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1. Answer any fifteen questions in PART- A. All questions carry
equal marks. (15 X 2 =30)
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2. Answer all questions, choosing any two sub-divisions
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each question under Part-B. All questions carry equal marks.
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(5 X 14 = 70) ( 7 + 7)
\[\underline{PART\ -\ A}\]
\[1.\ \color{green}{If\ A =\begin{bmatrix}
9 & 10 \\
13 & 20\\
\end{bmatrix}\ and\ B =\begin{bmatrix}
3 & 5 \\
8 & 9\\
\end{bmatrix}},\ \hspace{15cm}\]
\[\color {green}{\ find}\ 2A\ +\ B\ \hspace{10cm}\]
\[\color {blue}{Solution:}\ \hspace{18cm}\]
\[ 2A = 2\begin{bmatrix}
9 & 10 \\
13 & 20\\
\end{bmatrix}\ \hspace{13cm}\]
\[2A = \begin{bmatrix}
18 & 20 \\
26 & 40\\
\end{bmatrix}\ ———- (1)\ \hspace{8cm}\]
\[B = \begin{bmatrix}
3 & 5 \\
8 & 9\\
\end{bmatrix}\ ———- (2)\ \hspace{8cm}\]
\[2A\ +\ B = \begin{bmatrix}
18 & 20 \\
26 & 40\\
\end{bmatrix}\ +\ \begin{bmatrix}
3 & 5 \\
8 & 9\\
\end{bmatrix}\ \hspace{8cm}\]
\[=\begin{bmatrix}
18\ +\ 3 & 20\ +\ 5 \\
26\ +\ 8 & 40\ +\ 9\\
\end{bmatrix}\ \hspace{8cm}\]
\[2A\ +\ B = \begin{bmatrix}
21 & 25 \\
34 & 49\\
\end{bmatrix}\ \hspace{10cm}\]
\[2. \ \color{green}{Find\ the\ value\ of\ x\ if\ \begin{vmatrix}
2 & 3 \\
4 & x \\
\end{vmatrix}\ = 0}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{5cm}\ \begin{vmatrix}
2 & 3 \\
4 & x \\
\end{vmatrix}\ = 0 \hspace{15cm}\]
\[2(x)\ -\ 4 (3) = 0\ \hspace{13cm}\]
\[2x\ -\ 12\ =\ 0\ \hspace{13cm}\]
\[2x\ =\ 12\ \hspace{13cm}\]
\[x\ =\ 6\ \hspace{13cm}\]
\[3.\ \color{green}{Find\ the\ cofactor\ of\ -\ 2\ in}\ \hspace{20cm}\]
\[\hspace{5cm}\ \color{green}{ \begin{vmatrix}
1 & – 1 & 1 \\
2 & 3 & -\ 3 \\
6 & -\ 2 & -\ 1 \\
\end{vmatrix}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[cofactor\ of\ -\ 2\ =\ (-1)^{3\ +\ 2}\ \begin{vmatrix}
1 & 1 \\
2 & -\ 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (-\ 3\ -\ 2)\ \hspace{15cm}\]
\[= (-1) (-\ 5)\ \hspace{15cm}\]
\[\boxed{cofactor\ of\ -\ 2 =\ 5}\ \hspace{15cm}\]
\[4.\ \color{green}{Find\ the\ inverse\ matrix\ of\ \begin{pmatrix}
5 & 2 \\
-4 & 3 \\
\end{pmatrix}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\color {blue}{Solution:}\ Let\ A\ =\begin{pmatrix}
5 & 2 \\
-4 & 3 \\
\end{pmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =\ 5(3)\ -\ 2(-4) \hspace{13cm}\]
\[=\ 15\ +\ 8 \hspace{12cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ =\ 23\ \neq {0}\ \hspace{13cm}\]
\[\therefore\ A^{-1}\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 5 = (-1)^{1\ +\ 1}\ (3)\ \hspace{15cm}\]
\[= (-1)^2 (3)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = 3\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{1\ +\ 2}\ (-4)\ \hspace{15cm}\]
\[= (-1)^3 (-4)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 4\ \hspace{15cm}\]
\[cofactor\ of\ -4 = (-1)^{2\ +\ 1}\ (2)\ \hspace{15cm}\]
\[= (-1)^3 (2)\ \hspace{15cm}\]
\[cofactor\ of\ – 4 =\ -\ 2\ \hspace{15cm}\]
\[cofactor\ of\ 3\ = (-1)^{2\ +\ 2}\ (5)\ \hspace{15cm}\]
\[= (-1)^4 (5)\ \hspace{15cm}\]
\[cofactor\ of\ 3 =\ 5\ \hspace{15cm}\]
\[\therefore\ cofactor\ matrix\ = \begin{pmatrix}
3 & 4 \\
-2 & 5 \\
\end{pmatrix}\ \hspace{15cm}\]
\[Adj.\ A = \begin{pmatrix}
3 & -2 \\
4 & 5 \\
\end{pmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{1}{23}\ \begin{pmatrix}
3 & -2 \\
4 & 5 \\
\end{pmatrix}\ \hspace{2cm}\]
\[5.\ \color{green}{Convert\ 240^0\ to\ equivalent\ radians\ value}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ 240^0\ \times \frac{\pi\ radians}{180^0}\]
\[=\ \hspace{2cm}\ \frac{4}{3}\ \pi\ radians\]
\[6.\ \color{green}{Write\ any\ two\ characteristics\ of\ the\ function\ y\ =\ e^x}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ 1.\ \textbf{Exponential Growth}\ \hspace{15cm}\]
\[\hspace{3cm}\ The\ function\ y\ =\ e^x\ exhibits\ exponential\ growth.\ As\ x\ increases,\ y\ increases\ very\ rapidly\]
\[\hspace{3cm}\ Conversely,\ as x\ decreases,\ y\ approaches\ zero\ but\ never\ reaches\ it\]
\[\hspace{2cm}\ 2.\ \textbf{Derivative and Integral properties}\ \hspace{15cm}\]
\[\hspace{3cm}\ The\ function\ y\ =\ e^x\ \text{is unique in that it is its own derivative and integral}\]
\[\hspace{3cm}\ \frac{d}{dx}\ e^x\ =\ e^x\ and\ \int{e^x}\ dx\ =\ e^x\ +\ c\]
\[7.\ \color{green}{Evaluate\ cos\ {75}^0}\ \hspace{18cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{Given that}\ 75^0\ =\ 30^0\ +\ 45^0\ \text{we use angle sum formula for cosine}\]
\[\hspace{2cm}\ W.\ K.\ T\ cos( A + B )\ =\ cos A\ Cos B\ -\ sin A sin B\]
\[\hspace{3cm}\ cos( 75^0 )\ =\ cos(30^0\ +\ cos(45^0)\]
\[\hspace{3cm}\ =\ c os\ 30^0\ cos\ 45^0\ -\ sin\ 30^0\ sin\ 45^0\]
\[\hspace{3cm}\ =\ \frac{\sqrt{3}}{2}\ \frac{1}{\sqrt{2}}\ -\ \frac{1}{2}\ \frac{1}{\sqrt{2}}\]
\[\hspace{4cm}\ =\ \frac{\sqrt{3}\ -\ 1}{2\ \sqrt{2}}\]
\[8.\ \color{green}{If\ tan\ A\ =\ \frac{1}{2},\ find\ tan\ 2A}\ \hspace{18cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ tan\ A\ =\ \frac{1}{2}\ \hspace{18cm}\]
\[\hspace{2cm}\ W.\ K.\ T\ tan\ 2A\ =\ \frac{2\ tan\ A}{1\ -\ tan^2\ A}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{2\ \times\ \frac{1}{2}}{1\ -\ (\frac{1}{2})^2}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{1\ -\ \frac{1}{4}}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\frac{4\ -\ 1}{4}}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\frac{3}{4}}\ \hspace{10cm}\]
\[\hspace{4cm}\ =\ \frac{4}{3}\ \hspace{10cm}\]
\[9.\ \color{green}{If\ \overrightarrow{a}\ =\ 5\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\ 3\overrightarrow{k}\ and\ \overrightarrow{b}\ = -\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 5\overrightarrow{k},\ find\ 4\overrightarrow{a}\ +\ 6\overrightarrow{b}}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}= 5\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\ 3\overrightarrow{k}\]
\[\overrightarrow{b}\ =\ -\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 5\overrightarrow{k}\]
\[4\overrightarrow{a}\ +\ 6\overrightarrow{b}\ =\ 4(5\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\ 3\overrightarrow{k})\ +\ 6( -\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 5\overrightarrow{k})\]
\[=\ 20\overrightarrow{i}\ +\ 8\overrightarrow{j}\ -\ 12\overrightarrow{k}\ -\ 18\overrightarrow{i}\ -\ 12\overrightarrow{j}\ +\ 30\overrightarrow{k}\]
\[4\overrightarrow{a}\ +\ 6\overrightarrow{b}\ =\ 2\overrightarrow{i}\ -\ 4\overrightarrow{j}\ +\ 18\overrightarrow{k}\ \hspace{5cm}\]
\[10.\ \color{green}{Find\ the\ Direction\ cosines\ of\ the\ vector\ \overrightarrow{i}\ + 2\overrightarrow{j}\ -\ 3\overrightarrow{k}}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ \overrightarrow{i}\ + 2\overrightarrow{j}\ -\ 3\overrightarrow{k}\]
\[r =\overrightarrow{|a|} = \sqrt{(1)^2 + (2)^2 + (-3)^2 }\]
\[ = \sqrt{(1\ +\ 4\ +\ 9) }\]
\[r =\sqrt{14}\]
\[ Direction\ cosines\ are \frac{1}{\sqrt(14)}, \frac{2}{\sqrt(14)}, \frac{-3}{\sqrt(14)} \]
\[11.\ \color{green}{Show\ that\ the\ vectors\ \overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k} and\ -\ 2\overrightarrow{i}\ +\ 6\overrightarrow{j}\ +\ 4\overrightarrow{k} are\ mutually\ perpendicular}\ \hspace{2cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}= \overrightarrow{i} – 3\overrightarrow{j} + 5\overrightarrow{k} \]
\[\overrightarrow{b}= – 2\overrightarrow{i}+ 6\overrightarrow{j}+4\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i} – 3\overrightarrow{j} + 5\overrightarrow{k}) .(- 2\overrightarrow{i}+ 6\overrightarrow{j}+4\overrightarrow{k})\]
\[=\ 1(-2)\ +\ -3(6)\ +\ 5 (4)\]
\[=\ -\ 2\ -\ 18\ +\ 20\]
\[=\ 0\]
\[\overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors.\]
\[12.\ \color{green}{Find\ the\ values\ of\ \overrightarrow{i}\ . \ \overrightarrow{j}\ and\ \overrightarrow{i}\ \times\ \overrightarrow{j}}\ \hspace{18cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ \overrightarrow{i}\ . \ \overrightarrow{j}\ =\ 0\ and\ \overrightarrow{i}\ \times\ \overrightarrow{j}\ =\ \overrightarrow{k}\ \hspace{15cm}\]
\[13.\ \color{green}{Calculate\ the\ arithmetic\ mean\ of\ 14,\ 26,\ 28,\ 20,\ 32,\ 30}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ n\ =\ 6\ \hspace{20cm}\]
\[\Sigma x_i\ =\ 14\ +\ 26\ +\ +\ 28\ +\ 20\ +\ 32\ +\ 30\]
\[\Sigma x_i\ =\ 150\]
\[\bar{x}\ =\ \frac{\Sigma x_i}{n}\]
\[=\ \frac{150}{6}\]
\[\bar{x}\ =\ 25\]
\[14.\ \color{green}{The\ arithmetic\ mean\ of\ 10\ numbers\ is\ 20\ ,\ find\ the\ total\ of\ the}\ \hspace{10cm}\]\[\color{green}{numbers}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ n\ =\ 10\ and\ \bar{x}\ =\ 20\ \hspace{17cm}\]
\[\bar{x} \ =\ \frac{\Sigma X}{n}\]
\[20\ =\ \frac{\Sigma X}{10}\]
\[\therefore\ \Sigma X\ =\ 200\]
\[15.\ \color{green}{\text{If the standard deviation of a data is 6.8,}}\ \color{green}{find\ its\ variance}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ \sigma\ =\ 6.8\ \hspace{20cm}\]
\[\hspace{2cm}\ W.\ K.\ T\ variance\ =\ \sigma^2\]
\[\hspace{3cm}\ =\ (6.8)^2\]
\[\hspace{2cm}\ \boxed{variance\ =\ 46.24}\]
\[16.\ \color{green}{Write\ the\ normal\ equations\ of\ the\ straight\ line\ y\ =\ a\ x\ +\ b}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
The Normal equations are
\[a\ \Sigma x_i\ +\ nb\ =\ \Sigma y_i\ ——–\ (1)\]
\[a\ \Sigma x_i^2\ +\ b\ \Sigma\ x_i\ =\ \Sigma x_i\ y_i\ ——–\ (2)\]
\[17.\ \color{green}{A\ card\ is\ picked\ randomly\ from\ a\ pack\ of\ 52\ cards.}\ \hspace{15cm}\]\[\color {green} {Find\ the\ probability\ of\ getting\ a\ King}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
In a standard deck of 52 playing cards, there are 4 Kings (one from each suit: hearts, diamonds, clubs, and spades.
The probability PP of drawing a King from a deck of 52 cards can be calculated as follows:
\[P(King)\ =\ \frac{Number\ of\ Kings}{Total\ number\ of\ cards}\ =\ \frac{4}{52}\ =\ \frac{1}{13}\]
\[\text{So, the probability of picking a King from a standard deck of 52 cards is:}\ \boxed{\frac{1}{13}}\]
\[18.\ \color{green}{A\ die\ is\ rolled\ once,}\ \hspace{15cm}\]\[\color {green} {Find\ the\ probability\ of\ getting\ an\ odd\ number.}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{A standard die has 6 faces numbered from 1 to 6. The odd numbers on a die are 1, 3, and 5.}\]
\[\text{The probability P of rolling an odd number can be calculated as follows:}\]
\[P(King)\ =\ \frac{Number\ of\ Kings}{Odd\ number}\ =\ \frac{3}{6}\ =\ \frac{1}{2}\]
\[\text{So, the probability of rolling an odd number on a standard die is:}\ \boxed{\frac{1}{2}}\]
\[19.\ \color{green}{If\ P(A)\ =\ 0.5,\ P(B)\ =\ 0.3,\ and\ A\ \cap\ B\ is\ empty,}\ \hspace{10cm}\]\[\color {green} {find\ P(A\ \cup\ B)}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{Given P(A) = 0.5, P(B) = 0.3 and} A \cap\ B\ is\ empty\ (\text{ which means events A and B are mutually exclusive),}\]
\[we\ can\ find\ P(A\ \cup\ B)\ \text{using the formula for the union of two mutually exclusive events:}\]
\[P(A\ \cup\ B\ =\ P(A)\ +\ P(B)\]
\[\hspace{2cm}\ =\ 0.5\ +\ 0.3\]
\[\boxed{P(A\ \cup\ B\ =\ 0.8}\]
\[20.\ \color{green}{f\ P(A)\ =\ \frac{1}{3},\ P(B)\ =\ \frac{3}{4},\ P(A\ \cap\ B)\ =\ \frac{1}{6},}\ \hspace{10cm}\]\[\color{green}{find\ P(A/B)\ and\ P(B/A)}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[To\ find\ the\ conditional\ probabilities\ P(A/B)\ and\ P(B/A),\ \text{We use the definitions of conditional probability.}\]
\[1.\ P(A/B)\ =\ \frac{P(A\ \cap\ B)}{P(B)}\]
\[2.\ P(B/A)\ =\ \frac{P(A\ \cap\ B)}{P(A)}\]
\[Given\ \hspace{15cm}\]
\[\bullet\ P(A)\ =\ \frac{1}{3}\ \hspace{10cm}\]
\[\bullet\ P(B)\ =\ \frac{3}{4}\ \hspace{10cm}\]
\[\bullet\ P(A\ \cap\ B)\ =\ \frac{1}{6}\ \hspace{10cm}\]
\[P(A/B)\ =\ \frac{P(A\ \cap\ B)}{P(B)}\ =\ \frac{\frac{1}{6}} {\frac{3}{4}}\ =\ \frac{1}{6}\ \times\ \frac{4}{3}\ =\ \frac{2}{9}\]
\[P(B/A)\ =\ \frac{P(A\ \cap\ B)}{P(A)}\ =\ \frac{\frac{1}{6}} {\frac{1}{3}}\ =\ \frac{1}{6}\ \times\ \frac{3}{1}\ =\ \frac{1}{2}\]
\[\underline{PART\ -\ B}\]
\[21\ \hspace{1cm}\ (a)\ \hspace{1cm} \color{green}{Verify\ (AB)^T\ =\ B^T\ A^T\ If\ A =\begin{bmatrix}
1 & 0 & 3 \\
2 & 1 & – 1 \\
1 & -1 & 1 \\
\end{bmatrix}\ and\ \ B =\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 2 \\
1 & 2 & 0 \\
\end{bmatrix}}\ \hspace{12cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Let\ A =\begin{bmatrix}
1 & 0 & 3 \\
2 & 1 & – 1 \\
1 & -1 & 1 \\
\end{bmatrix}\ and\ \ B =\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 2 \\
1 & 2 & 0 \\
\end{bmatrix}\ \hspace{12cm}\]
\[AB =\begin{bmatrix}
1 & 0 & 3 \\
2 & 1 & – 1 \\
1 & -1 & 1 \\
\end{bmatrix}\ \begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 2 \\
1 & 2 & 0 \\
\end{bmatrix}\ \hspace{12cm}\]
\[ = \begin{bmatrix}
1\ ×\ 1\ +\ 0 ×\ 0\ +\ 3\ ×\ 1& 1\ ×\ 0\ +\ 0 ×\ 1\ +\ 3\ ×\ 2 & 1\ ×\ 2\ +\ 0 ×\ 2\ +\ 3\ ×\ 0\\
2\ ×\ 1\ +\ 1 ×\ 0\ +\ -1\ ×\ 1\ & 2\ ×\ 0\ +\ 1 ×\ 1\ +\ -1\ ×\ 2\ & 2\ ×\ 2\ +\ 1 ×\ 2\ +\ -1\ ×\ 0\\
1\ ×\ 1\ +\ -1 ×\ 0\ +\ 1\ ×\ 1\ & 1\ ×\ 0\ +\ -1 ×\ 1\ +\ -1\ ×\ 2\ & 1\ ×\ 2\ +\ -1 ×\ 2\ +\ 1\ ×\ 0\\
\end{bmatrix}\ \hspace{9cm}\]
\[ = \begin{bmatrix}
1\ +\ 0\ +\ 3\ & 0\ +\ 0\ +\ 6 & 2\ +\ 0 +\ 0\\
2\ +\ 0\ +\ -1\ & 0\ +\ 1\ -\ 2\ & 4\ +\ 2\ +\ 0\\
1\ +\ 0\ +\ 1\ & 0\ -\ 1\ -\ 2\ & 2\ -\ 2\ +\ 0\\
\end{bmatrix}\ \hspace{9cm}\]
\[ AB = \begin{bmatrix}
4 & 6 & 2 \\
1 & -1 & 6 \\
2 & -3 & 0 \\
\end{bmatrix}\ \hspace{12cm}\]
\[(AB)^T\ = \begin{bmatrix}
4 & 1 & 2 \\
6 & -1 & – 3 \\
2 & 6 & 0 \\
\end{bmatrix}\ ————– (1)\ \hspace{10cm}\]
\[Let\ B^T =\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 2 \\
2 & 2 & 0 \\
\end{bmatrix}\ and\ \ A^T\ =\begin{bmatrix}
1 & 2 & 1 \\
0 & 1 & – 1 \\
3 & -1 & 1 \\
\end{bmatrix}\ \hspace{12cm}\]
\[B^T\ A^T\ =\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 2 \\
2 & 2 & 0 \\
\end{bmatrix}\ \begin{bmatrix}
1 & 2 & 1 \\
0 & 1 & – 1 \\
3 & -1 & 1 \\
\end{bmatrix}\ \hspace{12cm}\]
\[ = \begin{bmatrix}
1\ ×\ 1\ +\ 0 ×\ 0\ +\ 1\ ×\ 3& 1\ ×\ 2\ +\ 0 ×\ 1\ +\ 1\ ×\ -1 & 1\ ×\ 1\ +\ 0 ×\ -1\ +\ 1\ ×\ 1\\
0\ ×\ 1\ +\ 1 ×\ 0\ +\ -2\ ×\ 3\ & 0\ ×\ 2\ +\ 1 ×\ 1\ +\ 2\ ×\ -1\ & 0\ ×\ 1\ +\ 1 ×\ -1\ +\ -2\ ×\ 1\\
2\ ×\ 1\ +\ 2 ×\ 0\ +\ 0\ ×\ 3\ & 2\ ×\ 2\ +\ 2 ×\ 1\ +\ 0\ ×\ -1\ & 2\ ×\ 1\ +\ 2 ×\ -1\ +\ 0\ ×\ 1\\
\end{bmatrix}\ \hspace{9cm}\]
\[ = \begin{bmatrix}
1\ +\ 0\ +\ 3\ & 2\ +\ 0\ +\ -1 & 1\ +\ 0 +\ 1\\
0\ +\ 0\ +\ 6\ & 0\ +\ 1\ -\ 2\ & 0\ -\ 1\ -\ 2\\
2\ +\ 0\ +\ 0\ & 4\ +\ 2\ +\ 0\ & 2\ -\ 2\ +\ 0\\
\end{bmatrix}\ \hspace{9cm}\]
\[B^T\ A^T\ = \begin{bmatrix}
4 & 1 & 2 \\
6 & -1 & – 3 \\
2 & 6 & 0 \\
\end{bmatrix}\ ————– (2)\ \hspace{10cm}\]
\[From\ (1)\ and\ (2)\ (AB)^T\ =\ B^T\ A^T\ \hspace{10cm}\]
\[\hspace{2cm}\ (b)\ \hspace{1cm} \color{green}{Solve\ the\ following\ equations\ using\ Cramers\ Rule}\ \hspace{20cm}\]
\[\hspace{4cm} \color{green}{3\ x\ +\ y\ -\ z\ =\ 2,\ 2\ x\ -\ y\ +\ 2\ z\ =\ 6\ and\ 2\ x\ +\ y\ -\ 2\ z\ =\ -\ 2}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[3\ x\ +\ y\ -\ z\ =\ 2\ ——————-(1)\ \hspace{6cm}\]
\[2\ x\ -\ y\ +\ 2\ z\ =\ 6\ \hspace{15cm}\]
\[2\ x\ +\ y\ -\ 2\ z\ =\ -\ 2\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix}
3 & 1 & -1 \\
2 & -1 & 2\\
2 & 1 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta =\ 3\begin{vmatrix}
-1 & 2 \\
1 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & 2 \\
2 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & -1\\
2 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta =\ 3(2\ -\ 2)\ – 1 (-\ 4\ -\ 4)\ -\ 1(2\ +\ 2)\
\hspace{9cm}\]
\[\Delta\ =\ 3(0)\ – 1 (-8)\ -\ 1(4)\
\hspace{13cm}\]
\[\Delta =\ 0\ +\ 8\ -\ 4\
\hspace{14cm}\]
\[\boxed{\Delta\ =\ 4}\
\hspace{17cm}\]
\[\Delta_x = \begin{vmatrix}
2 & 1 & -1 \\
6 & -1 & 2 \\
-2 & 1 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x\ =\ 2\begin{vmatrix}
-1 & 2 \\
1 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
6 & 2 \\
-2 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
6 & -1\\
-2 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x\ =\ 2(2\ -\ 2) – 1 (-12\ +\ 4)\ -\ 1(6\ -\ 2)\
\hspace{9cm}\]
\[\Delta_x\ =\ 2(0)\ – 1 (-8)\ -\ 1(4)\
\hspace{13cm}\]
\[\Delta_x =\ 0\ +\ 8\ -\ 4\
\hspace{14cm}\]
\[\boxed{\Delta_x\ =\ 4}\
\hspace{17cm}\]
\[\Delta_y = \begin{vmatrix}
3 & 2 & -1 \\
2 & 6 & 2 \\
2 & -2 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y\ =\ 3\begin{vmatrix}
6 & 2 \\
-2 & -2\\
\end{vmatrix}\ -\ 2\begin{vmatrix}
2 & 2 \\ 2 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & 6\\
2 & -2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y\ =\ 3(-\ 12\ +\ 4)\ -\ 2 (-4\ -\ 4)\ -\ 1(-4\ -\ 12)\
\hspace{9cm}\]
\[\Delta_y\ =\ 3(-8)\ -\ 2 (-8)\ -\ 1(-16)\
\hspace{13cm}\]
\[\Delta_y\ =\ -\ 24\ +\ 16\ +\ 16\
\hspace{14cm}\]
\[\boxed{\Delta_y\ =\ 8}\
\hspace{17cm}\]
\[\Delta_z = \begin{vmatrix}
3 & 1 & 2 \\
2 & -1 & 6 \\
2 & 1 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z\ =\ 3\begin{vmatrix}
-1 & 6 \\
1 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & 6 \\
2 & -2 \\
\end{vmatrix}\ +\ 2\begin{vmatrix}
2 & – 1\\
2 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z\ =\ 3(2\ -\ 6)\ -\ 1 (-\ 4\ -\ 12)\ +\ 2(2\ +\ 2)\
\hspace{9cm}\]
\[\Delta_z\ =\ 3(-4)\ -\ 1 (-16)\ +\ 2(4)\
\hspace{13cm}\]
\[\Delta_z\ =\ -\ 12\ +\ 16\ +\ 8\
\hspace{14cm}\]
\[\boxed{\Delta_z\ =\ 12}\
\hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta} =\ \frac{4}{4} =\ 1\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{8}{4} =\ 2\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta} =\ \frac{12}{4} =\ 3\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x\ =\ 1,\ y\ =\ 2\ and\ z = 3\ in\ equation (1)\ \hspace{18cm}\]
\[LHS\ =\ 3(1)\ +\ 2 -\ 3\]\[ =\ 3\ +\ 2\ –
=\ 2\]\[ = RHS\]
\[\hspace{2cm}\ (c)\ \hspace{1cm} \color{green}{Find\ the\ inverse\ of\ \begin{bmatrix}
1 & -1 & 1 \\
2 & -3 & -3 \\
6 & -2 & -1 \\
\end{bmatrix}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Let\ A\ =\begin{bmatrix}
1 & -1 & 1 \\
2 & -3 & -3 \\
6 & -2 & -1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =1\begin{vmatrix}
-3 & -3 \\
-2 & -1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -3 \\
6 & -1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -3\\
6 & -2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[ =1(3\ -\ 6)\ + 1 (-2\ +\ 18) + 1(-4\ +\ 18)\
\hspace{9cm}\]
\[ =1(-3)\ + 1 (16) + 1(14)\
\hspace{13cm}\]
\[ = -3\ + 16 + 14\
\hspace{14cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ = 27\ \neq\ 0\
\hspace{17cm}\]
\[\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ \begin{vmatrix}
-3 & -3 \\
-2 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^2 (3 – 6)\ \hspace{15cm}\]
\[= (1) (-3)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = -3\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{1\ +\ 2}\ \begin{vmatrix}
2 & -3 \\
6 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (-2 + 18)\ \hspace{15cm}\]
\[= (-1) (16)\ \hspace{15cm}\]
\[cofactor\ of\ -1 = -16\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 3}\ \begin{vmatrix}
2 & -3 \\
6 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (-4 + 18)\ \hspace{15cm}\]
\[= (1) (14)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = 14\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{2\ +\ 1}\ \begin{vmatrix}
-1 & 1 \\
-2 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (1+ 2)\ \hspace{15cm}\]
\[= (-1) (3)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = -3\ \hspace{15cm}\]
\[cofactor\ of\ -3 = (-1)^{2\ +\ 2}\ \begin{vmatrix}
1 & 1 \\
6 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (-1- 6)\ \hspace{15cm}\]
\[= (1) (-7)\ \hspace{15cm}\]
\[cofactor\ of\ -3 = -7\ \hspace{15cm}\]
\[cofactor\ of\ -3 = (-1)^{2\ +\ 3}\ \begin{vmatrix}
1 & -1 \\
6 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (-2+ 6)\ \hspace{15cm}\]
\[= (-1) (4)\ \hspace{15cm}\]
\[cofactor\ of\ -3 = -4\ \hspace{15cm}\]
\[cofactor\ of\ 6 = (-1)^{3\ +\ 1}\ \begin{vmatrix}
-1 & 1 \\
-3 & -3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (3 + 3)\ \hspace{15cm}\]
\[= (1) (6)\ \hspace{15cm}\]
\[cofactor\ of\ 6 = 6\ \hspace{15cm}\]
\[cofactor\ of\ -2 = (-1)^{3\ +\ 2}\ \begin{vmatrix}
1 & 1 \\
2 & -3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (-3- 2)\ \hspace{15cm}\]
\[= (-1) (-5)\ \hspace{15cm}\]
\[cofactor\ of\ -2 = 5\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{3\ +\ 3}\ \begin{vmatrix}
1 & -1 \\
2 & -3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^6 (-3+ 2)\ \hspace{15cm}\]
\[= (1) (-1)\ \hspace{15cm}\]
\[cofactor\ of\ -1 = -1\ \hspace{15cm}\]
\[Cofactor\ matrix=\begin{bmatrix}
-3 & -16 & 14 \\
-3 & -7 & -4 \\
6 & 5 & -1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A=\begin{bmatrix}
-3 & -3 & 6 \\
-16 & -7 & 5 \\
14 & -4 & -1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{1}{27}\ \begin{bmatrix}
-3 & -3 & 6 \\
-16 & -7 & 5 \\
14 & -4 & -1 \\
\end{bmatrix}\ \hspace{2cm}\]
\[22\ \hspace{1cm}\ (a)\ \hspace{1cm} \color{green}{If\ cos\ \theta\ =\ \frac{3}{5},\ then\ find\ \text{the values of other five trigonometric}}\ \hspace{10cm}\]\[\color{green}{ratios}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ cos\ \theta\ =\ \frac{3}{5}\ \hspace{18cm}\]
\[\hspace{2cm}\ sin^2θ\ +\ cos^2θ\ =\ 1\ \hspace{10cm}\]
\[\hspace{2cm}\ sin^2θ\ =\ 1\ -\ cos^2θ\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ 1\ -\ (\frac{3}{5})^2\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ 1\ -\ \frac{9}{25}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{25\ -\ 9}{25}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{16}{25}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{sin\ \theta\ =\ \pm\ \frac{4}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ tan\ θ\ =\ \frac{sin\ \theta}{cos\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{\pm\ \frac{4}{5}}{\frac{3}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{tan\ \theta\ =\ \pm\ \frac{4}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ cot\ θ\ =\ \frac{1}{tan\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\pm\ \frac{4}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{cot\ \theta\ =\ \pm\ \frac{5}{4}}\ \hspace{10cm}\]
\[\hspace{2cm}\ sec\ θ\ =\ \frac{1}{cos\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\frac{3}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{sec\ \theta\ =\ \frac{5}{3}}\ \hspace{10cm}\]
\[\hspace{2cm}\ cosec\ θ\ =\ \frac{1}{sin\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\pm\ \frac{4}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{sec\ \theta\ =\ \pm\ \frac{5}{4}}\ \hspace{10cm}\]
\[\hspace{2cm}\ (b)\ \hspace{1cm} \color{green}{If\ A\ +\ B\ =\ {45}^0 \ Prove\ that\ (1\ +\ tan\ A)\ (1\ +\ tan\ B)\ =\ 2}\ \hspace{15cm}\]\[\hspace{3cm}\ \color{green} {Hence\ deduce\ the\ value\ of\ tan\ 22\ {\frac{1}{2}}^0}\ \hspace{13cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ A\ +\ B\ =\ 45^{0}\ \hspace{18cm}\]
\[Taking\ Tan\ on\ both\ sides\ \hspace{10cm}\]
\[tan\ (A\ +\ B)\ =\ tan\ 45^{0}\ \hspace{10cm}\]
\[\frac{tan A\ +\ tan B}{1\ -\ tan A\ tan B}\ =\ 1\ \hspace{10cm}\]
\[tan\ A\ +\ tan\ B\ =\ 1\ -\ tan\ A\ tan\ B\ \hspace{10cm}\]
\[tan\ A\ +\ tan\ B\ +\ tan\ A\ tan\ B\ =\ 1\ ————\ (1)\ \hspace{10cm}\]
\[L.\ H.\ S\ =\ (1\ +\ tan\ A)\ (1\ +\ tan\ B)\ \hspace{10cm}\]
\[=\ 1\ +\ tan\ B\ +\ tan\ A\ +\ tan\ A\ tan\ B\ \hspace{10cm}\]
\[=\ 1\ +\ 1\ \hspace{5cm}\ using\ (1)\]
\[=\ 2\ =\ R.\ H.\ S\ \hspace{10cm}\]
\[(1\ +\ tan\ A)\ (1\ +\ tan\ B)\ =\ 2\ \hspace{10cm}\]
\[Put\ B\ =\ A\ \hspace{10cm}\]
\[A\ +\ A\ =\ 45^{0}\ \hspace{10cm}\]
\[2A\ =\ 45^{0}\ \hspace{10cm}\]
\[A\ =\ 22\ {\frac{1}{2}}^0\ \hspace{10cm}\]
\[(1\ +\ tan\ 22\ {\frac{1}{2}}^0)\ (1\ +\ tan\ 22\ {\frac{1}{2}}^0)\ =\ 2\ \hspace{10cm}\]
\[(1\ +\ tan\ 22\ {\frac{1}{2}}^0)^2\ =\ 2\ \hspace{10cm}\]
\[1\ +\ tan\ 22\ {\frac{1}{2}}^0\ =\ \sqrt{2}\ \hspace{10cm}\]
\[\boxed {tan\ 22\ {\frac{1}{2}}^0\ =\ \sqrt{2}\ -\ 1}\hspace{10cm}\]
\[\hspace{2cm}\ (c)\ \hspace{1cm} \color{green}{Show\ that\ \frac{sin\ A\ +\ sin\ 2A}{1\ +\ cos\ A\ +\ cos\ 2A}\ =\ tan\ A}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ LHS\ =\ \frac{sin\ A\ +\ sin\ 2A}{1\ +\ Cos\ A\ +\ cos\ 2A}\ \hspace{18cm}\]
\[=\ \frac{sin\ A\ +\ 2\ sin\ A\ Cos\ A}{2\ cos^2\ A\ +\ Cos\ A}\ \hspace{2cm}\ W.\ K.\ T.\ Sin\ 2A\ =\ 2\ Sin\ A\ Cos\ A\ , and\ 1\ +\ cos\ 2A\ =\ 2\ cos^2\ A\]
\[=\ \frac{Sin\ A(1\ +\ 2\ Cos\ A)}{Cos\ A(1\ +\ 2\ Cos\ A)}\ \hspace{15cm}\]
\[=\ \frac{Sin\ A}{CosS\ A}\ \hspace{15cm}\]
\[=\ tan\ A\ =\ R.H.S\ \hspace{15cm}\]
\[23\ \hspace{1cm}\ (a)\ \hspace{1cm} \color{green}{Prove\ that\ the\ points}\ \hspace{15cm}\]\[\color{green}{5\overrightarrow{i}\ + 6\overrightarrow{j}\ +\ 7\overrightarrow{k}, 6\overrightarrow{i}\ +\ 7\overrightarrow{j}\ +\ 5\overrightarrow{k} and\ 7\overrightarrow{i}\ +\ 5 \overrightarrow{j}\ +\ 6\overrightarrow{k}\ form\ an\ equilateral\ triangle}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ \hspace{17cm}\]
\[\overrightarrow{OA}\ =\ 5\overrightarrow{i}\ + 6\overrightarrow{j}\ +\ 7\overrightarrow{k}\]
\[\overrightarrow{OB}\ =\ 6\overrightarrow{i}\ +\ 7\overrightarrow{j}\ +\ 5\overrightarrow{k}\]
\[\overrightarrow{OC}\ =\ 7\overrightarrow{i}\ +\ 5 \overrightarrow{j}\ +\ 6\overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=\ 6\overrightarrow{i}\ +\ 7\overrightarrow{j}\ +\ 5\overrightarrow{k}\ -\ (5\overrightarrow{i}\ + 6\overrightarrow{j}\ +\ 7\overrightarrow{k})\]
\[=\ 6\overrightarrow{i}\ +\ 7\overrightarrow{j}\ +\ 5\overrightarrow{k}\ -\ 5\overrightarrow{i}\ -\ 6\overrightarrow{j}\ – 7\overrightarrow{k}\]
\[\overrightarrow{AB}\ =\ \overrightarrow{i}\ +\ \overrightarrow{j}\ – 2\overrightarrow{k}\]
\[AB =\overrightarrow{|AB|} = \sqrt{(1)^2 + (1)^2 +(-2)^2 }\]
\[ = \sqrt{1 + 1 +4 }\]
\[AB = \sqrt{6}\]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=\ 7\overrightarrow{i}\ +\ 5 \overrightarrow{j}\ +\ 6\overrightarrow{k}\ -\ (6\overrightarrow{i}\ +\ 7\overrightarrow{j}\ +\ 5\overrightarrow{k})\]
\[=\ 7\overrightarrow{i}\ +\ 5 \overrightarrow{j}\ +\ 6\overrightarrow{k}\ – 6\overrightarrow{i}\ – 7\overrightarrow{j}\ – 5\overrightarrow{k}\]
\[\overrightarrow{BC}= \overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ \overrightarrow{k}\]
\[BC =\overrightarrow{|BC|} = \sqrt{(1)^2 + (-2)^2 +(1)^2 }\]
\[ = \sqrt{1 + 4 + 1}\]
\[BC = \sqrt{6}\]
\[\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}\]
\[=\ 7\overrightarrow{i}\ +\ 5 \overrightarrow{j}\ +\ 6\overrightarrow{k}\ -\ (5\overrightarrow{i}\ + 6\overrightarrow{j}\ +\ 7\overrightarrow{k})\]
\[=\ 7\overrightarrow{i}\ +\ 5 \overrightarrow{j}\ +\ 6\overrightarrow{k}\ – 5\overrightarrow{i}\ -\ 6\overrightarrow{j}\ -\ 7\overrightarrow{k}\]
\[\overrightarrow{AC}\ =\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\]
\[AC =\overrightarrow{|AC|} = \sqrt{(2)^2 + (-1)^2 +(-1)^2 }\]
\[ = \sqrt{4 + 1 + 1}\]
\[AC = \sqrt{6}\]
\[AB = BC = AC = \sqrt{6}\]
\[The\ given\ triangle\ is\ an\ equilateral\ triangle\]
\[\hspace{2cm}\ (b)\ \hspace{1cm} \color{green}{Prove\ that\ \overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k},\ \overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k}\ and\ 7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k} are\ mutually\ orthogonal.}\ \hspace{13cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}= \overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k}\]
\[ \overrightarrow{b}= \overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k} \]
\[\overrightarrow{c}= 7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k}) .(\overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k})\]
= 1(1) + 2(1) + 1(-3)
= 0
\[\overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors.\]
\[ \overrightarrow{b}.\overrightarrow{c}= (\overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k}) .( 7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})\]
= 1(7) + 1(-4) – 3(1)
= 7 – 4 – 3
= 0
\[\overrightarrow{b} and\ \overrightarrow{c}\ are\ perpendicular\ vectors.\]
= 7(1) -4 (2) + 1 (1)
\[ \overrightarrow{c}.\overrightarrow{a}= (7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k}) .(\overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k})\]
= 7 – 8 + 1
= 0
\[\overrightarrow{c} and\ \overrightarrow{a}\ are\ perpendicular\ vectors.\]
\[ The\ three\ vectors\ are\ mutually\ perpendicular.\]
\[\hspace{2cm}\ (c)\ \hspace{1cm} \color{green}{Find\ the\ area\ of\ the\ triangle\ formed\ by\ the\ points\ whose\ position\ vectors\ are}\ \hspace{8cm}\]\[\color{green}{\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k}\ , 2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\ and\ -\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 3\overrightarrow{k}}\ \hspace{7cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{OA}\ =\ \overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k}\]
\[\overrightarrow{OB}\ =\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\]
\[\overrightarrow{OC}\ =\ -\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\ – (\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k})\]
\[=\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\ -\ \overrightarrow{i}\ – 3\overrightarrow{j}\ – 2\overrightarrow{k}\]
\[\boxed{\overrightarrow{AB}\ =\ \overrightarrow{i}\ – 4\overrightarrow{j}\ -\ \overrightarrow{k}}\]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=\ -\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\ – (2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k})\]
\[=\ -\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\ – 2\overrightarrow{i}\ +\ \overrightarrow{j}\ -\overrightarrow{k}\]
\[\boxed{\overrightarrow{BC}\ =\ -\ 3\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k}}\]
\[\overrightarrow{AB}×\overrightarrow{BC} =\ \begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k}\\
1 & -4 & -1\\
-3 & 3 & 2\\
\end{vmatrix}\]
\[ = \overrightarrow{i}( -8 + 3)\ -\ \overrightarrow{j}(2 – 3)\ +\ \overrightarrow{k}(3 – 12)\]
\[ = \overrightarrow{i}(-5) -\overrightarrow{j}(-1)+\overrightarrow{k}(-9)\]
\[\boxed{\overrightarrow{AB}× \overrightarrow{BC}\ =\ -5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 9\overrightarrow{k}}\]
\[|\overrightarrow{AB} × \overrightarrow{BC}| = \sqrt{(-5 )^2\ +\ (1)^2\ + (-9)^2 }=\sqrt{25 + 1 + 81 }=\sqrt{107}\]
\[ Area\ of \ triangle = \frac{1}{2}|\overrightarrow{AB} × \overrightarrow{BC}|\]
\[\boxed{Area\ of \ triangle\ =\frac{\sqrt{107}}{2}\ sq. units}\]
\[24\ \hspace{1cm}\ (a)\ \hspace{1cm} \color{green}{Find\ the\ arithmetic\ mean\ of\ the\ following\ data}\ \hspace{10cm}\]
| Class-interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| Frequency | 3 | 5 | 16 | 18 | 12 | 7 | 4 |
\[\color {blue} {Soln:}\ \hspace{20cm}\]
| Marks | No. of students fi | Mid-value xi | fixi |
| 0-10 | 3 | 5 | 15 |
| 10-20 | 5 | 15 | 75 |
| 20-30 | 16 | 25 | 400 |
| 30-40 | 18 | 35 | 630 |
| 40-50 | 12 | 45 | 540 |
| 50-60 | 7 | 55 | 385 |
| 60-70 | 4 | 65 | 260 |
| N = 65 |
\[\therefore\ \bar{x}\ =\ \frac{\Sigma f_i\ x_i}{N}\ =\ \frac{2305}{65}\ =\ 35.46\]
\[\hspace{2cm}\ (b)\ \hspace{1cm} \color{green}{\text{Calculate the standard deviation for the following data}}\ \hspace{10cm}\]
| Items | 5 | 15 | 25 | 35 |
| Frequency | 2 | 1 | 1 | 3 |
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ n\ =\ 4\ \hspace{20cm}\]
\[\sigma\ =\ \sqrt{\frac{\Sigma f_i\ d_i^2}{N}\ -\ (\frac{\Sigma f_i\ d_i}{N}})^2, \ where\ d_i\ =\ x_i\ -\ A\ and\ N\ =\ \Sigma {f_i}\]
\[A\ =\ \bar{x}\ =\ \frac{5\ +\ 15\ +\ 25\ +\ 35}{4}\ =\ \frac{80}{4}\ =\ 20\]
| xi | fi | di = xi – A | di 2 | fi di | fi di 2 |
| 5 | 2 | – 15 | 225 | – 30 | 450 |
| 15 | 1 | – 5 | 25 | – 5 | 25 |
| 25 | 1 | 5 | 25 | 5 | 25 |
| 35 | 3 | 15 | 225 | 45 | 675 |
| N=7 |
\[\sigma\ =\ \sqrt{\frac{\Sigma f_i\ d_i^2}{N}\ -\ (\frac{\Sigma f_i\ d_i}{N}})^2\]
\[=\ \sqrt{\frac{1175}{7}\ -\ (\frac{15}{7}})^2\]
\[=\ \sqrt{\frac{1175}{7}\ -\ \frac{225}{7\ \times\ 7}}\]
\[=\ \sqrt{\frac{8225\ -\ 225}{7\ \times\ 7}}\]
\[=\ \sqrt{\frac{8000}{49}}\]
\[=\ \sqrt{163.265}\]
\[\sigma\ =\ 12.79\]
\[\hspace{2cm}\ (c)\ \hspace{1cm} \color{green}{Using\ the\ method\ of\ least\ square\ ,\ fit\ a\ straight\ line\ to\ the\ following\ data}\ \hspace{8cm}\]
| x | 0 | 1 | 2 | 3 | 4 |
| y | 10 | 14 | 19 | 26 | 30 |
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Let\ the\ line\ be\ y\ =\ a\ x\ +\ b\ ——–\ (1)\]
\[The\ Normal\ equations\ are\]
\[a\ \Sigma x_i\ +\ nb\ =\ \Sigma y_i\ ——–\ (2)\]
\[a\ \Sigma x_i^2\ +\ b\ \Sigma\ x_i\ =\ \Sigma x_i\ y_i\ ——–\ (3)\]
| xi | yi | xi 2 | xi yi |
| 0 | 10 | 0 | 0 |
| 1 | 14 | 1 | 14 |
| 2 | 19 | 4 | 38 |
| 3 | 26 | 9 | 78 |
| 4 | 30 | 16 | 120 |
\[(2)\ becomes\ a\ (10)\ +\ 5b\ =\ 99\]
\[2\ a\ +\ b\ =\ 19.8\ ——–\ (4)\]
\[(3)\ becomes\ a\ (30)\ +\ b\ (10)\ =\ 250\]
\[3\ a\ +\ b\ =\ 25\ ——–\ (5)\]
\[Solving\ (4)\ and\ (5)\]
\[2\ a\ +\ b\ =\ 19.8\]
\[3\ a\ +\ b\ =\ 25\]
———————————-
\[a\ =\ 5.2\]
\[put\ a\ =\ 5.2\ in\ (4)\]
\[2(5.2)\ +\ b\ =\ 19.8\]
\[10.4\ +\ b\ =\ 19.8\]
\[b\ =\ 19.8\ -\ 10.4\]
\[Eqn\ (1)\ becomes\ y\ =\ 5.2\ x\ +\ 9.4\]
\[25\ \hspace{1cm}\ (a)\ \hspace{1cm} \color{green}{Three\ coins\ are\ tossed\ simultaneously.\ Find\ the\ probability\ of\ getting}\ \hspace{5cm}\]\[\color{green}{(i)\ exactly\ one\ head}\ \hspace{13cm}\]\[\color{green}{(ii)\ exactly\ two\ heads}\ \hspace{13cm}\]\[\color{green}{(iii)\ at least\ two\ heads}\ \hspace{13cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{When three coins are tossed simultaneously, each coin has two possible outcomes: heads (H) or tails (T).}\]\[\therefore\ \text{the total number of possible outcomes when three coins are tossed is:}\]
\[2^3\ =\ 8\]
\[\text{Total number of outcomes = 8}\ \hspace{10cm}\]
\[\text{{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}}\ \hspace{10cm}\]
\[(i)\ \bf{\color{green}{exactly\ one\ head}}\ \hspace{13cm}\]
\[\text{To find the probability of getting exactly one head, we need to count the}\ \hspace{5cm}\]\[\text{outcomes where there is exactly one head.}\ \hspace{10cm}\]
\[\text{The favourable number of outcomes are:}\ \hspace{10cm}\]
\[\text{{HTT, THT, TTH}}\ \hspace{10cm}\]
\[P(Exactly\ one\ head)\ =\ \frac{\text{Number of favourable outcomes}}{\text{Total Number of outcomes}}\ =\ \frac{3}{8}\]
\[(ii)\ \bf{\color{green}{exactly\ two\ heads}}\ \hspace{13cm}\]
\[\text{To find the probability of getting exactly two heads, we need to count the}\ \hspace{5cm}\]\[\text{outcomes where there are exactly two heads.}\ \hspace{10cm}\]
\[\text{The favourable number of outcomes are:}\ \hspace{10cm}\]
\[\text{{HHT, HTH, THH}}\ \hspace{10cm}\]
\[P(Exactly\ two\ heads)\ =\ \frac{\text{Number of favourable outcomes}}{\text{Total Number of outcomes}}\ =\ \frac{3}{8}\]
\[(ii)\ \bf{\color{green}{exactly\ two\ heads}}\ \hspace{13cm}\]
\[\text{To find the probability of getting exactly one head, we need to count the}\ \hspace{5cm}\]\[\text{outcomes where there are exactly two heads.}\ \hspace{10cm}\]
\[\text{The favourable number of outcomes are:}\ \hspace{10cm}\]
\[\text{{HHT, HTH, THH}}\ \hspace{10cm}\]
\[P(Exactly\ two\ heads)\ =\ \frac{\text{Number of favourable outcomes}}{\text{Total Number of outcomes}}\ =\ \frac{3}{8}\]
\[(iii)\ \bf{\color{green}{at\ least\ two\ heads}}\ \hspace{13cm}\]
\[\text{To find the probability of getting atleast two heads, we need to count the}\ \hspace{5cm}\]\[\text{outcomes where there are either two or three heads.}\ \hspace{10cm}\]
\[\text{The favourable number of outcomes are:}\ \hspace{10cm}\]
\[\text{{HHT, HTH, THH, HHH}}\ \hspace{10cm}\]
\[P(at\ least\ two\ heads)\ =\ \frac{\text{Number of favourable outcomes}}{\text{Total Number of outcomes}}\ =\ \frac{4}{8}\ =\ \frac{1}{2}\]
\[\hspace{2cm}\ (b)\ \hspace{1cm} \color{green}{Two\ dice\ are\ thrown\ simultaneously\ Find\ the\ probability}\ \hspace{10cm}\\\ \color{green}{of\ getting\ a\ sum\ 6\ or\ same\ number\ on\ both\ dice}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[When\ two\ dice\ are\ thrown\ simultaneously,\ each\ die\ has\ 6\ faces,\ so\ there\ are\ a\ \hspace{2cm}\\\ total\ of\ 6\ \times\ 6\ =\ 36\ possible\ outcomes\ \hspace{12cm}\]
\[1.\ \bf{\color{green}{probability\ of\ getting\ a\ sum\ of\ 6}}\ \hspace{10cm}\]
\[\text{The possible outcomes where the sum of the two dice is 6 are:}\ \hspace{5cm}\]
\[\bullet\ (1,\ 5)\ \hspace{15cm}\]
\[\bullet\ (2,\ 4)\ \hspace{15cm}\]
\[\bullet\ (3,\ 3)\ \hspace{15cm}\]
\[\bullet\ (4,\ 2)\ \hspace{15cm}\]
\[\bullet\ (5,\ 1)\ \hspace{15cm}\]
\[\text{There are 5 such outcomes.}\ \hspace{10cm}\]
\[2.\ \bf{\color{green}{probability\ of\ getting\ the\ same\ number\ on\ both\ dice}}\ \hspace{7cm}\]
\[\text{The possible outcomes where the numbers on both dice are the same are:}\ \hspace{5cm}\]
\[\bullet\ (1,\ 1)\ \hspace{15cm}\]
\[\bullet\ (2,\ 2)\ \hspace{15cm}\]
\[\bullet\ (3,\ 3)\ \hspace{15cm}\]
\[\bullet\ (4,\ 4)\ \hspace{15cm}\]
\[\bullet\ (5,\ 5)\ \hspace{15cm}\]
\[\bullet\ (6,\ 6)\ \hspace{15cm}\]
\[\text{There are 6 such outcomes.}\ \hspace{10cm}\]
\[3.\ \bf{\color{green}{probability\ of\ either\ getting\ a\ \ sum\ of\ 6\ or\ the\ same\ number\ on\ both\ dice}}\ \hspace{2cm}\]
\[A\ denotes\ the\ event\ of\ getting\ a\ sum\ of\ 6\ \hspace{10cm}\]
\[B\ denotes\ the\ event\ of\ getting\ the\ same\ number\ on\ both\ dice\ \hspace{5cm}\]
\[we\ needed\ to\ find\ P(A\ \cup\ B) =\ ?\]
\[P(A\ \cup\ B)\ =\ P(A)\ +\ P(B)\ -\ P(A\ \cap\ B)\ \hspace{5cm}\]
\[P(A)\ =\ \frac{5}{36}\ \hspace{10cm}\]
\[P(B)\ =\ \frac{6}{36}\ \hspace{10cm}\]
\[P(P(A\ \cap\ B))\ =\ \frac{1}{36}\ \hspace{10cm}\]
\[P(A\ \cup\ B)\ =\ \frac{5}{36}\ +\ \frac{6}{36}\ -\ \frac{1}{36}\ \hspace{5cm}\]
\[=\ \frac{10}{36}\ =\ \frac{5}{18}\ \hspace{3cm}\]
\[\text{So, the probability of getting a sum of 6 or the same number on both dice is:}\ =\ \frac{5}{18}\ \hspace{1cm}\]
\[\hspace{2cm}\ (c)\ \hspace{1cm} \color{green}{A\ problem\ in\ statistics\ is\ given\ to\ two\ students\ A\ and\ B.\ The}\ \hspace{10cm}\\\ \color{green}{probability\ of\ A\ solves\ the\ problem\ is\ \frac{1}{2}\ and\ that\ of\ B\ solves\ the}\ \hspace{8cm}\\\ \color{green}{problem\ is\ \frac{2}{3}.\ If\ the\ students\ solve\ the\ problems\ independently.}\ \hspace{8cm}\\\ \color{green}{find\ the\ probability\ that\ the\ problem\ is\ solved.}\ \hspace{12cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Given,\ P(A)\ =\ \frac{1}{2}\ and\ P(B)\ =\ \frac{2}{3}\ \hspace{10cm}\\\ \text{Probability that the problem is solved = Probability that A solves the problem or B solves the problem}\]
\[=\ P(A\ \cup\ B)\ \hspace{5cm}\]
\[=\ P(A)\ +\ P(B)\ -\ (P(A\ \cap\ B)\ \hspace{2cm}\]
\[=\ P(A)\ +\ P(B)\ -\ P(A)\ \cdot \ P(B)\ Since A\ and\ B\ are\ independent\ \hspace{2cm}\]
\[=\ \frac{1}{2}\ +\ \frac{2}{3}\ -\ \frac{1}{2}\ \cdot \ \frac{2}{3}\ \hspace{2cm}\]
\[=\ \frac{1}{2}\ +\ \frac{2}{3}\ -\ \frac{1}{3}\ \hspace{2cm}\]
\[=\ \frac{3\ +4\ -\ 2}{6}\ \hspace{2cm}\]
\[=\ \frac{5}{6}\ \hspace{2cm}\]
\[\text{Thus, the probability that the problem is solved by either student A, student B, or both is:}\ =\ \frac{5}{6}\]
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