3 Marks Question

Question 13 Marks

On her birthday Seema decided to donate some money to children of an orphanage home. If there were 8 children less, everyone would have got ₹ 10 more. However if there were 16 children more, everyone would have got ₹ 10 less. Using matrix method, find the number of children and the amount distributed by Seema.

Answer

Let the number of children be $x$ and the amount distributed by Seema for one student be ₹ $y.$
$\text{So,}\quad (x-8)(y+10)=xy$
$\Rightarrow 5x-4y=40\quad\ldots \text{(i)}$
$\text{and}\quad (x+16)(y-10)=xy$
$\Rightarrow 5x-8y=-80\quad\ldots\text{(ii)}$
Here $A=\left[\begin{array}{ll}5 & -4 \\ 5 & -8\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right], B=\left[\begin{array}{c}40 \\ -80\end{array}\right]$
$AX=B \Rightarrow X=A^{-1}B$
$A^{-1}=-\frac{1}{20}\left[\begin{array}{ll}-8 & 4 \\ -5 & 5\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}32 \\ 30\end{array}\right]$
$\Rightarrow x=32,y=30$
No. of students $=32$
Amount given to each student $=$ ₹ $30.$

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Question 23 Marks

Ishan wants to donate a rectangular plot of land for a school in his village. When he was asked to give dimensions of the plot, he told that if its length is decreased by 50 m and breadth is increased by 50 m, then its area will remain the same. But if length is decreased by 10 m and breadth is decreased by 20 m, then its area will decrease by $5300 \text{m}^2.$ Using matrices, find the dimensions of the plot.

Answer

Let length be $\text{x m}$ and breadth be $\text{y m}.$
$\therefore \quad(x-50)(y+50)=x y$
$\Rightarrow 50x-50y=2500 \text{ or }x-y=50$
$\text{and }(x-10)(y-20)=xy-5300$
$\Rightarrow 2x+y=550$
$\left(\begin{array}{cc}1 & -1 \\ 2 & 1\end{array}\right)\binom{x}{y}=\binom{50}{550}$
$\Rightarrow\binom{x}{y}=\frac{1}{3}\left(\begin{array}{cc}1 & 1 \\ -2 & 1\end{array}\right)\binom{50}{550}$
$\Rightarrow x=\frac{1}{3}(600)=200 \text{ m,}$
$y=\frac{1}{3}(450)=150\text{ m}$

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Question 33 Marks

Let $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{cc}4 & -6 \\ -2 & 4\end{array}\right]$. Then compute $A B$. Hence, solve the following system of equations: $2 x+y=4,3 x+2 y=1.$

Answer

$\begin{array}{l}A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right] \\ \text { and } B=\left[\begin{array}{cc}4 & -6 \\ -2 & 4\end{array}\right]\end{array}$
$\begin{array}{r}\text { then } A B=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\left[\begin{array}{cc}4 & -6 \\ -2 & 4\end{array}\right] \\ A B=\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]=2 I\end{array}$
$\Rightarrow \quad A\left(\frac{1}{2} B\right)=I$
On multiplying by $A^{-1}$
$\begin{aligned}A^{-1} & =\frac{1}{2} B \\A^{-1} & =\left[\begin{array}{cc}2 & -3 \\-1 & 2\end{array}\right]\end{aligned}$
The given system of equations are equivalent to $A^{\prime} X=C,$
where $X=\left[\begin{array}{l}x \\ y\end{array}\right]$
$\text{and}\quad C=\left[\begin{array}{l}4 \\1\end{array}\right], A^{\prime}=\left[\begin{array}{ll}2 & 1 \\3 & 2\end{array}\right]$
$\begin{array}{rlrl}X & =\left(A^{\prime}\right)^{-1} C=\left(A^{-1}\right)^{\prime} C \\ \Rightarrow \quad\left[\begin{array}{l}x \\ y\end{array}\right] & =\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]\left[\begin{array}{l}4 \\ 1\end{array}\right] \\ \Rightarrow \quad\left[\begin{array}{l}x \\ y\end{array}\right] & =\left[\begin{array}{c}7 \\ -10\end{array}\right]\end{array}$
$\therefore x=7$
$\text{and }y=-10$

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Question 43 Marks

If $A=\left[\begin{array}{cc}1 & -2 \\ 2 & 1\end{array}\right]$ then using $A^{-1}$, solve the following system of equations : $x-2 y=-1,2 x+y=2$.

Answer

$|A|=5$
$\operatorname{adj} A=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]$
$A^{-1}=\frac{\operatorname{adj} A}{|A|}=\frac{1}{5}\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]$
Given system of equations is $AX=B,$ where
$A=\left[\begin{array}{cc}1 & -2 \\ 2 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$
$B=\left[\begin{array}{c}-1 \\ 2\end{array}\right]$
$\left[\begin{array}{cc}1 & -2 \\ 2 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}-1 \\ 2\end{array}\right]$
$\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{5}\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{r}-1 \\ 2\end{array}\right]$
$\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{5}\left[\begin{array}{c}-1+4 \\ 2+2\end{array}\right]=\left[\begin{array}{l}\frac{3}{5} \\ \frac{4}{5}\end{array}\right]$
$x=\frac{3}{5}$ and $y=\frac{4}{5}$

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Question 53 Marks

Using properties of determinants, prove the following:
$\left|\begin{array}{ccc}1+a^2-b^2 & 2 a b & -2 b \\2 a b & 1-a^2+b^2 & 2 a \\2 b & -2 a & 1-a^2-b^2\end{array}\right|=\left(1+a^2+b^2\right)^3 .$

Answer

$\begin{array}{l}\Delta=\left|\begin{array}{ccc}1+a^2-b^2 & 2 a b & -2 b \\ 2 a b & 1-a^2+b^2 & 2 a \\ 2 b & -2 a & 1-a^2-b^2\end{array}\right| \\ \text { Using } C_1 \rightarrow C _1-b C _{3'} C_2 \rightarrow C _2+a C _3\end{array}$
$=\left|\begin{array}{ccc}1+a^2+b^2 & 0 & -2 b \\ 0 & 1+a^2+b^2 & 2 a \\ b\left(1+a^2+b^2\right) & -a\left(1+a^2+b^2\right) & 1-a^2-b^2\end{array}\right|$
Taking $\left(1+a^2+b^2\right)$ common from $C _1$ and $C _2$
$=\left(1+a^2+b^2\right)^2\left|\begin{array}{ccc}1 & 0 & -2 b \\0 & 1 & 2 a \\b & -a & 1-a^2-b^2\end{array}\right|$
$\begin{aligned} \text { Using } & R_3 \rightarrow R_3+a R_2-b R_1 \\ & =\left(1+a^2+b^2\right)^2\left|\begin{array}{ccc}1 & 0 & -2 b \\ 0 & 1 & 2 a \\ 0 & 0 & 1+a^2+b^2\end{array}\right|\end{aligned}$
Taking $\left(1+a^2+b^2\right)$ common from $R_3$
$=\left(1+a^2+b^2\right)^3\left|\begin{array}{ccc}1 & 0 & -2 b \\0 & 1 & 2 a \\0 & 0 & 1\end{array}\right|$
Expanding with respect to $R_3$
$\begin{aligned}\Delta & =\left(1+a^2+b^2\right)^3(1) \\\text { or } \Delta & =\left(1+a^2+b^2\right)^3\end{aligned}$

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Question 63 Marks

Using properties of determinants, prove that:
$\left|\begin{array}{ccc}1+a & 1 & 1 \\1 & 1+b & 1 \\1 & 1 & 1+c\end{array}\right|=a b c+b c+c a+a b.$

Answer

$\begin{array}{l}R_1 \rightarrow \frac{1}{a} R_1, R_2 \rightarrow \frac{1}{b} R_2, R_3 \rightarrow \frac{1}{c} R_3 \\ \therefore \quad \text { LHS }=a b c\left|\begin{array}{ccc}\frac{1}{a}+1 & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1\end{array}\right| \\ R_1 \rightarrow R_1+R_2+R_3\end{array}$
$=a b c\left|\begin{array}{ccc}1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\\frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\\frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1\end{array}\right|$
Taking $\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ common from $R_1$
$\begin{array}{l}=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left|\begin{array}{ccc}1 & 1 & 1 \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1\end{array}\right| \\ C_2 \rightarrow C_2-C_1 \text { and } C_3 \rightarrow C_3-C_1\end{array}$
$=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left|\begin{array}{lll}1 & 0 & 0 \\ \frac{1}{b} & 1 & 0 \\ \frac{1}{c} & 0 & 1\end{array}\right|$
Expand along $R_1$
$\begin{array}{l}=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(1) \\ =a b c+b c+c a+a b=\text { RHS }\end{array}$

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Question 73 Marks

Using properties of determinants, prove the following:
$\left|\begin{array}{ccc}a+b+c & -c & -b \\-c & a+b+c & -a \\-b & -a & a+b+c\end{array}\right|=2(a+b)(b+c)(c+a) \text {. }$

Answer

$\text{LHS}=\left|\begin{array}{ccc}a+b+c & -c & -b \\ -c & a+b+c & -a \\ -b & -a & a+b+c\end{array}\right|$
$\text{C}_{2} \rightarrow \text{C}_{2}+\text{C}_{1'}\text{C}_{3} \rightarrow \text{C}_{3}+\text{C}_{1}$
$=\left|\begin{array}{ccc}a+b+c & a+b & a+c \\ -c & a+b & -(a+c) \\ -b & -(a+b) & (a+b)\end{array}\right|$
$=(a+b)(a+c)\left|\begin{array}{crr}a+b+c & 1 & 1 \\ -c & 1 & -1 \\ -b & -1 & 1\end{array}\right|$
$\text{C}_{3} \rightarrow \text{C}_{3}+\text{C}_{2}$
$=(a+b)(a+c)\left|\begin{array}{ccc}a+b+c & 1 & 2 \\ -c & 1 & 0 \\ -b & -1 & 0\end{array}\right|$
$=2(a+b)(a+c)(c+b)=\text{RHS.}$

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Question 83 Marks

Using properties of determinants, prove that:
$\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 3 & 4+3 p & 2+4 p+3 q \\ 4 & 7+4 p & 2+7 p+4 q\end{array}\right|=1$

Answer

$\text{Taking}\quad\text{L.H.S.}=\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 3 & 4+3 p & 2+4 p+3 q \\ 4 & 7+4 p & 2+7 p+4 q\end{array}\right|$
$\text{Applying }R_{2} \rightarrow R_{2}-3R_{1}$
$=\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 0 & 1 & -1+p \\ 4 & 7+4 p & 2+7 p+4 q\end{array}\right|$
$\text{Applying } R_{3} \rightarrow R_{3}-4R_{1}$
$=\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 0 & 1 & -1+p \\ 0 & 3 & -2+3 p\end{array}\right|$
$\text{Applying } R_{3} \rightarrow R_{3}-3R_{2}$
$=\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 0 & 1 & -1+p \\ 0 & 0 & 1\end{array}\right|$
$\text{Expanding along }R_{1}=$ $\left|\begin{array}{cc}1 & -1+p \\ 0 & 1\end{array}\right|$
$=1=\text{R.H.S. Hence Proved.}$

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Question 93 Marks

Using properties of determinants, solve for $x$ :
$\left|\begin{array}{lll}a+x & a-x & a-x \\a-x & a+x & a-x \\a-x & a-x & a+x\end{array}\right|=0$

Answer

$\begin{array}{l}\left|\begin{array}{lll}a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x\end{array}\right|=0 \\ \text { Applying } R_1 \rightarrow R_1+R_2+R_3 \\ \qquad\left|\begin{array}{ccc}3 a-x & 3 a-x & 3 a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x\end{array}\right|=0\end{array}$
$\text{Applying }C_2 \rightarrow C_2-C_1$
$\left|\begin{array}{ccc}3 a-x & 0 & 3 a-x \\a-x & 2 x & a-x \\a-x & 0 & a+x\end{array}\right|=0$
$\text{Applying }C_3 \rightarrow C_3-C_1$
$\left|\begin{array}{ccc}3 a-x & 0 & 0 \\a-x & 2 x & 0 \\a-x & 0 & 2 x\end{array}\right|=0$
$\begin{array}{l}\text { Expanding along } C_3 \\ \qquad 4 x^2(3 a-x)=0 \\ \therefore x=0,3 a\end{array}$

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Applied Maths 3 Marks Question

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