full question & answer→Question 435 Marks
Prove that:
$\begin{vmatrix}\text{a}-\text{b}-\text{c}&2\text{a}&2\text{a}\\2\text{b}&\text{b}-\text{c}-\text{a}&2\text{b}\\2\text{c}&2\text{c}&\text{c}-\text{a}-\text{b} \end{vmatrix}=(\text{a}+\text{b}+\text{c})^3$
Answer$\begin{vmatrix}\text{a}-\text{b}-\text{c}&2\text{a}&2\text{a}\\2\text{b}&\text{b}-\text{c}-\text{a}&2\text{b}\\2\text{c}&2\text{c}&\text{c}-\text{a}-\text{b} \end{vmatrix}=(\text{a}+\text{b}+\text{c})^3$
$\text{L.H.S}=\begin{vmatrix}\text{a}-\text{b}-\text{c}&2\text{a}&2\text{a}\\2\text{b}&\text{b}-\text{c}-\text{a}&2\text{b}\\2\text{c}&2\text{c}&\text{c}-\text{a}-\text{b} \end{vmatrix}$
Apply: $\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3$
$\begin{vmatrix}\text{a}+\text{b}+\text{c}&\text{a}+\text{b}+\text{c}&\text{a}+\text{b}+\text{c}\\2\text{b}&\text{b}-\text{c}-\text{a}&2\text{b}\\2\text{c}&2\text{c}&\text{c}-\text{a}-\text{b} \end{vmatrix}$
Take (a + b + c) common from $R_1$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&1&1\\2\text{b}&\text{b}-\text{c}-\text{a}&2\text{b}\\2\text{c}&2\text{c}&\text{c}-\text{a}-\text{b} \end{vmatrix}$
Apply: $C_2 \rightarrow C_2-C_1, C_3 \rightarrow C_3-C_1$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&0&0\\2\text{b}&-\text{b}-\text{c}-\text{a}&0\\2\text{c}&2\text{c}&-\text{c}-\text{a}-\text{b} \end{vmatrix}$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&0&0\\2\text{b}&\text{b}+\text{c}+\text{a}&0\\2\text{c}&2\text{c}&\text{c}+\text{a}+\text{b} \end{vmatrix}$
$=(\text{a}+\text{b}+\text{c})^3$
$=\text{R.H.S}$
View full question & answer→Question 445 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\sin\text{x}&\cos\text{x}&\sin\text{y}\\-\cos\text{x}&\sin\text{x}&-\cos\text{y} \end{vmatrix}$
Answer$\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\sin\text{x}&\cos\text{x}&\sin\text{y}\\-\cos\text{x}&\sin\text{x}&-\cos\text{y} \end{vmatrix}$
$=\frac{1}{\sin\text{y}\cos\text{y}}\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y})\\\sin\text{x}\sin\text{y}&\cos\text{x}\sin\text{y}&\sin^2\text{y}\\-\cos\text{x}\cos\text{y}&\sin\text{x}\cos\text{y}&-\cos^2\text{y} \end{vmatrix}$
[Applying $R_2 \rightarrow$ siny $R_2$ and $R_3 \rightarrow$ cosy $R_3$ ]
$=\frac{1}{\sin\text{y}\cos\text{y}}\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\sin\text{x}\sin\text{y}-\cos\text{x}\cos\text{y}&\cos\text{x}\sin\text{y}+\sin\text{x}\sin\text{y}&\sin^2\text{y}-\cos^2\text{y}\\-\cos\text{x}\cos\text{y}&\sin\text{x}\cos\text{y}&-\cos^2\text{y} \end{vmatrix}$
$[$Applying $R_2→ R_2+ R_3]$
$=\frac{1}{\sin\text{y}\cos\text{y}}\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\-\cos\text{x}\cos\text{y}&\sin\text{x}\cos\text{y}&-\cos^2\text{y} \end{vmatrix}$
$=0$
View full question & answer→Question 455 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sin^2\text{A}&\cot\text{A}&1\\\sin^2\text{B}&\cot\text{B}&1\\\sin^2\text{C}&\cot\text{C}&1\end{vmatrix}$
Answer$\begin{vmatrix}\sin^2\text{A}&\cot\text{A}&1\\\sin^2\text{B}&\cot\text{B}&1\\\sin^2\text{C}&\cot\text{C}&1\end{vmatrix}$
$=\begin{vmatrix}\sin^2\text{A}-\sin^2\text{B}&\cot\text{A}-\cot\text{B}&0\\\sin^2\text{B}&\cot\text{B}&1\\\sin^2\text{C}-\sin^2\text{B}&\cot\text{C}-\cot\text{B}&0\end{vmatrix} [$Applying $R_1 → R_1 - R_2$ and $R_3 → R_3 - R_2]$
$=\begin{vmatrix}\sin(\text{A}+\text{B})\sin(\text{A}-\text{B})&\frac{\cos\text{A}\sin\text{B}-\cos\text{B}\sin\text{A}}{\sin\text{A}\sin\text{B}}&0\\\sin^2\text{B}&\cot\text{B}&1\\\sin(\text{C}+\text{B})\sin(\text{C}-\text{B})&\frac{\cos\text{C}\sin\text{B}-\cos\text{B}\sin\text{C}}{\sin\text{B}\sin\text{C}}&0\end{vmatrix}$
$=\begin{vmatrix}\sin(\pi-\text{C})\sin(\text{A}-\text{B})&\frac{-\sin(\text{A}-\text{B})}{\sin\text{A}\sin\text{B}}&0\\\sin^2\text{B}&\cot\text{B}&1\\\sin(\pi-\text{A})\sin(\text{C}-\text{B})&\frac{-\sin(\text{C}-\text{B})}{\sin\text{B}\sin\text{C}}&0\end{vmatrix}$ $[\because\text{A}+\text{B}+\text{C}=\pi]$
$=\begin{vmatrix}\sin\text{C}\sin(\text{A}-\text{B})&\frac{-\sin(\text{A}-\text{B})}{\sin\text{A}\sin\text{B}}&0\\\sin^2\text{B}&\frac{\cos\text{B}}{\sin\text{B}}&1\\\sin\text{A}\sin(\text{C}-\text{B})&\frac{-\sin(\text{C}-\text{B})}{\sin\text{A}\sin\text{B}}&0\end{vmatrix}$
$=\frac{\sin(\text{A}-\text{B})\sin(\text{C}-\text{B})}{\sin\text{B}}\begin{vmatrix}\sin\text{C}&\frac{-1}{\sin\text{A}}&0\\\sin^2\text{B}&\cos\text{B}&1\\\sin\text{A}&\frac{-1}{\sin\text{C}}&0\end{vmatrix}$
$=\frac{\sin(\text{A}-\text{B})\sin(\text{C}-\text{B})}{\sin\text{B}\sin\text{A}\sin\text{C}}\begin{vmatrix}\sin\text{C}\sin\text{A}&-1&0\\\sin^2\text{B}&\cos\text{B}&1\\\sin\text{A}\sin\text{C}&-1&0\end{vmatrix} [$Applying $R_1 → \sin A R_1$ and $R_3 → \sin C R_3]$
$=\frac{\sin(\text{A}-\text{B})\sin(\text{C}-\text{B})}{\sin\text{B}\sin\text{A}\sin\text{C}}\begin{vmatrix}0&0&0\\\sin^2\text{B}&\cos\text{B}&1\\\sin\text{A}\sin\text{C}&-1&0\end{vmatrix} [$Applying $R_1 → R_1 - R_3]$
$=0$
View full question & answer→Question 465 Marks
Prove that:
$\begin{vmatrix}\text{a}^2&\text{bc}&\text{ac}+\text{c}^2\\\text{a}^2+\text{ab}&\text{b}^2&\text{ac}\\\text{ab}&\text{b}^2+\text{ac}&\text{c}^2\end{vmatrix}=4\text{a}^2\text{b}^2\text{c}^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^2&\text{bc}&\text{ac}+\text{c}^2\\\text{a}^2+\text{ab}&\text{b}^2&\text{ac}\\\text{ab}&\text{b}^2+\text{ac}&\text{c}^2\end{vmatrix}$
$=\text{abc}\begin{vmatrix}\text{a}&\text{c}&\text{a}+\text{c}\\\text{a}+\text{b}&\text{b}&\text{a}\\\text{b}&\text{b}+\text{c}&\text{c}\end{vmatrix}$
$[$Taking out a, b and c common from $C_1, C_2$ and $C_3]$
$=\text{abc}\begin{vmatrix}\text{a}&\text{c}&0\\\text{a}+\text{b}&\text{b}&-2\text{b}\\\text{b}&\text{b}+\text{c}&-2\text{b}\end{vmatrix}$
$[$Applying $C_3→ C_3 - C_2 - C_1]$
$=(\text{abc})(-2\text{b})\begin{vmatrix}\text{a}&\text{c}&0\\\text{a}+\text{b}&\text{b}&1\\\text{b}&\text{b}+\text{c}&1\end{vmatrix}$
$[$Taking $(-2b)$ common from $C_3]$
$=(\text{abc})(-2\text{b})\begin{vmatrix}\text{a}&\text{c}&0\\\text{a}&-\text{c}&0\\\text{b}&\text{b}+\text{c}&1\end{vmatrix}$
$[$Applying $R_2 → R_2 - R_1]$
$=(\text{abc})(-2\text{b})\times1\begin{vmatrix}\text{a}&\text{c}\\\text{a}&-\text{c}\end{vmatrix}$
$[$expanding along $C_3]$
$=(\text{abc})(-2\text{b})(-2\text{ac})$
$=4\text{a}^2\text{b}^2\text{c}^2$
$=\text{R.H.S}$
View full question & answer→Question 475 Marks
Prove that:
$\begin{vmatrix}\text{a}^2&2\text{ab}&\text{b}^2\\\text{b}^2&\text{a}^2&2\text{ab}\\2\text{ab}&\text{b}^2&\text{a}^2\end{vmatrix}=(\text{a}^3+\text{b}^3)^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^2&2\text{ab}&\text{b}^2\\\text{b}^2&\text{a}^2&2\text{ab}\\2\text{ab}&\text{b}^2&\text{a}^2\end{vmatrix}$
$=\text{a}^2\begin{vmatrix}\text{a}^2&2\text{ab}\\\text{b}^2&\text{a}^2\end{vmatrix}-(2\text{ab})\begin{vmatrix}\text{b}^2&2\text{ab}\\2\text{ab}&\text{a}^2\end{vmatrix}+\text{b}^2\begin{vmatrix}\text{b}^2&\text{a}^2\\2\text{ab}&\text{b}^2\end{vmatrix}$ [Expanding]
$=\text{a}^2(\text{a}^4-2\text{ab}^3)-(2\text{ab})(\text{b}^2\text{a}^2-4\text{a}^2\text{b}^2)+\text{b}^2(\text{b}^4-2\text{a}^3\text{b})$
$=\text{a}^6-2\text{a}^3\text{b}^3-2\text{a}^3\text{b}^3+8\text{a}^3\text{b}^3+\text{b}^6-2\text{a}^3\text{b}^3$
$=\text{a}^6+2\text{a}^3\text{b}^3+(\text{b}^3)^2$
$=(\text{a}^3+\text{b}^3)^2$
$=\text{R.H.S}$
View full question & answer→Question 485 Marks
Solve the following determinant equations:
$\begin{vmatrix}\text{x}+\text{a}&\text{b}&\text{c}\\\text{a}&\text{x}+\text{b}&\text{c}\\\text{a}&\text{b}&\text{x}+\text{c}\end{vmatrix}=0$
AnswerLet $\begin{vmatrix}\text{x}+\text{a}&\text{b}&\text{c}\\\text{a}&\text{x}+\text{b}&\text{c}\\\text{a}&\text{b}&\text{x}+\text{c}\end{vmatrix}$
$=\begin{vmatrix}\text{x}+\text{a}+\text{b}+\text{c}&\text{b}&\text{c}\\\text{x}+\text{a}+\text{b}+\text{c}&\text{x}+\text{b}&\text{c}\\\text{x}+\text{a}+\text{b}+\text{c}&\text{b}&\text{x}+\text{c}\end{vmatrix} [$Applying $C_1 → C_1 + C_2 + C_3]$
$=(\text{x}+\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{b}&\text{c}\\1&\text{x}+\text{b}&\text{c}\\1&\text{b}&\text{x}+\text{c}\end{vmatrix}$
$=(\text{x}+\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{b}&\text{c}\\0&\text{x}&0\\1&\text{b}&\text{x}+\text{c}\end{vmatrix}$ $[$Applying $R_2 → R_2 - R_1]$
$=(\text{x}+\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{b}&\text{c}\\0&\text{x}&0\\1&0&\text{x}\end{vmatrix} [$Applying $R_3 → R_3 - R_1]$
$=(\text{x}+\text{a}+\text{b}+\text{c})(\text{x}^2-0)=0$ [Given]
$\Rightarrow\text{x}^2=0$ or $\text{x}+\text{a}+\text{b}+\text{c}=0$
$\Rightarrow\text{x}=0$ or $\text{x}=-(\text{a}+\text{b}+\text{c})$
View full question & answer→Question 495 Marks
For what value of x the matrix A is singular?
$\text{A}=\begin{vmatrix}1+\text{x}&7\\3-\text{x}&8 \end{vmatrix}$
Answer$\text{A}=\begin{vmatrix}1+\text{x}&7\\3-\text{x}&8 \end{vmatrix}$
A matrix A is called singular if |A| = 0
Now expanding along the first row |A|
$=(\text{x}-1)\begin{vmatrix}\text{x}-1&1\\1&\text{x}-1 \end{vmatrix}-1\begin{vmatrix}1&1\\1&\text{x}-1 \end{vmatrix}+1\begin{vmatrix}1&1\\\text{x}-1&-1 \end{vmatrix}$
$=(\text{x}-1)\big[(\text{x}-1)^2-1\big]-1[\text{x}-1-1]+1[1-\text{x}+1]$
$=(\text{x}-1)(\text{x}^2+1-2\text{x}-1)-1(\text{x}-2)+1(2-\text{x})$
$=(\text{x}-1)(\text{x}^2-2\text{x})-\text{x}+2+2-\text{x}$
$=(\text{x}-1)\times(\text{x})\times(\text{x}-2)+(4-2\text{x})$
$=(\text{x}-1)\times(\text{x})\times(\text{x}-2)+2(2-\text{x})$
$=(\text{x}-1)\times(\text{x})\times(\text{x}-2)-2(\text{x}-2)$
$=(\text{x}-2)[\text{x}(\text{x}-1)-2]$ (Taking (x - 2) common)
Since A is a singular matrix, so |A| = 0
i.e., $(\text{x}-2)(\text{x}^2-\text{x}-2)=0$
either $(\text{x}-2)=0$ or $\text{x}^2-\text{x}-2=0$
$\text{x}=2$ or $\text{x}^2-2\text{x}+\text{x}-2=0$
$\text{x}(\text{x}-2)+1(\text{x}-2)=0$
$(\text{x}-2)(\text{x}+1)=0$
$\text{x}=2,-1$
$\text{x}=2\text{ or}-1$
View full question & answer→Question 505 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\text{a}+\text{b}&2\text{a}+\text{b}&3\text{a}+\text{b}\\2\text{a}+\text{b}&3\text{a}+\text{b}&4\text{a}+\text{b}\\4\text{a}+\text{b}&5\text{a}+\text{b}&6\text{a}+\text{b} \end{vmatrix}$
Answer$\begin{vmatrix}\text{a}+\text{b}&2\text{a}+\text{b}&3\text{a}+\text{b}\\2\text{a}+\text{b}&3\text{a}+\text{b}&4\text{a}+\text{b}\\4\text{a}+\text{b}&5\text{a}+\text{b}&6\text{a}+\text{b} \end{vmatrix}$
Apply: $C_3 → C_3 - C_2$
$\begin{vmatrix}\text{a}+\text{b}&2\text{a}+\text{b}&\text{a}\\2\text{a}+\text{b}&3\text{a}+\text{b}&\text{a}\\4\text{a}+\text{b}&5\text{a}+\text{b}&\text{a} \end{vmatrix}$
Apply: $C_2 → C_2 - C_1$
$\begin{vmatrix}\text{a}+\text{b}&\text{a}&\text{a}\\2\text{a}+\text{b}&\text{a}&\text{a}\\4\text{a}+\text{b}&\text{a}&\text{a} \end{vmatrix}$
$=0$
$\because\text{C}_3=\text{C}_2$
View full question & answer→