Gina, Sam, and Robby all rented movies from the same video store. They each rented some dramas, comedies, and documentaries. Gina rented 11 movies total. Sam rented twice as many dramas, three times as many comedies, and twice as many documentaries Gina. He rented 27 movies total. If Robby rented 19 movies total with the same number of dramas, twice as many comedies, and twice as many documentaries as Gina, how many movies of each type did Gina rent? Answers:
3 dramas, 5 comedies, and 3 documentaries
2 dramas, 6 comedies, and 3 documentaries
1 dramas, 4 comedies, and 6 documentaries
4 dramas, 3 comedies, and 4 documentaries

Answers

Answer 1

Answer:

Answer:

Gina rent 3 dramas, 5 comedies, and 3 documentaries.

Step-by-step explanation:

Let Gina rented x movies of dramas , y movies of comedies and z movies of documentaries then , Gina rented total 11 movies.

$\Rightarrow x+y+z=11$ .........(1)

Also given Sam rented twice as many dramas, three times as many comedies, and twice as many documentaries Gina, thus he rented 2x movies of dramas , 3y movies of comedies and 2z movies of documentaries. also, he rented a total of 27 movies.

$\Rightarrow 2x+3y+2z=27$ ............(2)

Also, Robby rented the same number of dramas, twice as many comedies, and twice as many documentaries as Gina, thus, he rented x movies of dramas , 2y movies of comedies and 2z movies of documentaries also, he rented a total of 19 movies.

$\Rightarrow x+2y+2z=19$ ............(3)

Solving the three equation using matrix form,

$\left[\begin{array}{ccc}1&1&1\n2&3&2\n1&2&2\end{array}\right] \left[\begin{array}{c}x\ny\nz\end{array}\right]=\left[\begin{array}{c}11\n27\n19\end{array}\right]$

This, system is in form of AX= b,

Where, $A=\left[\begin{array}{ccc}1&1&1\n2&3&2\n1&2&2\end{array}\right]$ , $X=\left[\begin{array}{c}x\ny\nz\end{array}\right]$ , $b=\left[\begin{array}{c}11\n27\n19\end{array}\right]$

Pre-mutiply by A inverse both sides,

$X=A^(-1)b$ ............(P)

First finding inverse,

$\mathrm{Augment\:with\:a}\:3x3\:\mathrm{identity\:matrix}$

$=\begin{bmatrix}1&1&1&\mid \:&1&0&0\n 2&3&2&\mid \:&0&1&0\n 1&2&2&\mid \:&0&0&1\end{bmatrix}$

$\mathrm{Swap\:matrix\:rows:}\:R_1\:\leftrightarrow \:R_2$

$=\begin{bmatrix}2&3&2&\mid \:&0&1&0\n 1&1&1&\mid \:&1&0&0\n 1&2&2&\mid \:&0&0&1\end{bmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_2\:\mathrm{\:by\:performing}\:R_2\:\leftarrow \:R_2-(1)/(2)\cdot \:R_1$

$=\begin{bmatrix}2&3&2&\mid \:&0&1&0\n 0&-(1)/(2)&0&\mid \:&1&-(1)/(2)&0\n 1&2&2&\mid \:&0&0&1\end{bmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_3\:\mathrm{\:by\:performing}\:R_3\:\leftarrow \:R_3-(1)/(2)\cdot \:R_1$

$=\begin{bmatrix}2&3&2&\mid \:&0&1&0\n 0&-(1)/(2)&0&\mid \:&1&-(1)/(2)&0\n 0&(1)/(2)&1&\mid \:&0&-(1)/(2)&1\end{bmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_3\:\mathrm{\:by\:performing}\:R_3\:\leftarrow \:R_3+1\cdot \:R_2$

$=\begin{bmatrix}2&3&2&\mid \:&0&1&0\n 0&-(1)/(2)&0&\mid \:&1&-(1)/(2)&0\n 0&0&1&\mid \:&1&-1&1\end{bmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_1\:\mathrm{\:by\:performing}\:R_1\:\leftarrow \:R_1-2\cdot \:R_3$

$=\begin{bmatrix}2&3&0&\mid \:&-2&3&-2\n 0&-(1)/(2)&0&\mid \:&1&-(1)/(2)&0\n 0&0&1&\mid \:&1&-1&1\end{bmatrix}$

$\mathrm{Multiply\:matrix\:row\:by\:constant:}\:R_2\:\leftarrow \:-2\cdot \:R_2$

$=\begin{bmatrix}2&3&0&\mid \:&-2&3&-2\n 0&1&0&\mid \:&-2&1&0\n 0&0&1&\mid \:&1&-1&1\end{bmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_1\:\mathrm{\:by\:performing}\:R_1\:\leftarrow \:R_1-3\cdot \:R_2$

$=\begin{bmatrix}2&0&0&\mid \:&4&0&-2\n 0&1&0&\mid \:&-2&1&0\n 0&0&1&\mid \:&1&-1&1\end{bmatrix}$

$\mathrm{Multiply\:matrix\:row\:by\:constant:}\:R_1\:\leftarrow (1)/(2)\cdot \:R_1$

$=\begin{bmatrix}1&0&0&\mid \:&2&0&-1\n 0&1&0&\mid \:&-2&1&0\n 0&0&1&\mid \:&1&-1&1\end{bmatrix}$

Thus, $A^(-1)=\begin{pmatrix}2&0&-1\n -2&1&0\n 1&-1&1\end{pmatrix}$

Put values in equation (P),

$X=A^(-1)b$

$\left[\begin{array}{c}x\ny\nz\end{array}\right]=\left[\begin{array}{c,c,c}2&0&-1\n -2&1&0\n 1&-1&1\end{array}\right]\left[\begin{array}{c}11\n27\n19\end{array}\right]$

$\left[\begin{array}{c}x\ny\nz\end{array}\right]=\left[\begin{array}{c,c,c}2\cdot \:11+0\cdot \:27+\left(-1\right)\cdot \:19\n \left(-2\right)\cdot \:11+1\cdot \:27+0\cdot \:19\n 1\cdot \:11+\left(-1\right)\cdot \:27+1\cdot \:19\end{array}\right]$